Chapter 3, Problem 50P

To determine

The time taken to overtake speeder.

Answer to Problem 50P

The time taken to overtake is 15.8 sec.

Explanation of Solution

Given:

Unmarked police car, travelling the constant 95km/h, is passed by the speeder traveling 145km/h.

Calculation:

The relative velocity of the police car with respect to the speeder:

  $\begin{array}{l}{v}_{ps}=v_{pg}-v_{sg}\\ \text{where,}\\ v_{pg}=\text{velocity of the police car with respect to the ground}\\ {\text{v}}_{sg}\text{=velocity of the speeder with respect to the ground}\end{array}$

Substituting the known value:

  $\begin{array}{l}{v}_{ps}=95\left(\frac{1000}{3600}\right)-145\left(\frac{1000}{3600}\right)\\ =-13.88\text{m/s}\end{array}$

The negative sign denotes the police car is behind the speeder.

The distance travelled by the police car:

  $\begin{array}{l}d=v_{ps}t\\ d=\left(-13.88\right)\left(1\right)\\ =-13.88\text{m}\end{array}$

The acceleration is increased to 2m/s²:

The displacement is:

  $\begin{array}{l}x=x_o+v_{ps}t+\frac{1}{2}at{\text{'}}^2\\ 0=-13.88+\left(-13.88\right)t\text{'}+\frac{1}{2}\left(2\right)t{\text{'}}^2\\ t\text{'}=14.825\text{s}\text{,}\text{-9}\text{.036s}\end{array}$

Hence, the total time required to overtake the speeder:

  $\begin{array}{l}t"=t\text{'}+t\\ t"=14.825+1\\ =15.825\text{s}\end{array}$

Hence, the total time required to overtake the speeder is 15.8s.