# Distance Question

1. Sep 6, 2006

### Asteldoth

The question I'm stuck on is as follows:
A cheetah sees an antelope and sneaks up on it until it is 35m away. It then charges after the terrified animal at 31 m/s. The antelope simultaneously flees away from the predator at a rate of 20 m/s. What distance does the cheetah have to covrer in order to catch the antelope?

I'm at a complete loss here. If someone could show me how to figure this out, I'd be very greatful. I don't need the answer, just to be guided through the problem. Thanks in advance!

2. Sep 6, 2006

### chroot

Staff Emeritus
Do you know the general form of equations of motion with constant velocity?

$s(t) = s_0 + v_0 t$

Write two such equations, one for the cheetah, one for the antelope. Set them equal, meaning the two animals reach the same distance, and solve for t.

Finally, plug that value for t back into the cheetah's equation, and you can find the distance it travelled.

- Warren

3. Sep 6, 2006

### Asteldoth

I have equations like that... What does the S stand for?

This is only my third week in the class and my teacher is pretty much useless.

4. Sep 6, 2006

### chroot

Staff Emeritus
s, in this case, is the distance of the animal from the origin. You can also use x or any other letter you want; I typically use s.

The origin of the coordinate system is best chosen to be the cheetah's starting position. At time t=0, the cheetah is at position s=0, and the antelope is at position s=35.

- Warren

5. Sep 6, 2006

### Asteldoth

Gotcha! Thanks a lot! :D