AERN 45350 Avionics Name: _______________________________ 1 | P a g e Homework Set One (40 Points) Due: 25 September 2015 General Instructions: Answer the following questions, submitting your answers on Blackboard. SHOW YOUR WORK on any math problems. Consider the following voltage signal: V t 12sin377t 1. What is the peak voltage of the signal [Volts]? 2. What is the angular frequency [rad/sec]? 3. What is the frequency of the signal [Hz]? 4. What is the period of the signal [sec/cycle]? In a heterodyne receiver, the intermediate frequency of the receiver is 21.4 MHz. 5. What is the local oscillator frequency (f1) if the tuned frequency (f2) is 120.9 MHz? 6. If the local oscillator frequency (f1) is 145.7 MHz, what is the tuned frequency (f2)? The gain of a power amplifier is 5. 7. If 30W are coming in, what is the power going out? 8. What is the gain in decibels (dB)? The attenuation of a voltage attenuator is 10. 9. If 120V are coming in, what is the voltage going out? 10. What is the loss in decibels (dB)? 11. What is the component of the ILS that provides the extended centerline of the runway? 12. What is the component of the ILS that provides vertical guidance to the runway? 13. If the aircraft is on the correct trajectory, the airplane will arrive at the outer marker on the ILS corresponding to intercepting what? 14. If the aircraft is on the correct trajectory, the airplane will arrive at the middle marker on the ILS corresponding to reaching what? 15. All marker beacons transmit at what frequency? 16. Why doesn’t this cause problems (all marker beacons transmitting on the same frequency)? 17. What are the four components to an ILS? 18. What is the most common ILS category? 19. Which ILS category requires aircraft with the “auto-land” feature? 20. An attenuator leads to a power ratio of 0.5. What is that in decibels (dB)?

AERN 45350 Avionics Name: _______________________________ 1 | P a g e Homework Set One (40 Points) Due: 25 September 2015 General Instructions: Answer the following questions, submitting your answers on Blackboard. SHOW YOUR WORK on any math problems. Consider the following voltage signal: V t 12sin377t 1. What is the peak voltage of the signal [Volts]? 2. What is the angular frequency [rad/sec]? 3. What is the frequency of the signal [Hz]? 4. What is the period of the signal [sec/cycle]? In a heterodyne receiver, the intermediate frequency of the receiver is 21.4 MHz. 5. What is the local oscillator frequency (f1) if the tuned frequency (f2) is 120.9 MHz? 6. If the local oscillator frequency (f1) is 145.7 MHz, what is the tuned frequency (f2)? The gain of a power amplifier is 5. 7. If 30W are coming in, what is the power going out? 8. What is the gain in decibels (dB)? The attenuation of a voltage attenuator is 10. 9. If 120V are coming in, what is the voltage going out? 10. What is the loss in decibels (dB)? 11. What is the component of the ILS that provides the extended centerline of the runway? 12. What is the component of the ILS that provides vertical guidance to the runway? 13. If the aircraft is on the correct trajectory, the airplane will arrive at the outer marker on the ILS corresponding to intercepting what? 14. If the aircraft is on the correct trajectory, the airplane will arrive at the middle marker on the ILS corresponding to reaching what? 15. All marker beacons transmit at what frequency? 16. Why doesn’t this cause problems (all marker beacons transmitting on the same frequency)? 17. What are the four components to an ILS? 18. What is the most common ILS category? 19. Which ILS category requires aircraft with the “auto-land” feature? 20. An attenuator leads to a power ratio of 0.5. What is that in decibels (dB)?

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Consider the spacing and peak widths of the light patterns produced by diffraction and interference of light.a) The spacing of the minima in the patterns produced by a green laser will be: greater than, less than or equal to bthat of a red laser. Justify your answer

Consider the spacing and peak widths of the light patterns produced by diffraction and interference of light.a) The spacing of the minima in the patterns produced by a green laser will be: greater than, less than or equal to bthat of a red laser. Justify your answer

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ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

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A laser pulse of energy 2.00 mJ has a pulse length of 3.00 ns. If this laser pulse is focused to a spot size of radius r =1.25 mm, find the peak electric field amplitude associated with this pulse. Assume that the pulse is uniform in time and uniform in intensity across the area of the spot.

A laser pulse of energy 2.00 mJ has a pulse length of 3.00 ns. If this laser pulse is focused to a spot size of radius r =1.25 mm, find the peak electric field amplitude associated with this pulse. Assume that the pulse is uniform in time and uniform in intensity across the area of the spot.

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Political science. Executive power has waxed and waned over time. While executive power is at its peak in war time, the Court has not always afforded the executive the same deference in the exercise of this power. In an essay trace the historical evolution of the president’s wartime powers with particular attention to the changing context surrounding each decision.

Political science. Executive power has waxed and waned over time. While executive power is at its peak in war time, the Court has not always afforded the executive the same deference in the exercise of this power. In an essay trace the historical evolution of the president’s wartime powers with particular attention to the changing context surrounding each decision.

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F7.10 The flame spread rate through porous solids increases with concurrent wind velocity. decreases with concurrent wind velocity. is independent of concurrent wind velocity. F7.11 Surface tension accelerates opposed-flow flame spread over liquid fuels. True False F7.12 Opposed-flow flame spread rates over a solid surface are typically much smaller than 1 mm/s. around 1mm/s. much greater than 1 mm/s. F7.13 Upward flame spread rate over a vertical surface is typically between 10 and 1000 mm/s. True False F7.14 The Steiner tunnel test described in ASTM standard E 84 is used to assess the fire performance of interior finish materials based on lateral flame spread over a vertical sample. True False F8.1 Describe the triad of fire growth. F8.2 Liquid pool fires reach steady burning conditions within seconds after ignition. True False F8.3 The heat of gasification of liquid fuels is typically less than 1 kJ/g. between 1 and 3 kJ/g. greater than 3 kJ/g. F8.4 The heat flux from the flame to the surface of real burning objects can usually be determined with sufficient accuracy so that reasonable burning rate predictions can be made. True False F8.5 The mass burning flux generally associated with extinction is 0.5 g/m2s. 5 g/m2s. 50 g/m2s. F8.6 The mass burning flux of a liquid pool fire is a function of only the pool diameter. only the fuel type. pool diameter and fuel type. F8.7 The energy release rate of real objects can be measured in an oxygen bomb calorimeter. an oxygen consumption calorimeter. a room/corner test. F8.8 The peak energy release rate of typical domestic upholstered furniture can be as high as 3000 kW. True False F8.9 Draw a typical curve of the mass burning flux of a char forming fuel as a function of time. F8.10 A fast fire as defined in NFPA 72B grows proportionally to t2 and reaches an energy release rate of 1 MW in 75 sec. 150 sec. 300 sec. F9.1 Air entrainment into turbulent pool fire flames is due to buoyancy. True False F9.2 The frequency of vortex shedding in turbulent pool fire flames increases with pool diameter. decreases with pool diameter. is independent of pool diameter. F9.3 The height of turbulent jet flames for a given fuel type and orifice size is independent of energy release rate. True False F9.4 The exit velocities of fuel vapors leaving a solid or liquid pool fire surface are responsible for entrainment of air in the plume. True False F9.5 The height of a turbulent pool fire flame is a function of only energy release rate. only pool diameter. energy release rate and pool diameter. F9.6 Turbulent pool fire flame heights fluctuate in time within a factor of 2. True False F9.7 The Q* value for jet fires is 102 or greater. 104 or greater. 106 or greater. F9.8 The temperature in the continuous flame region of moderate size turbulent pool fires is approximately 820°C. True False F9.9 The temperature at the maximum flame height of a turbulent pool fire flame is approximately 1200°C. 800°C. 300°C. F9.10 The adiabatic flame temperature of hydrocarbon fuels is 1700-2000°C. 2000-2300°C. 2300-2600°C. F10.1 The stoichiometric air to fuel mass ratio of hydrocarbon fuels is of the order of 1.5 g/g. 15 g/g. 150 g/g. F10.2 Give two examples of products of incomplete combustion that occur in fires. F10.3 Slight amounts of products of incomplete combustion are generated in overventilated fires. True False F10.4 The CO yield of a fire is a function of only the fuel involved. only the ventilation conditions. the fuel and the ventilation conditions. F10.5 A carboxyhemoglobin level of 40% in the blood is usually lethal. True (doubt) False F10.6 Carbon monoxide is the leading killer of people in fires. True False F10.7 HCN is a narcotic gas. an irritant gas. a fuel vapor. F10.8 The hazard to humans from narcotic gases is a function of only the concentration of the gas. only the duration of exposure. the product of concentration and duration of exposure. F10.9 The effects on lethality of CO, HCN, and reduced O2 are additive. True False F10.10 Irritant gases typically cause post-exposure fatalities. True False F10.11 Visibility through smoke improves with increasing optical density. True False F10.12 Heat stress occurs when the skin is exposed to a heat flux of 1 kW/m2. the skin reaches a temperature of 45°C. the body’s core temperature reaches 41°C.

F7.10 The flame spread rate through porous solids increases with concurrent wind velocity. decreases with concurrent wind velocity. is independent of concurrent wind velocity. F7.11 Surface tension accelerates opposed-flow flame spread over liquid fuels. True False F7.12 Opposed-flow flame spread rates over a solid surface are typically much smaller than 1 mm/s. around 1mm/s. much greater than 1 mm/s. F7.13 Upward flame spread rate over a vertical surface is typically between 10 and 1000 mm/s. True False F7.14 The Steiner tunnel test described in ASTM standard E 84 is used to assess the fire performance of interior finish materials based on lateral flame spread over a vertical sample. True False F8.1 Describe the triad of fire growth. F8.2 Liquid pool fires reach steady burning conditions within seconds after ignition. True False F8.3 The heat of gasification of liquid fuels is typically less than 1 kJ/g. between 1 and 3 kJ/g. greater than 3 kJ/g. F8.4 The heat flux from the flame to the surface of real burning objects can usually be determined with sufficient accuracy so that reasonable burning rate predictions can be made. True False F8.5 The mass burning flux generally associated with extinction is 0.5 g/m2s. 5 g/m2s. 50 g/m2s. F8.6 The mass burning flux of a liquid pool fire is a function of only the pool diameter. only the fuel type. pool diameter and fuel type. F8.7 The energy release rate of real objects can be measured in an oxygen bomb calorimeter. an oxygen consumption calorimeter. a room/corner test. F8.8 The peak energy release rate of typical domestic upholstered furniture can be as high as 3000 kW. True False F8.9 Draw a typical curve of the mass burning flux of a char forming fuel as a function of time. F8.10 A fast fire as defined in NFPA 72B grows proportionally to t2 and reaches an energy release rate of 1 MW in 75 sec. 150 sec. 300 sec. F9.1 Air entrainment into turbulent pool fire flames is due to buoyancy. True False F9.2 The frequency of vortex shedding in turbulent pool fire flames increases with pool diameter. decreases with pool diameter. is independent of pool diameter. F9.3 The height of turbulent jet flames for a given fuel type and orifice size is independent of energy release rate. True False F9.4 The exit velocities of fuel vapors leaving a solid or liquid pool fire surface are responsible for entrainment of air in the plume. True False F9.5 The height of a turbulent pool fire flame is a function of only energy release rate. only pool diameter. energy release rate and pool diameter. F9.6 Turbulent pool fire flame heights fluctuate in time within a factor of 2. True False F9.7 The Q* value for jet fires is 102 or greater. 104 or greater. 106 or greater. F9.8 The temperature in the continuous flame region of moderate size turbulent pool fires is approximately 820°C. True False F9.9 The temperature at the maximum flame height of a turbulent pool fire flame is approximately 1200°C. 800°C. 300°C. F9.10 The adiabatic flame temperature of hydrocarbon fuels is 1700-2000°C. 2000-2300°C. 2300-2600°C. F10.1 The stoichiometric air to fuel mass ratio of hydrocarbon fuels is of the order of 1.5 g/g. 15 g/g. 150 g/g. F10.2 Give two examples of products of incomplete combustion that occur in fires. F10.3 Slight amounts of products of incomplete combustion are generated in overventilated fires. True False F10.4 The CO yield of a fire is a function of only the fuel involved. only the ventilation conditions. the fuel and the ventilation conditions. F10.5 A carboxyhemoglobin level of 40% in the blood is usually lethal. True (doubt) False F10.6 Carbon monoxide is the leading killer of people in fires. True False F10.7 HCN is a narcotic gas. an irritant gas. a fuel vapor. F10.8 The hazard to humans from narcotic gases is a function of only the concentration of the gas. only the duration of exposure. the product of concentration and duration of exposure. F10.9 The effects on lethality of CO, HCN, and reduced O2 are additive. True False F10.10 Irritant gases typically cause post-exposure fatalities. True False F10.11 Visibility through smoke improves with increasing optical density. True False F10.12 Heat stress occurs when the skin is exposed to a heat flux of 1 kW/m2. the skin reaches a temperature of 45°C. the body’s core temperature reaches 41°C.

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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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Describe at least two strategies that a companies can use to overcome the challenges associated with transportation infrastructure congestion.

Describe at least two strategies that a companies can use to overcome the challenges associated with transportation infrastructure congestion.

Transportation and inventory economics are serious network design deliberations. In … Read More...
Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

Chapter 4 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Advice for the Quarterback A quarterback is set up to throw the football to a receiver who is running with a constant velocity directly away from the quarterback and is now a distance away from the quarterback. The quarterback figures that the ball must be thrown at an angle to the horizontal and he estimates that the receiver must catch the ball a time interval after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is and that the horizontal position of the quaterback is . Use for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. Part A Find , the vertical component of the velocity of the ball when the quarterback releases it. Express in terms of and . Hint 1. Equation of motion in y direction What is the expression for , the height of the ball as a function of time? Answer in terms of , , and . v r D  tc y = 0 x = 0 g v0y v0y tc g y(t) t g v0y ANSWER: Incorrect; Try Again Hint 2. Height at which the ball is caught, Remember that after time the ball was caught at the same height as it had been released. That is, . ANSWER: Answer Requested Part B Find , the initial horizontal component of velocity of the ball. Express your answer for in terms of , , and . Hint 1. Receiver’s position Find , the receiver’s position before he catches the ball. Answer in terms of , , and . ANSWER: Football’s position y(t) = v0yt− g 1 2 t2 y(tc) tc y(tc) = y0 = 0 v0y = gtc 2 v0x v0x D tc vr xr D vr tc xr = D + vrtc Typesetting math: 100% Find , the horizontal distance that the ball travels before reaching the receiver. Answer in terms of and . ANSWER: ANSWER: Answer Requested Part C Find the speed with which the quarterback must throw the ball. Answer in terms of , , , and . Hint 1. How to approach the problem Remember that velocity is a vector; from solving Parts A and B you have the two components, from which you can find the magnitude of this vector. ANSWER: Answer Requested Part D xc v0x tc xc = v0xtc v0x = + D tc vr v0 D tc vr g v0 = ( + ) + D tc vr 2 ( ) gtc 2 2 −−−−−−−−−−−−−−−−−−−  Typesetting math: 100% Assuming that the quarterback throws the ball with speed , find the angle above the horizontal at which he should throw it. Your solution should contain an inverse trig function (entered as asin, acos, or atan). Give your answer in terms of already known quantities, , , and . Hint 1. Find angle from and Think of velocity as a vector with Cartesian coordinates and . Find the angle that this vector would make with the x axis using the results of Parts A and B. ANSWER: Answer Requested Direction of Velocity at Various Times in Flight for Projectile Motion Conceptual Question For each of the motions described below, determine the algebraic sign (positive, negative, or zero) of the x component and y component of velocity of the object at the time specified. For all of the motions, the positive x axis points to the right and the positive y axis points upward. Alex, a mountaineer, must leap across a wide crevasse. The other side of the crevasse is below the point from which he leaps, as shown in the figure. Alex leaps horizontally and successfully makes the jump. v0  v0x v0y v0  v0x v0y v0xx^ v0yy^   = atan( ) v0y v0x Typesetting math: 100% Part A Determine the algebraic sign of Alex’s x velocity and y velocity at the instant he leaves the ground at the beginning of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Algebraic sign of velocity The algebraic sign of the velocity is determined solely by comparing the direction in which the object is moving with the direction that is defined to be positive. In this example, to the right is defined to be the positive x direction and upward the positive y direction. Therefore, any object moving to the right, whether speeding up, slowing down, or even simultaneously moving upward or downward, has a positive x velocity. Similarly, if the object is moving downward, regardless of any other aspect of its motion, its y velocity is negative. Hint 2. Sketch Alex’s initial velocity On the diagram below, sketch the vector representing Alex’s velocity the instant after he leaves the ground at the beginning of the jump. ANSWER: ANSWER: Typesetting math: 100% Answer Requested Part B Determine the algebraic signs of Alex’s x velocity and y velocity the instant before he lands at the end of the jump. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Typesetting math: 100% Hint 1. Sketch Alex’s final velocity On the diagram below, sketch the vector representing Alex’s velocity the instant before he safely lands on the other side of the crevasse. ANSWER: Answer Requested ANSWER: Answer Requested Typesetting math: 100% At the buzzer, a basketball player shoots a desperation shot. The ball goes in! Part C Determine the algebraic signs of the ball’s x velocity and y velocity the instant after it leaves the player’s hands. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s initial velocity On the diagram below, sketch the vector representing the velocity of the basketball the instant after it leaves the player’s hands. ANSWER: Typesetting math: 100% ANSWER: Correct Part D Determine the algebraic signs of the ball’s x velocity and y velocity at the ball’s maximum height. Type the algebraic signs of the x velocity and the y velocity separated by a comma (examples: +,- and 0,+). Hint 1. Sketch the basketball’s velocity at maximum height Typesetting math: 100% On the diagram below, sketch the vector representing the velocity of the basketball the instant it reaches its maximum height. ANSWER: ANSWER: Answer Requested PSS 4.1 Projectile Motion Problems Learning Goal: Typesetting math: 100% To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 9.00 and launch angle 30.0 (above the horizontal) travels a horizontal distance of = 17.0 before hitting the ground. From what height was the rock thrown? Use the value = 9.810 for the free-fall acceleration. PROBLEM-SOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: and . Thus, the problem becomes one of two-dimensional kinematics. The kinematic equations are , . is the same for the horizontal and vertical components of the motion. Find from one component, and then use that value for the other component. ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly. Visualize Part A Which diagram represents an accurate sketch of the rock’s trajectory? Hint 1. The launch angle In a projectile’s motion, the angle of the initial velocity above the horizontal is called the launch angle. ANSWER: m/s  d m g m/s2 ax = 0 ay = −g xf = xi +vixt, yf = yi +viyt− g(t 1 2 )2 vfx = vix = constant, and vfy = viy − gt t t v i Typesetting math: 100% Typesetting math: 100% Correct Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile’s acceleration, , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn’t change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? ANSWER: ay At ground level below the point where the rock is launched At the point where the rock strikes the ground At the peak of the trajectory At the point where the rock is released At ground level below the peak of the trajectory Typesetting math: 100% Correct It’s best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it’s generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below: Solve Part C Find the height from which the rock was launched. Express your answer in meters to three significant figures. yi yi Typesetting math: 100% Hint 1. How to approach the problem The time needed to move horizontally to the final position = 17.0 is the same time needed for the rock to rise from the initial position to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for . Knowing this time will allow you to use the equations of motion for the vertical direction to solve for . Hint 2. Find the time spent in the air How long ( ) is the rock in the air? Express your answer in seconds to three significant figures. Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time the rock spent in the air? ANSWER: Hint 2. Find the x component of the initial velocity What is the x component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: t xf = d m yi t yi t t xf = xi + vixt yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt vix = 7.79 m/s Typesetting math: 100% Hint 3. Find the y component of the initial velocity What is the y component of the rock’s initial velocity? Express your answer in meters per second to three significant figures. ANSWER: ANSWER: Answer Requested Assess Part D A second rock is thrown straight upward with a speed 4.500 . If this rock takes 2.181 to fall to the ground, from what height was it released? Express your answer in meters to three significant figures. Hint 1. Identify the known variables What are the values of , , , and for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: t = 2.18 s viy = 4.50 m/s yi = 13.5 m m/s s H yf viy t a Typesetting math: 100% Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, . ANSWER: ANSWER: Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and free-fall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, 4.500 , and fell the same vertical distance of 13.5 , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly. ± Arrow Hits Apple An arrow is shot at an angle of above the horizontal. The arrow hits a tree a horizontal distance away, at the same height above the ground as it was shot. Use for the magnitude of the acceleration due to gravity. Part A , , , = 0,4.500,2.181,-yf viy t a 9.810 m, m/s, s, m/s2 H yi = H yf = yi + viyt− g(t 1 2 )2 vfy = viy − gt = − 2g( − ) v2f y v2i y yf yi H = 13.5 m viy = m/s m  = 45 D = 220 m g = 9.8 m/s2 Typesetting math: 100% Find , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures. Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables and for the initial components of velocity. Then use kinematics to relate them and solve for . What is the vertical component of the initial velocity? Express your answer symbolically in terms of and . Hint 1. Find Find the horizontal component of the initial velocity. Express your answer symbolically in terms of and given symbolic quantities. ANSWER: Hint 2. Find What is the vertical component of the initial velocity? Express your answer symbolically in terms of . ANSWER: ANSWER: ta vy0 vx0 ta vy0 ta D vx0 vx0 ta vx0 = D ta vy0 vy0 vx0 vy0 = vx0 vy0 = D ta Typesetting math: 100% Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find in terms of and . Give your answer in terms of and . Hint 1. Find When applied to the y-component of velocity, in this problem the formula for with constant acceleration is What is , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of only. ANSWER: ANSWER: Hint 3. Put the algebra together to find symbolically. If you have an expression for the initial vertical velocity component in terms in terms of and , and another in terms of and , you should be able to eliminate this initial component to find an expression for Express your answer symbolically in terms of given variables. ANSWER: ta vy0 g vy0 g vy(ta) v(t) −g vy(t) = vy0 − g t vy(ta ) vy0 vy(ta) = −vy0 ta = 2vy0 g ta D ta g ta ta2 t2 = a 2D g Typesetting math: 100% ANSWER: Answer Requested Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree. Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures. Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters. Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: Answer Requested ANSWER: ta = 6.7 s tf = 1.1 s td = 5.6 s Typesetting math: 100% Answer Requested Video Tutor: Ball Fired Upward from Accelerating Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. Part A Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? Hint 1. Determine how long the ball is in the air How will doubling the initial upward speed of the ball change the time the ball spends in the air? A kinematic equation may be helpful here. The time in the air will ANSWER: be cut in half. stay the same. double. quadruple. Typesetting math: 100% Hint 2. Determine the appropriate kinematic expression Which of the following kinematic equations correctly describes the horizontal distance between the ball and the cart at the moment the ball lands? The cart’s initial horizontal velocity is , its horizontal acceleration is , and is the time elapsed between launch and impact. ANSWER: ANSWER: Correct The ball will spend twice as much time in the air ( , where is the ball’s initial upward velocity), so it will land four times farther behind the cart: (where is the cart’s horizontal acceleration). Video Tutor: Ball Fired Upward from Moving Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point. d v0x ax t d = v0x t d = 1 2 axv0x t2 d = v0x t+ 1 2 axt2 d = 1 2 axt2 the same distance twice as far half as far four times as far by a factor not listed above t = 2v0y/g v0y d = 1 2 axt2 ax Typesetting math: 100% Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance. Hint 1. How to approach the problem While the crate is on the plane, it shares the plane’s velocity. What is the crate’s velocity immediately after it is released? Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate’s horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that’s not very realistic in this situation.) ANSWER: Correct At the moment it is released, the crate shares the plane’s horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell. Score Summary: Your score on this assignment is 0%. Before the plane is directly over the target After the plane has flown over the target When the plane is directly over the target Typesetting math: 100% You received 0 out of a possible total of 0 points. Typesetting math: 100%

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Why does customer service not increase proportionately to increases in total cost when a logistical system is being designed?

Why does customer service not increase proportionately to increases in total cost when a logistical system is being designed?

Typical inventory increases as the number of warehouses in a … Read More...