Chapter 12.8, Problem 7P

To determine

To calculate: How many seconds will it take the car to travel $1800$ m at acceleration of $25m/s^2$ .

The distance of a car travels from rest varies jointly as it acceleration and the square of the time of motion. A car travels $250m$ from rest in $5$ seconds at an acceleration of $20m/s^2$ .

Answer to Problem 7P

The car travels distance in: $12$ sec.

Explanation of Solution

Given information:

How many seconds will it take the car to travel $1800$ m at acceleration of $25m/s^2$ .

The distance of a car travels from rest varies jointly as it acceleration and the square of the time of motion. A car travels $250m$ from rest in $5$ seconds at an acceleration of $20m/s^2$ .

Formula used:

For varies jointly:

  $d=K{a}_1{t}_1{}^2$

Calculation:

How many seconds will it take the car to travel $1800$ m at acceleration of $25m/s^2$ .

The distance of a car travels from rest varies jointly as it acceleration and the square of the time of motion. A car travels $250m$ from rest in $5$ seconds at an acceleration of $20m/s^2$ .

Recall that for varies jointly:

  $d=K{a}_1{t}_1{}^2$

The values are:

  $d=250,a=20,t=5$

Therefore, the value of $k$ is,

  $\begin{array}{c}d=K{a}_1{t}_1{}^2\\ 250=K\times 20\times 5^2\\ K=\frac{250}{20\times 25}\\ K=\frac{1}{2}\end{array}$

Now to find the value of $r$ ,

  $d_2=K{a}_2{t}_2{}^2$

The values are:

  $d=1800,K=\frac{1}{2},a=25$

Hence, when:

  $\begin{array}{c}{d}_2=K{a}_2{t}_2{}^2\\ 1800=\frac{1}{2}\times 25\times t_2{}^2\\ \frac{1800\times 2}{25}=t_2{}^2\\ 144=t_2{}^2\\ \sqrt{144}=t_2\end{array}$

So,

  $\begin{array}{c}\sqrt{144}=t_2\\ 12=t_2\end{array}$

Thus, The car travels distance in: $12$ sec.