Posted by **alyson** on Tuesday, July 8, 2008 at 6:14pm.

i REALLY don't get this problem. PLEASE HELP!!!

the position of a runner during a 3 second time interval the runners position changes from x1 = 50m to x2 = 30.5m what was the runners average velocity?

find the average speed as well as the average velocity.

- physics -
**Damon**, Tuesday, July 8, 2008 at 6:29pm
The velocity is both magnitude and direction. Let's consider velocity in the +x direction positive.

Then the average velocity would be

V = (x2-x1)/t = (30.5-50)/3 = -6.5 m/s

which is negative.

The speed however is the magnitude of that vector or +6.5 m/s

- physics -
**alyson**, Tuesday, July 8, 2008 at 6:35pm
thanks a lot!!!!

- physics -
**Damon**, Tuesday, July 8, 2008 at 6:32pm
If you have had the "unit vector" in your physics subject so far, then a unit vector in the x direction would be i, in the y direction j and in the z direction k

In this case the velocity is -6.5 in the x direction so you would write the velocity vector V as

V = -6.5 i

- physics -
**Damon**, Tuesday, July 8, 2008 at 6:33pm
or V = -6.5 i + 0 j + 0 k

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