physics
posted by babidi on .
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.

There is no question here. If you submit it again, show your work.

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? 
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.