Need Help With Kinematics, Speeding Car

Hi

I have been trying to work out an answer to the question below. I have found a good example of a question answered on here that has given me a lot of help in understanding. I still can't work it out. The only problem I have with this question is the 14seconds to reach 2.5 m/s². Which is different from the time overall taken.

Here's the question:

A police car is at rest, reading the speeds of passing cars. The police car is passed by a car breaking the speed limit at 30 m/s (uniform velocity). It takes 14 seconds later before the police car moves at a constant acceleration of 2.5 m/s² . How long will it take before the police car catches up with the speeding car?

What I've been doing is:

For the police officer:

d = ½ . a . t²

For the speeder:

d = v . t

The police officer can be considered to have passed the speeder when they have travelled an equal distance, so, I'm substituting the second equation into the first, giving us.

v . t = ½ . a . t²

It's here I get lost. This is because I have the time 't' and the time (14s). How do I work this from this point. It will probably end up as a quadratic equation to be solved by rule. It's just I can't even get to that point.

Thanks for any help

All help will be much appreciated as I feel I'm nearly there

One thing is certain that if the first car is speeding the police are going to have to break their own rules to catch up to the speeding car, so for the first 14 seconds police car was not moving, for the next period it will be still falling behind until it is able to accelerate to the same speed, after that the police car has to keep on accelerating until he has caught up the distance between them at that time. So I would look at the problem and solve it in the 3 sections.

Thanks a lot Robittybob

I'm going to assume the same as you say, that the police car was not moving for the first 14 seconds. It could also be assumed that it took 14seconds for the police car to get to a constant acceleration of 2.5m/s2.

For me I'm going to assume uniform acceleration AFTER 14 seconds passed. This way I can work out the time it took for both to cover the SAME distances (once their distances covered are equal then we know the police have caught up). I will then add the 14 seconds back on, as the speeder got a 14 second head start.

Hopefully this will work

Thanks again

Let us know what the answer is later.