Posted by **Sari** on Wednesday, September 2, 2009 at 9:58pm.

1

A runner dashes from the starting line (x = 0) to a point 130 m away and then turns around and runs to a point 11 m away from the starting point in 21 seconds. To the nearest tenth of a m/s what is the average speed?

2

What is the runner's average velocity in the previous problem?

- Please Help, Physics -
**MathMate**, Wednesday, September 2, 2009 at 10:04pm
1. average speed

Total distance

= 130 + 130 - 11

= 249 m

Total time = 21 s.

Average speed

= total distance / total time

=?? m/s

2. average velocity

Total displacement

=130-(130-11)

= 11

Total time

= 21 s.

Average velocity = total displacement/total time

= ?? m/s

- Please Help, Physics -
**bobpursley**, Wednesday, September 2, 2009 at 10:04pm
avg velocity=changedisplacement/time=11m/21s

avg speed=changedistance/time=(130+130-11)/21 m/s

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