# Someone please help me in this kinematics question. I tried 4 times?

An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 120 km/h.

Precisely 1.00 sec. after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.90 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Here's my working:

I don't know where I went the wrong but the answer on the internet is 6.1s.

### 1 Answer

- Andrew SmithLv 71 month agoBest Answer
The first thing I looked for is if you had correctly converted to m/s 475/18 for the police car.. yes

120/3.6 = 100/3 for the other car yes.

So far so good.

The difference in their positions after 1 sec is also correct. = 125/18 m

I am having a little trouble following each step after this so I will firstly do it my way.

From the time the police car starts to accelerate it must gain as much speed above that of the speedster as it is currently below it to be as far behind as it was at the moment it started to accelerate.

ie 1.9 * t = 2*(100/3 - 475/18)

t = 2 * (100/3 - 475/18) / 1.9 = 7.3 s

If I recheck with the original figures just in case of a conversion error

t = 2*(120-95)/3.6 * 1.9) = 7.3

So you have made no errors. It really did take 7.3 s from the moment the police car stepped onto the accelerator. At this time he is STILL that distance of 125/18 m behind.

Clearly it is absolutely impossible for the answer of 6.1 to be valid. It takes MORE than 7.3 from starting to accelerate and more than 8.3 s in total.

I have now checked all your other steps and as far as I can see they are all correct.