Question

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

Short answer

The average force on the car is 3135 N.

Step-by-step solution

Step 1. Given data and assumption

During the motion, the car has a constant acceleration.The force on the vehicle is equal to the product of the mass and acceleration.

Given data:

The mass of the car is $m=950\mathrm{kg}$.

The time to move the car is $t=8.0\mathrm{s}$.

The initial speed of the car is

$\begin{array}{c}{v}_{\mathrm{i}}=95\frac{\mathrm{km}}{\mathrm{h}}\\ =26.39\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

The final speed of the car is $v_{\mathrm{f}}=0$as the car stops at the end.

Assumption:

Let a be the acceleration of the car.

Step 2. Calculation of the average force

The acceleration of the car is

$\begin{array}{c}a=\frac{v_f-v_i}{t}\\ =\frac{0-26.39\frac{\mathrm{m}}{\mathrm{s}}}{8.0\mathrm{s}}\\ =-3.30\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, from Newton’s second law, you get that the average force on the car is

$\begin{array}{c}F=ma\\ =\left(950\mathrm{kg}\right)\times \left(-3.30\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\\ =-3135\mathrm{N}\end{array}$

(Negative value of the average force means that the average force is against the direction of the initial velocity.)

Hence, the average force on the car is 3135 N.