Chapter 8.7, Problem 39PS

Solution

To calculate: To determine how far the police must travel before they are able to stop a car. The idea is that they should both travel the same distance.

The time taken for the police car to catch up the other car is $\text{0}\text{.32 s}$ .

Given information: The car is moving at a rate of $\text{0}\text{.8 mi/min}$ . The distance of the police car is given by $\text{d = 2}{\text{.5t}}^{\text{2}}$ .

Calculation:

Since the car is moving at a rate of $\text{0}\text{.8 mi/min}$ .

The distance passes is

  $\begin{array}{c}\text{v = }\frac{\text{s}}{\text{t}}\\ \text{s = vt}\end{array}$ .

The distance of the police car is given by $\text{d = 2}{\text{.5t}}^{\text{2}}$ .

Now solving the equation,

  $\begin{array}{c}\text{2}{\text{.5t}}^{\text{2}}\text{ = 0}\text{.8t}\\ \text{2}{\text{.5t}}^{\text{2}}\text{ - 0}\text{.8t = 0}\\ \text{t(2}\text{.5t - 0}\text{.8) = 0}\end{array}$

By solving to get $\text{t = 0}$ and $\text{t = 0}\text{.32}$ .

Since time cannot be zero, the time taken for the police car to catch the other car is $\text{t = 0}\text{.32 s}$ .