# Distance under constant acceleration

1. Nov 6, 2006

### Alpha[X]²

A body starts from rest with an acceleration of 11m/s².

What is it's velocity after 6 seconds. (I got 66 for that) and how far did it travel?

What forumla do I have to use for this?

2. Nov 6, 2006

There is a well-known expression which represents displacement under constant acceleration. Try to find it. (Either here or google it up.)

3. Nov 6, 2006

### Alpha[X]²

Is displacement the same as distance?

4. Nov 6, 2006

Yes it is, in this case.

5. Nov 6, 2006

### Alpha[X]²

What about the first part of the question?

I tried to average out the velocity and then using the d=vt formula I got 198m.

6. Nov 6, 2006

Yes, you did it correctly. Remember that you could calculate the displacement averaging out the velocity only because the acceleration is constant.

The other formula would be: d = (1/2)at2. Check that the solution is the same.

Try to show that v(average)*t = (1/2)at2

Last edited: Nov 6, 2006