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

Question

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 [latex]24.0 \mathrm{~m} / \mathrm{s}[/latex] 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 [latex]3.00 \mathrm{~m} / \mathrm{s}^{2}[/latex]. (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 kine matics equations of position: one for the trooper and the other for the car. Choose [latex]t=0[/latex] to correspond to the time the trooper takes up the chase, when the car is at [latex]x_{ ext {car }}=24.0 \mathrm{~m}[/latex] because of its head start [latex](24.0 \mathrm{~m} / \mathrm{s} imes 1.00 \mathrm{~s})[/latex]. The trooper catches up with the car when their positions are the same. This suggests setting [latex]x_{ ext {trooper }}=x_{ ext {car }}[/latex] and solving for time, which can then be used to find the trooper's speed in part (b).

Accepted Answer

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

Write the equation for the car's displacement:

[latex] \Delta x_{ ext {car }}=x_{ ext {car }} - x_{0}=v_{0} t+\frac{1}{2} a_{ ext {car }} t^{2}[/latex]

Take [latex]x_{0}=24.0 \mathrm{~m}, v_{0}=24.0 \mathrm{~m} / \mathrm{s}[/latex], and [latex]a_{\mathrm{car}}=0[/latex]. Solve for [latex]x_{\mathrm{car}}[/latex] :

[latex]x_{\mathrm{car}}=x_{0}+v t=24.0 \mathrm{~m}+(24.0 \mathrm{~m} / \mathrm{s}) t[/latex]

Write the equation for the trooper's position, taking [latex]x_{0}=0, v_{0}=0[/latex], and [latex]a_{ ext {trooper }}=3.00 \mathrm{~m} / \mathrm{s}^{2}[/latex] :

[latex]x_{ ext {trooper }}=\frac{1}{2} a_{ ext {trooper }} t^{2}={ }_{2}^{1}\left(3.00 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}=\left(1.50 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}[/latex]

Set [latex]x_{ ext {trooper }}=x_{ ext {car }}[/latex], and solve the quadratic equation. (The quadratic formula appears in Appendix A, Equation A.8.)

[latex]x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}[/latex] [A.8]

Only the positive root is meaningful.

[latex]\begin{aligned}& \left(1.50 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}=24.0 \mathrm{~m}+(24.0 \mathrm{~m} / \mathrm{s}) t &&\& \left(1.50 \mathrm{~m} / \mathrm{s}^{2} ight) t^{2}-(24.0 \mathrm{~m} / \mathrm{s}) t-24.0 \mathrm{~m}=0& \ & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad t=16.9 \mathrm{~s}\end{aligned}[/latex]

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

Substitute the time into the trooper's velocity equation:

[latex]\begin{aligned}v_{ ext {trooper }} &=v_{0}+a_{ ext {trooper }} t=0+\left(3.00 \mathrm{~m} / \mathrm{s}^{2} ight)(16.9 \mathrm{~s}) \&=50.7 \mathrm{~m} / \mathrm{s}\end{aligned}[/latex]

REMARKS The trooper, traveling about twice as fast as the car, must swerve or apply the brakes strongly to avoid a collision! This problem can also be solved graphically by plotting position vs. 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.

Question 2.5: CAR CHASE GOAL Solve a problem involving two objects, one mo...

Related Answered Questions

Verified Answer:

Verified Answer:

CATCHING A FLY BALL GOAL Apply the definition of instantaneous acceleration. PROBLEM A baseball player moves in a straight line path in order to catch a fly ball hit to the outfield. His velocity as a func tion of time is shown in Figure 2.11a. Find his instantaneous acceleration at points ...

Verified Answer:

A ROCKET GOES BALLISTIC GOAL Solve a problem involving a powered ascent followed by free-fall motion. PROBLEM A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s². It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height ...

Verified Answer:

Verified Answer:

NOT A BAD THROW FOR A ROOKIE! GOAL Apply the kinematic equations to a freely falling object with a nonzero initial velocity. PROBLEM A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer: