Question

12) Driving home from school one day, you spot a ball rolling out into the street (FIGURE 5-27). You brake for 1.20 s, slowing your 950-kg car from 16.0 m>s to 9.50 m>s. What was the average forceexerted on your car during braking and How far did you travel while braking?

Answer 1

We are given the following information

Mass of car = 950 kg

Initial speed of car = 16.0 m/s

Final speed of car = 9.50 m/s

Time = 1.20 s

The average force exerted on the car during braking can be found using Newton's 2nd law of motion

$F=m\cdot a$

Where m is the mass of the car and a is the acceleration of the car.

The acceleration of the car is given by

$\begin{gathered} a=\frac{v_f-v_i}{t} \\ a=\frac{9.50-16.0}{1.20} \\ a=-5.4167\; \; \frac{m}{s^2} \end{gathered}$

The negative sign indicates deacceleration since the car is stopping.

So, the force is

$\begin{gathered} F=m\cdot a \\ F=950\cdot5.4167 \\ F=5145.865\; \; N \end{gathered}$

Therefore, an average force of 5145.865 N was exerted on your car during braking.

The distance traveled by the car while braking can be found as

$s=v_i\cdot t+\frac{1}{2}\cdot a\cdot t^2$

Let us substitute the given values

$\begin{gathered} s=16.0\cdot1.20+\frac{1}{2}\cdot(-5.4167)\cdot(1.20)^2 \\ s=19.20-3.90 \\ s=15.3\; m \end{gathered}$

Therefore, the car traveled a distance of 15.3 m while braking.