Physics
posted by Zoe on .
A person walks first at a constant speed of 4.70 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 2.80 m/s.
(a) What is her average speed over the entire trip?
(b) What is her average velocity over the entire trip?
I thought that I could just take the average of the two speeds in part A but it isn't the answer. I don't know how to solve for it with no distance given for either A or B.

The average speed does not give the real average. You would need to take the harmonic mean of the speed, i.e.
mean speed = 1/(1/4.7 + 1/2.8)/2= 3.5 m/s
You can also find the average speed by assuming a distance of one (metre).
Total distance = 2
Total time = 1/4.7 + 1/2.8 = 0.57 sec.
Average speed = total distance / total time
= 2/0.57 = 3.5 m/s. 
C.) What is her average velocity over the entire trip?
I was wondering how it is that we find out the velocity? I know that velocity would be change in X/change in time, but for the way the problem was set up, I'm not sure how to go about solving it. 
If he makes a round trip, the distance is 2*D, where D is the distance for the single trip.
At the end of the round trip, his displacement is zero, because he ends up in the same place as before, irrespective of the history of what he did in the mean time.
Velocity = change in displacement / change in time = ?