Jack and Jill exercise in a 25.0-m-long swimming pool. Jack swims nine lengths of the pool in 2 minutes and 22.3 seconds, while Jill, the faster swimmer, covers ten lengths in the same time interval. Find the average velocity and average speed of each swimmer.

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Jack and Jill exercise in a 25.0-m-long swimming pool. Jack swims nine lengths of the pool in 2 minutes and 22.3 seconds, while Jill, the faster swimmer, covers ten lengths in the same time interval. Find the average velocity and average speed of each swimmer.

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Hmmm, Lall appears to be slightly confused by the change in direction, which occurs each time the swimmers turn at the ends of the pool and requires an ACCELERATION. As Jill has completed an even number of lengths and hence returned to the starting point her average velocity must be zero. However, Jack has completed an odd number of lengths so must have a non-zero average velocity. i.e his DISPLACEMENT from the starting point divided by TIME taken is greater than zero. I hope this helps.
so i was wrong so i must delete the post above...
So would you go with 25.0 m of the pool and multiply it by 9 since he did it that many times to get the displacement then divide it by the 2 minutes and 22.3 seconds (about 142 seconds)?

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https://akk.li/pics/anne/jpg that should help
velocity is a vector quantity, whereas speed is an algebraic quantity. For example, If I swim two time the length of the pool then the average velocity is the zero vector |dw:1444183986824:dw|
whereas the average speed is: \[\frac{{2L}}{{\Delta t}}\] where \L\) is the length of the pool and \( \Delta t\) is time needed to swim the pool two times
oops... where \(L\) is...
So I would take 25.0/2 mins and 22.3 secs for the the speed?
In the case of Jake, his velocity is a vector whose magnitude is: \(25/142.3=...m/sec\) whereas the average speed is: \( (25 \cdot 9)/142.3=...m/sec\)
I meant average velocity, of course
So 1.58 m/s for the speed .18 m/s for the velocity

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