Second image has background info to help solve the practice problem CAL CONTENTEXAMPLE 2.9 Pursuit!In this example we will, for the first time, look at the constant accelerated motion of two objects movingsimultaneously. Clearly, we will need to describe the position and velocity of each separately and then use theresulting equations to answer any questions about their motion. Suppose that a motorist is traveling at a con-stant velocity of 15 m/s and passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h).A police officer on a motorcycle stopped at the corner then starts off in pursuit with a constant accelerationof 3.0 m/s² (Figure 2.22a). (a) How much time elapses before the officer catches up with the car? (b) What isthe officer's speed at that point? (c) What is the total distance the officer has traveled at that point?BARCHindoespinnlamin sed1340 252500 SCHOOLCROSSINGOorXpap = 3.0 m/s²ved bas jas mail net 25 owiadt as teong as ooiwt lev(a)A FIGURE 2.22 (a) A diagram of the problem. (b) Graphs of position as a function of time forthe police officer and the motorist.xcxp = xo + Upot + apt² = 0 + 0 + (3.0 m/s²),²= (3.0 m/s²)r²,xc = xo +ucot + act² = 0 + (15 m/s)t + 0 = (15 m/s)t.Uco 15 m/s.=SOLUTIONSET UP Both objects move in a straight line, which we'll designate asthe x axis. Then all positions, velocities, and accelerations have onlyx components. We'll omit the subscripts x in our solution, but alwayskeep in mind that when we say "velocity," we really mean "x compo-nent of velocity," and so on.th The motorcycle and the car both move with constant acceleration,so we can use the formulas we have developed. We have two differentobjects in motion; Figure 2.22a shows our choice of coordinate system.The origin of coordinates is at the corner where the officer is stationed;both objects are at this point at time t = 0, so xo = 0 for both. Let xpbe the police officer's position and xc be the car's position, at any timet. At the instant when the officer catches the car, the two objects are atthe same position. We need to apply the constant-acceleration equations and the officer's position isto each, and find the time when xp and xc are equal. We denote theinitial x components of velocities as Upo and Uco and the accelerations0, Uco 15as ap and ac. From the data given, we have Upo= 0, Uco = 15 m/s,ap= 3.0 m/s², and ac = 0.m/s,Part (a): At the time the officer catches the car, both must be at thesame position, so at this time, xc = xp. Equating the preceding twoexpressions, we have(3.0 m/s²)² = (15 m/s)r,t = 0, 10 s.2.4 Motion with Constant Acceleration 45OSAOCx (m)16012080400ol(b)2Video Tutor SolutionMotoristOfficer18 10 12 (s)146sertal odiThere are two times when the two vehicles have the same x co-ordinate; the first is the time (t = 0) when the car passes the parkedmotorcycle at the corner, and the second is the time when the officercatches up.SOLVE Applying Equation 2.10 to each object, we find that Figure 2.22b shows graphs of xp and xc as functions of time. At firstREFLECT A graphical description of the motion is helpful.xc is greater because the motorist is ahead of the officer. But the cartravels with constant velocity, while the officer accelerates, closing thegap between the two. At the point where the curves cross, the officer hascaught up to the motorist. We see again that there are two times whenthe two positions are the same. Note that the two vehicles don't have thesame speed at either of these times.Part (b): From Equation 2.6, we know that the officer's velocity Up atany time t is given byUp = Upo + apt = 0 + (3.0 m/s²)r,so when t = 10 s, Up= 30 m/s. When the officer overtakes the car, sheis traveling twice as fast as the motorist is.Part (c): When t= 10 s, the car's position isxc= (15 m/s) (10 s) = 150 m,bas xp = (3.0 m/s²) (10 s)² = 150 m.This result verifies that, at the time the officer catches the car, they havegone equal distances and are at the same position.Practice Problem: If the officer's acceleration is 5.0 m/s², what dis-tance does she travel before catching up with the car? What is her veloc-ity when she has caught up? During what time interval is she movingmore slowly than the car? Answers: 90 m, 30 m/s, 0 to 3 s. Name:G#Page 44-Practice Problem 2.9:If the officer's acceleration is 5.0m/s2, what distance does she travel before catching upwith the car? What is her velocity when she has caught up? During what time interval isshe moving more slowly than the car? Answer: 90m, 30m/s, 0 to 3s.

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Description: Second image has background info to help solve the practice problem CAL CONTENTEXAMPLE 2.9 Pursuit!In this example we will, for the first time, look at the constant accelerated motion of two objects movingsimultaneously. Clearly, we will need to desc

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Second image has background info to help solve the practice problem CAL CONTENTEXAMPLE 2.9 Pursuit!In this example we will, for the first time, look at the constant accelerated motion of two objects movingsimultaneously. Clearly, we will need to describe the position and velocity of each separately and then use theresulting equations to answer any questions about their motion. Suppose that a motorist is traveling at a con-stant velocity of 15 m/s and passes a school-crossing corner where the speed limit is 10 m/s (about 22 mi/h).A police officer on a motorcycle stopped at the corner then starts off in pursuit with a constant accelerationof 3.0 m/s² (Figure 2.22a). (a) How much time elapses before the officer catches up with the car? (b) What isthe officer's speed at that point? (c) What is the total distance the officer has traveled at that point?BARCHindoespinnlamin sed1340 252500 SCHOOLCROSSINGOorXpap = 3.0 m/s²ved bas jas mail net 25 owiadt as teong as ooiwt lev(a)A FIGURE 2.22 (a) A diagram of the problem. (b) Graphs of position as a function of time forthe police officer and the motorist.xcxp = xo + Upot + apt² = 0 + 0 + (3.0 m/s²),²= (3.0 m/s²)r²,xc = xo +ucot + act² = 0 + (15 m/s)t + 0 = (15 m/s)t.Uco 15 m/s.=SOLUTIONSET UP Both objects move in a straight line, which we'll designate asthe x axis. Then all positions, velocities, and accelerations have onlyx components. We'll omit the subscripts x in our solution, but alwayskeep in mind that when we say &#34;velocity,&#34; we really mean &#34;x compo-nent of velocity,&#34; and so on.th The motorcycle and the car both move with constant acceleration,so we can use the formulas we have developed. We have two differentobjects in motion; Figure 2.22a shows our choice of coordinate system.The origin of coordinates is at the corner where the officer is stationed;both objects are at this point at time t = 0, so xo = 0 for both. Let xpbe the police officer's position and xc be the car's position, at any timet. At the instant when the officer catches the car, the two objects are atthe same position. We need to apply the constant-acceleration equations and the officer's position isto each, and find the time when xp and xc are equal. We denote theinitial x components of velocities as Upo and Uco and the accelerations0, Uco 15as ap and ac. From the data given, we have Upo= 0, Uco = 15 m/s,ap= 3.0 m/s², and ac = 0.m/s,Part (a): At the time the officer catches the car, both must be at thesame position, so at this time, xc = xp. Equating the preceding twoexpressions, we have(3.0 m/s²)² = (15 m/s)r,t = 0, 10 s.2.4 Motion with Constant Acceleration 45OSAOCx (m)16012080400ol(b)2Video Tutor SolutionMotoristOfficer18 10 12 (s)146sertal odiThere are two times when the two vehicles have the same x co-ordinate; the first is the time (t = 0) when the car passes the parkedmotorcycle at the corner, and the second is the time when the officercatches up.SOLVE Applying Equation 2.10 to each object, we find that Figure 2.22b shows graphs of xp and xc as functions of time. At firstREFLECT A graphical description of the motion is helpful.xc is greater because the motorist is ahead of the officer. But the cartravels with constant velocity, while the officer accelerates, closing thegap between the two. At the point where the curves cross, the officer hascaught up to the motorist. We see again that there are two times whenthe two positions are the same. Note that the two vehicles don't have thesame speed at either of these times.Part (b): From Equation 2.6, we know that the officer's velocity Up atany time t is given byUp = Upo + apt = 0 + (3.0 m/s²)r,so when t = 10 s, Up= 30 m/s. When the officer overtakes the car, sheis traveling twice as fast as the motorist is.Part (c): When t= 10 s, the car's position isxc= (15 m/s) (10 s) = 150 m,bas xp = (3.0 m/s²) (10 s)² = 150 m.This result verifies that, at the time the officer catches the car, they havegone equal distances and are at the same position.Practice Problem: If the officer's acceleration is 5.0 m/s², what dis-tance does she travel before catching up with the car? What is her veloc-ity when she has caught up? During what time interval is she movingmore slowly than the car? Answers: 90 m, 30 m/s, 0 to 3 s. Name:G#Page 44-Practice Problem 2.9:If the officer's acceleration is 5.0m/s2, what distance does she travel before catching upwith the car? What is her velocity when she has caught up? During what time interval isshe moving more slowly than the car? Answer: 90m, 30m/s, 0 to 3s.

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