Trooper chasing speeding car with constant acceleration

[Zhalfirin88]

Homework Statement

A car traveling at a constant speed of 108 km/hr passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets in a chase after the car with a constant acceleration of 3.0 m/s². How far does the trooper travel before he overtakes the speeding car?

Homework Equations

$$\Delta x = v_0 t + \frac{1}{2} a t^2$$

$$108 \frac {km}{hr} = 30 \frac {m}{sec}$$

The Attempt at a Solution

Well, since they're both moving along the same axis, I set the equations equal to each other.

$$v_0 t = \frac{1}{2} a t^2$$

$$v_0 = \frac{1}{2} a t$$

$$\frac{2 v_0}{a} = t$$

and I got t = 20. Then,

$$\Delta x = \frac{1}{2} a t^2$$

$$\Delta x = \frac{1}{2} (30 \frac{m}{s}) (20sec)^2$$

and I got 600 m but that was wrong.

 

[tiny-tim]

Welcome to PF!

Hi Zhalfirin8! Welcome to PF!

… One second after the speeding car passes the billboard, the trooper sets in a chase after the car …
…
Well, since they're both moving along the same axis, I set the equations equal to each other.

$$v_0 t = \frac{1}{2} a t^2$$

$$v_0 = \frac{1}{2} a t$$

$$\frac{2 v_0}{a} = t$$

and I got t = 20. Then,

$$\Delta x = \frac{1}{2} a t^2$$

$$\Delta x = \frac{1}{2} (30 \frac{m}{s}) (20sec)^2$$

and I got 600 m but that was wrong.

erm … what happened to that one second?

 

[Sakha]

Try equating both displacements, and don't forget that the cop will have a delayed time (i.e t-1s)

EDIT: tiny-tim was faster, sorry for the repetitive post.

 

[Zhalfirin88]

What would that change? The cop would still catch up to the guy in 20 seconds right?

So the 2nd equation would be:

$$\Delta x = \frac{1}{2} (30 \frac{m}{s}) (21sec)^2$$

?

 

[tiny-tim]

(just got up :zzz: …)

What would that change?

It would change t (in one of the equations only) to either t + 1 or t - 1.

(and then you have two equations which use the same t, so you can solve them jointly)

Have a go! ​