Chapter 2 Problem 90

Problem

A police car at rest, passed by a speeder traveling at a constant 130km/h, takes off in hot pursuit. The police officer catches up to the speeder in 750 m, maintaining a constant acceleration.

(a) Qualitatively plot the position vs. time graph for both cars from the police car’s start to the catch-up point.

Calculate

(b) how long it took the police officer to overtake the speeder,

(c) the required police car acceleration, and

(d) the speed of the police car at the overtaking point.

Solution

Take the origin to be the location where the speeder passes the police car. The speeder’s constant speed is

$v_{\textrm {speeder}}=\left(130{\textrm {km/h}}\right)\left({\frac {1{m/s}}{3.6{\textrm {km/h}}}}\right)=36.1{\textrm {m/s}}$

$x_{\textrm {speeder}}=v_{\textrm {speeder}}t$

$x_{\textrm {police}}=v_{0}t+{\frac {1}{2}}at^{2}={\frac {1}{2}}a_{\textrm {police}}t^{2}$

(a)

(b)

$\left(750{\textrm {m}}\right)=\left(36.1{\textrm {m/s}}\right)t\rightarrow$ $t={\frac {\left(750{\textrm {m}}\right)}{\left(36.1{\textrm {m/s}}\right)}}=20.8{\textrm {s}}\approx 21{\textrm {s}}$

(c)

$\left(750{\textrm {m}}\right)={\frac {1}{2}}a_{\textrm {police}}\left(20.8{\textrm {s}}\right)^{2}\rightarrow$ $a_{\textrm {police}}={\frac {2\left(750{\textrm {m}}\right)}{\left(20.8{\textrm {s}}\right)^{2}}}=3.47{\textrm {m/s}}^{2}\approx 3.5{\textrm {m/s}}^{2}$

(d)

$v=v_{0}+at=0+\left(3.47{\textrm {m/s}}^{2}\right)\left(20.8{\textrm {s}}\right)=72.2{\textrm {m/s}}\approx 72{\textrm {m/s}}$