Posted by **babidi** on Sunday, February 7, 2010 at 12:59am.

A person walks first at a constant speed of 4.50 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s.

- physics -
**drwls**, Sunday, February 7, 2010 at 1:04am
There is no question here. If you submit it again, show your work.

- physics -
**babidi**, Sunday, February 7, 2010 at 2:47am
A person walks first at a constant speed of 4.50 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s.

(a) What is her average speed over the entire trip?

(b) What is her average velocity over the entire trip?

- physics -
**drwls**, Sunday, February 7, 2010 at 8:15am
Since different lengths of time are spent walking in each direction, you can't just take the average of the two numbers. If v is the average speed,

1/v = (1/2) (1/4.5 + 1/3.0)

= (1/2)(2/9 + 3/9) = 5/18

v = 18/5 = 3.6 m/s

How can you derive that?

Let L be the distance between A and B

v*T = 2L defines tha average speed v

T = L/V1 + L/V2 = L(1/V1 + 1/V2)

1/v = T/2L = (1/2)(1/V1 + 1/V2)

The average velocity is the distance between start and finish points divided by elapsed time. Consider where you ended up and the answer will be obvious.

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