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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