# Kinematics car chase question

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1. Mar 19, 2017

### vbrasic

1. The problem statement, all variables and given/known data
We have that a police car is travelling at $26.4$ m/s, and that at a time, $t=0$, a car speeds by at $37.5$ m/s. After $1$ s, the police car accelerates constantly at $2$ m/s2. We are asked to find the time, $t$, where the police car catches the speeder.

2. Relevant equations
$v_f=v_i+at$, and $s=v_0t+\frac{1}{2}at^2$. (Standard kinematics equations.)

3. The attempt at a solution
Granted that the police car is not accelerating for the first second, I chose to simply begin analyzing the motion at the $1$ s mark. In the first second, the police car moves, $26.4$ m, while the speeder moves, $37.5$ m. We have then, in this frame, that the speeder's displacement after $1$ s is given by, $(37.5-26.4)+37.5t$. That is, I simply state that the initial position of the speeder is given by the displacement between the two vehicles after $1$ s. The police car's displacement is $26.4t+t^2$. We simply rearrange then to solve the quadratic for $t$, such that $t=12.2$ s. My book says the answer is $13$ s. I'm not sure if this is simply due to a rounding error, or if there's something wrong with my logic.

2. Mar 19, 2017

### sunnnystrong

Use: X-Xo = VoT + 1/2AT^2
For the police car:
X-Xo = 1/2(2)(T-1)^2 + (26.4)T
For the car:
X-Xo = (37.5)T

Set them equal to each other & solve for T :)
The trick is realizing that the police car has two components for distance...

3. Mar 19, 2017

### TomHart

I got 12.02 s as an answer.

Edit: I thought your method looked good.

Edit2: Oops. We both forgot to add back in the 1 second, so the answer looks like it should be 13.02 s.

4. Mar 19, 2017

### TomHart

That will work too, but for the police car I think it should be:
x-xo = (0.5)(2)(t-1)2 + 26.4(t-1)
Edit: I'm sorry. If you are using xo = 0, then your equation is correct as it is written. That is probably what you intended.

Last edited: Mar 19, 2017
5. Mar 19, 2017

### vbrasic

Well that explains it. I knew I was missing something. Thanks.