## College Physics (4th Edition)

(a) On the graph, we can see that the two velocities are equal at $t = 11~s$ (b) The police car has not caught up with the speeder at $t = 16 ~s$ (c) At $t = 5~s$, the velocity of the motorcycle with respect to the police car is $12~m/s$ At $t = 10~s$, the velocity of the motorcycle with respect to the police car is $2~m/s$
(a) On the graph, we can see that the two velocities are equal at $t = 11~s$ (b) The area under the velocity versus time graph is the displacement. We can see that between $t=0~s$ and $t=16~s$, the area under the motorcycle's graph is greater than the area under the police car's graph. Therefore, the motorcycle's displacement is greater, so the police car has not caught up with the speeder at $t = 16 ~s$ (c) At $t = 5~s$, the velocity of the police car is $24~m/s$ and the velocity of the motorcycle is $36~m/s$. The velocity of the motorcycle with respect to the police car is $12~m/s$ At $t = 10~s$, the velocity of the police car is $34~m/s$ and the velocity of the motorcycle is $36~m/s$. The velocity of the motorcycle with respect to the police car is $2~m/s$