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Chapter 2

Q. 2.5

Car Chase

GOAL Solve a problem involving two objects, one moving at constant acceleration and the other at constant velocity.

PROBLEM A car traveling at a constant speed of 24.0 m/s passes a trooper hidden behind a billboard, as in Figure 2.17. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.00 m/s². (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time?

STRATEGY Solving this problem involves two simultaneous kinematics equations of position, one for the trooper and the other for the car. Choose t = 0 to correspond to the time the trooper takes up the chase, when the car is at x_{car}  =  24.0  m because of its head start (24.0 m/s × 1.00 s). The trooper catches up with the car when their positions are the same, which suggests setting x_{trooper}  =  x_{car} and solving for time, which can then be used to find the trooper’s speed in part (b).

F2.17

Step-by-Step

Verified Solution

(a) How long does it take the trooper to overtake the car?

Write the equation for the car’s displacement:

Δx_{car}  =  x_{car}  –  x_0  =  v_0t  +  \frac{1}{2} a_{car}t²

Take x_0  =  24.0  m,  v_0  =  24.0  m/s,  and  a_{car}  =  0. Solve for x_{car}:

x_{car}  =  x_0  +  vt  =  24.0  m  +  (24.0  m/s)t

Write the equation for the trooper’s position, taking x_0  =  0,  v_0  =  0, and a_{trooper}  =  3.00  m/s²:

x_{trooper}  =  \frac{1}{2} a_{trooper} t²  =  \frac{1}{2} (3.00  m/s²)t²  =  (1.50  m/s²)t²

Set x_{trooper}  =  x_{car}, and solve the quadratic equation. (The quadratic formula appears in Appendix A, Equation A.8.) Only the positive root is meaningful.

x  =  \frac{-b  \pm  \sqrt{b^2  –  4ac}}{2a}       (A.8)

(1.50  m/s²)t²  =  24.0  m  +  (24.0  m/s)t
(1.50  m/s²)t²  –  (24.0  m/s)t  –  24.0  m  =  0
t = 16.9 s

(b) Find the trooper’s speed at that time.

Substitute the time into the trooper’s velocity equation:

v_{trooper}  =  v_0  +  a_{trooper} t  =  0  +  (3.00  m/s²)(16.9  s)
= 50.7 m/s

REMARKS The trooper, traveling about twice as fast as the car, must swerve or apply his brakes strongly to avoid a collision! This problem can also be solved graphically by plotting position versus time for each vehicle on the same graph. The intersection of the two graphs corresponds to the time and position at which the trooper overtakes the car.