In this example we will, for the first time, look at the constant accelerated mo

Question

In this example we will, for the first time, look at the constant accelerated motion of two objects moving simultaneously. Clearly, we will need to describe the position and velocity of each separately and then use the resulting equations to answer any questions about their motion. Suppose that a motorist is traveling at a constant 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 acceleration of 3.0 m/s2 (Figure 1) . (a) How much time elapses before the officer catches up with the car? (b) What is the officer’s speed at that point? (c) What is the total distance the officer has traveled at that point?

SOLUTION

SET UP  Both objects move in a straight line, which we'll designate as the x axis. Then all positions, velocities, and accelerations have only x components. We'll omit the subscripts x in our solution, but always keep in mind that when we say “velocity,” we really mean “xcomponent of velocity,” and so on. The motorcycle and the car both move with constant acceleration, so we can use the formulas we have developed. We have two different objects in motion; (Figure 1)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 x0=0 for both. Let xP be the police officer’s position and xC be the car’s position, at any time t. At the instant when the officer catches the car, the two objects are at the same position. We need to apply the constant-acceleration equations to each, and find the time when xP and xC are equal. We denote the initial x components of velocities as vP0 and vC0 and the accelerations as aP and aC. From the data given, we have vP0=0, vC0=15m/s, aP=3.0m/s2, and aC=0.

SOLVE Applying x=x0+v0xt+12axt2 to each object, we find that

xPxC===x0+vP0t+12aPt2=0+0+12(3.0m/s2)t212(3.0m/s2)t2x0+vC0t+12aCt2=0+(15m/s)t+0=(15m/s)t

Part (a): At the time the officer catches the car, both must be at the same position, so at this time, xC=xP. Equating the preceding two expressions, we have

12(3.0m/s2)t2=(15m/s)t

or

t=0,10s

There are two times when the two vehicles have the same xcoordinate; the first is the time (t=0)when the car passes the parked motorcycle at the corner, and the second is the time when the officer catches up.

Part (b): We know that the officer’s velocity vP at any time t is given by

vP=vP0+aPt=0+(3.0m/s2)t

so when t=10s, vP=30m/s. When the officer overtakes the car, she is traveling twice as fast as the motorist is.

Part (c): When t=10s, the car’s position is

xC=(15m/s)(10s)=150m

and the officer’s position is

xP=12(3.0m/s2)(10s)2=150m

This result verifies that, at the time the officer catches the car, they have gone equal distances and are at the same position.

REFLECT A graphical description of the motion is helpful. (Figure 2)shows graphs of xP and xC as functions of time. At first xC is greater because the motorist is ahead of the officer. But the car travels with constant velocity, while the officer accelerates, closing the gap between the two. At the point where the curves cross, the officer has caught up to the motorist. We see again that there are two times when the two positions are the same. Note that the two vehicles don’t have the same speed at either of these times.

QUESTIONS:

If the officer’s acceleration is 5.2 m/s2 , what distance does she travel before catching up with the car?

What is her velocity when she has caught up?

During what time interval is she moving more slowly than the car?

0 to 5.77

s0 to 15.0

s0 to 10.0

?s0 to 2.88 s

Explanation / Answer

a = acceleration = 5.2 m/s2

V = speed of car = 15 m/s

t = time of travel

distance travelled by bike = distance travelled by car

(0.5) a t2 = V t

(0.5) (5.2) t2 = (15) t

t = 5.77 sec

distance travelled by bike = (0.5) a t2 = (0.5) (5.2) (5.77)2 = 86.6 m

velocity = V = at = 5.2 x 5.77 = 30 m/s

time interval = 0 to 2.88

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