Posted by **Zoe** on Thursday, October 8, 2009 at 12:55pm.

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.

- Physics -
**MathMate**, Thursday, October 8, 2009 at 1:35pm
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.

- Physics -
**Zoe**, Thursday, October 8, 2009 at 2:41pm
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.

- Physics -
**MathMate**, Thursday, October 8, 2009 at 2:45pm
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 = ?

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