Answer
(a) $F=5100N$ (b) $15$ meters
Work Step by Step
(a) To find the acceleration of the car, use a kinematics equation relating initial velocity, final velocity, acceleration, and time, which is $$a=\frac{\Delta v}{\Delta t}$$ Substituting known values of $\Delta t=1.20s$ and $\Delta v=9.50m/s-16.0m/s=-6.50m/s$ yields an acceleration of $$a=\frac{-6.50m/s}{1.20s}=-5.4m/s^2$$ Using the fact that $F=ma$ and substituting known values of $a=5.42m/s^2$ and $m=950kg$ yields a force of $$F=(5.4m/s^2)(950kg)=5100N$$ (b) To find the change in distance, use the kinematics formula relating initial velocity, final velocity, distance, and acceleration, which is $$\Delta x=\frac{1}{2}a\Delta t^2 + v_o\Delta t$$ Substituting known values of $a=-5.4m/s^2$, $v_o=16.0m/s$, and $v_f=9.50m/s$ yields a distance of $$\Delta x=\frac{1}{2}(-5.4m/s^2)(1.20s)^2+(16.0m/s)(1.20s)=15m$$
