Chapter 4, Problem 7P

What average force is required to stop a 950-kg car in 8.0 s if the car is traveling at 95 km/h?

To determine

The average force required to stop a car.

Answer to Problem 7P

Solution:

The force required to stop a car is $-3133.6\text{ N}$

Explanation of Solution

The concept we have to use to solve this problem is average acceleration and force. The average acceleration is the change in velocity with respect to the elapsed time. So the expression for the average acceleration is given as below:

$a_{\text{avg}}=\frac{v_f-v_i}{t}$ ………… (1)

Here, $v_f$ is the final velocity, $v_i$ is the initial velocity and $t$ is the final time.

The force on an object is calculated by using Newton’s second law of motion as given below:

$F_{\text{avg}}=m{a}_{\text{avg}}$ ………… (2)

Given:

Mass of the car $m=950\text{ kg}$

Final Time $t=8.0\text{ s}$

Initial speed $v_i=95\text{ km/h}$

Final speed $v_f=0\text{ km/h}$

Formula used:

$a_{\text{avg}}=\frac{v_f-v_i}{t}$

$F_{\text{avg}}=m{a}_{\text{avg}}$

Calculation:

Substitute, $v_i=95\text{ km/h}$, $v_f=0\text{ km/h}$ and $t=8.0\text{ s}$ in the equation (1)so the average acceleration is given as below:

$a_{\text{avg}}=\frac{v_f-v_i}{t}$

        $=\frac{0\text{ km/h-}95\text{ km/h}\left(\frac{1000\text{ m}}{3600\text{ s}}\right)}{8.0\text{ s}}$

        =-3.29 m/s²

Now we can find the magnitude of force exerted by car by substituting $m=950\text{ kg}$ and $a_{\text{avg}}=-3.29{\text{ m/s}}^2$ in equation (2)as given below:

$F_{\text{avg}}=m{a}_{\text{avg}}$

        $=\left(950\text{ kg}\right)\left(-3.29{\text{ m/s}}^{\text{2}}\right)$

        =-3133.6 N