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
[tex]F=m\cdot a[/tex]
Where m is the mass of the car and a is the acceleration of the car.
The acceleration of the car is given by
[tex]\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}[/tex]
The negative sign indicates deacceleration since the car is stopping.
So, the force is
[tex]\begin{gathered} F=m\cdot a \\ F=950\cdot5.4167 \\ F=5145.865\; \; N \end{gathered}[/tex]
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
[tex]s=v_i\cdot t+\frac{1}{2}\cdot a\cdot t^2[/tex]
Let us substitute the given values
[tex]\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}[/tex]
Therefore, the car traveled a distance of 15.3 m while braking.
