Chapter 3.5, Problem 51E

To determine

To write: a system of equations to solve the value of the time taken by the police car to catch up to the other car.

Answer to Problem 51E

The required time taken by the police car to catch up to the other car is 0.32 minutes.

Explanation of Solution

Given information:

Constant speed of the car $=r$

The diagram of the distance d (in miles) the police car travels as a function of time t (in minutes):

  

Formula used:

Distance formula:

The distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ .

Calculation:

The given diagram is

  

In the graph,

  $r=0.8$

  $d=2.5t^2$

Express the distance d the cars traveled until the police car catches the other car as the sytem:

  $\{\begin{array}{l}d=rt\\ d=2.5t^2\end{array}$

Compare the above two equations and write as:

  $rt=2.5t^2$

Substitute 0.8 for r in the above equation and simplify.

  $\begin{array}{c}0.8t=2.5t^2\\ 0.8t-0.8t=2.5t^2-0.8t\\ 2.5t^2-0.8t=0\\ t\left(2.5t-0.8\right)=0\end{array}$

Solve the above factors for t .

  $t=0$

And,

  $\begin{array}{c}2.5t-0.8=0\\ \frac{\cancel{2.5}t}{\cancel{2.5}}=\frac{0.8}{2.5}\\ t=\frac{8}{\cancel{10}}\times \frac{\cancel{10}}{25}\\ t=0.32\end{array}$

The value $t=0$ indicates the time when the police car has started.

Thus, $t=0.32$ minutes

Hence, the required time taken by the police car to catch up to the other car is 0.32 minutes.