# Catching a Crook

1. Sep 3, 2004

### epeake

The professor gave this problem in class today and i have tried all the formulas and apparently I am not doing something right. I am pretty sure the answer is about 1.5 seconds. This is the problem:

A crook is driving at a constant speed 85 mph. He zooms past a cop. It takes the cop 4 seconds to react. He goes from 0-95 in 14 seconds. How long does it take the cop to catch the crook.

2. Sep 3, 2004

### jamesrc

You need to find when the position of the two cars is the same. First consider the position of the crook as a function of time:

$$x_{\rm crook}(t) = v_{\rm crook} t$$

That one is easy because the crook travels at a constant speed.

Now the cop doesn't start until t=4 seconds. You are given the acceleration of the cop car (assume for this problem that the cop car accelerates at a constant rate and he accelerates until he catches the crook).

$$x_{\rm cop}(t) = \frac 1 2 a (t-4)^2$$

(this formula is only valid for t>4 seconds). Do you see why it's (t-4) instead of t in that formula?

If you set the positions equal to each other, you can solve for t. One caveat here, are you given whether or not the cop car has a maximum speed? Is, for example, 95 mph as fast as it can go? If so, it will affect the way you solve the problem. Try to figure out how. If you need any more help, please post your work on the problem so we can see how you're doing.

3. Sep 3, 2004

### epeake

Thanks thats awesome help. :)

Do you necessarily have to change the speed to meters?

4. Sep 3, 2004

### epeake

Would his acceleration speed be the 95 since that seems like when the cop becomes constant?

5. Sep 3, 2004

### jamesrc

Not if you're careful with the units. For example, mph/s, is a perfectly valid, if somewhat awkward way to express acceleration. There's no real reason to switch to meters here when you can stay in miles (or even mph*s though I'm not reccomending that).

6. Sep 3, 2004

### jamesrc

The acceleration would be:

$$a = \frac{(95-0)\rm mph}{14 \rm s}$$

which, as I was saying, you should convert to say, miles/s/s

7. Sep 3, 2004

### epeake

this is what i came up with.

42.46t = 1/2 3.03(14-4)^2
42.46t = 1/2 303
42.46t = 151.5
t = 151.5 / 42.46
t = 3.56 seconds

a = (42.46 - 0)
-----------
14s
a = 3.03

8. Sep 4, 2004

### jamesrc

Here's my solution:

$$x_{\rm crook}(t) = (85 \rm mph)t$$

$$x_{\rm cop}(t) = .5a(t-4)^2$$

$$a = \frac{95 \rm mph}{14} = 6.79 \frac{\rm mph}{\rm s}$$

Solve for t when:

$$x_{\rm crook} = x_{\rm cop}$$
$$85 t = .5*6.79*(t-4)^2$$

You solve that for t (using the quadratic formula) and get 32.5 seconds. (You actually get 2 roots for the equation, but one of them is ~.5 seconds which is invalid, because the equation we wrote for the cop's position is only valid after t = 4 seconds.)

Now let's check something: how fast would the cop be going if he was constantly accelerating for that long? Using v = v(0) + at, I get around 221 mph.

That's a little ridiculous. Let's assume the cop car has a max speed of 95 mph. (Intuitively, you should realize that this will make it take longer to catch the crook.)

The expression for the crook's position is the same as before. The cop's position is now given by:

$$x_{\rm cop} = X_{\rm acc} + (95 \rm mph)(t-18)$$

where Xacc is the distance travelled while accelerating. Really try to see why this expression is correct (and realize that it is only valid for t>18s).

Let's calculate Xacc first:

$$v^2 = v_o^2 + 2aX_{\rm acc}$$
$$X_{\rm acc} = \frac{(95 \rm mph)^2}{2\cdot 6.97 \frac{\rm mph}{\rm s}}$$
watch your units here to calculate:
$$X_{\rm acc} = 0.18 \rm miles$$

Now we can solve for t as before:

$$x_{\rm crook} = x_{\rm cop}$$
$$85 t = 0.18 + 95(t - 18)$$

For which, I get:

t ~ 171 s = 2 minutes and 11 seconds