Answered

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A jogger runs 4.0 km [W] in 0.50 h, then turns and runs 1.0 km [E] in 0.20 h, then 1.5 km [N] in 0.25 h, then 3.0 km [E] in 0.75 h and finally 1.5 km [S] in 0.30 h.  What is the average velocity?

Sagot :

AL2006
This is a sneaky trick question, to help you discover whether you know
one of the differences between velocity and speed.
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If you make a list of the distances and directions, and ignore the times,
you find these:

4 - west,  (3 + 1) - east . . . . .  zero in the east/west direction

1.5 - north,  1.5 - south . . . . . zero in the north/south direction

This jogger went out, had a nice jog around the neighborhood,and ended up exactly where he started.

Average velocity = (distance between start point and end point) / (time)

IF the question asked for average SPEED, then you would need the total distance, and divide it by the total time.  But it asks for VELOCITY, and that only involves the straight distance between the start point and the end point, regardless of the route taken in between.

The jogger ended up exactly where he started.  The distance between start and end points was zero.  Average velocity is  (zero) / (time) .  And that fraction is going to be Zero, no matter how long or how short the trip was, and no matter how much time it took.
After you draw the displacement vectors you will find the gogger returned to the start location
velocity = (displacement) /  ( t )      >>>>>>>>>>displacement = 0 ( jogger returns to its started point) , t = 0.5 h + 0.2h + 0.25h + 0.75h = 1.7 h
So v = 0/ 1.7 h = 0 
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