Recent content by ShearonR

Thanks for the help everyone, I will keep this in mind next time I am answering these kinds of questions! :)

Alright, I tried something out and I am not entirely sure if it even makes sense. For the speeding car: Δd=vΔt Δt=Δd/v For the police car: Δd=vfΔt-0.5aΔt2 Δt=√Δd/0.5a Next I did the math: vΔt=vfΔt-0.5aΔt2 30(Δd/30)=vf(√Δd/0.5a)-0.5a(√Δd/0.5a)2 60Δd=vf(√Δd/0.5a)-(√Δd2/0.5a)...

Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.

Homework Statement A car, traveling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed...