Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 <i>displacement</i> 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 = ?