Distance traveled = Average velocity x time

[Voltman]

A car traveling with a constant speed of 80 km/h passes a stationary motorcyle policeman. The policeman sets off in pursuit, accelerating to 80 km/h in 10.0 s and reaching a constant speed of 100km/h after a further 5.0 s. At what time will the policeman catch up with the car?


I am stuck in the above problem. I do not know what to do. Any help/complete working out solution in the above problem will be appreciated! Please provide hints for me to work this out.

 

[Gib Z]

Welcome to Physicsforums Voltman! We have a policy that we can not give help without at least some sort of minimal effort, as I'm sure you can understand why. There should have been a template given to you, you shouldn't have deleted it but rather filled it out, its quite useful actually!

I will try to give you some sort of a start. From the way the question was worded, let's assume that the accelerations of the policeman are constant. So in the first 10 seconds, what was his acceleration? How about in the next 5 seconds?

Are you familiar with the standard equations of motion? eg v=u + at. Using these equations, with our new knowledge of the accelerations for each part of the bikers motion so far, you can work out the distance the biker has traveled in these 15 seconds. Its also easy to work out the distance the car got as well, so we know the distance between them after 15 seconds, and their speeds. You can surely finish that problem.

 

[rl.bhat]

The distance traveled = Average velocity x time.
Using this formula find the distance traveled by the policeman in 15 s.
During 15 s find the distance traveled by the car.
Find the distance between the car and policeman. Let it be x.
Now both are moving with constant velocity. After time t policeman catches the car. During this period car travels s distance and policeman travels s + x distance.
So (100km/h)/(s + x) = (80km/h)/s. Find s and then t.