calculating distance between two moving vehicles

[Ax3]

"A speeding motorbike travels past a stationary police car at a speed of 35ms-1. The police car starts accelerating immediately at 4ms-2, and keeps accelerating at this rate until it has passed the bike.

- How far has does the police car travel before it overtakes the motorbike? (Ans=613m)
- At what time does the police car overtake the motorbike? (Ans=17.5s)

I have tried using a suvat equation, namely v=u+at to calculate the time taken for the police car to catch up to the motorbike (where i use v=35), however obviously the police car will never catch up to the bike if it only matches its velocity.. I'm really unsure of what to do next because trial and error using suvat equations has not provided me with much assistance.

Thanks in advance for any help! (Im just after some clarification as the steps i would need to take/variables i would need to identify in order to answer a question similar to this)

[Aden]

Let us first find the time it takes for the police car to catch up to the motorbike (ie. when their distances are equal).

To do this, we can use for the motorbike, since we know that ; and for the police car since we know that . Since we are trying to find the time when the police car catches up to the motorbike, we can say that both the variables in the two equations are the same. Knowing this, we have:




The time cannot equal to 0, so it must be 17.5 seconds.

For the distance, we can just sub in the 17.5 seconds into either equation:



Actually, there is probably a much easier way to do this question...

[superflya]

draw the velocity - time graph to help u visualise.

[Ax3]

Ohh shit i didnt assume the motorbikes velocity was constant. Thanks guys

[superflya]

haha it happens.

[Chavi]

I'd like to propose a slightly different version of this question:

A car travels at a constant speed of 80km/h and passes a stationary police bike.
The police car begins accelerating uniformly to 80km/s in 10s and then reaches a constant speed of 100km/h after a further 5s.
At what time + distance does the policeman catch up to the car?

[superflya]

by drawing the v-t graph, ull see that after 15 seconds the distance isn't equal. calculate the distance required and use x=vt to find the additional time btw the velocity after 15 seconds is the v of bike minus the v of car. there is another more equation involved way that I'd use but too lazy to use latex. this method should work fine.

[Aden]

I've done this question before, here:

http://vcenotes.com/forum/index.php/topic,21316.0.html

You will have to consider the police car's motion from , and ; not just from and .