Chapter 3, Problem 51P

To determine

The speed of the speeder

Answer to Problem 51P

The speed of the speeder is 114km/hr.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Time to overtake, }t=7\text{ sec}\\ \text{Acceleration, }a=2{\text{ m/s}}^2\\ \text{Velocity of police car, }{v}_p=95\text{ km/hr}\\ \text{Relative displacement, }{d}_{rel}=13.88\text{ m}\end{array}$

Calculation:

Considering the origin be the location where the speeder passes the police car.

Let the speed be v_s.

The distance covered in the 7s:

  $\begin{array}{l}\Delta x_s=v_S.t\\ =v_{s.7}\\ =7v_s\end{array}$

As given the police car is running at the original velocity, and 6 seconds under the acceleration.

Evaluating the distance travel by the police car:

  $\begin{array}{l}{v}_p=\text{speed of police car}\\ \Delta x_p=v_p\left(1\right)+v_p\left(6\right)+\frac{1}{2}{a}_p{\left(6s\right)}^2\\ \text{Since,}\\ v_p=95\text{km/hr}\\ \text{=95}\left(\frac{5}{18}\right)\text{m/s}\\ \text{=26}\text{.39m/s}\\ a_p=2{\text{m/s}}^2\\ \text{hence,}\\ \Delta x_p=26.39+\left(26.39\right)6+\frac{1}{2}\left(2\right){\left(6\right)}^2\\ =220.7\text{m}\end{array}$

The police car overtakes the speeder when two distances will be equal.

Hence,

  $\begin{array}{l}7v_s=220.7\\ v_s=31.53\text{m/s=114km/hr}\end{array}$