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3 years ago

Jack and jill exercise in a 25.0-m-long swimming pool

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1 answer:

Andrej [43]3 years ago

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Is this your question:

<u><em>Jack and jill exercise in a 25.0-m-long swimming pool. jack swims nine lengths of the pool in 2 minutes and 34.5 seconds, while jill, the faster swimmer, covers ten lengths in the same time interval. find the average velocity and average speed of each swimmer.</em></u>

If it is Here your answer:

<em><u>Average Speed of Jack : 1.46 m/s</u></em>

<em><u>Average Speed of Jill : 1.62 m/s</u></em>

<em><u>Average Velocity of Jack : 0</u></em>

<em><u>Average Velocity of Jill : 0.16 m/s</u></em>

Explanation:

Jack swims a distance of 9*25.0m=225.0 meters is 2 min and 34.5 seconds.

2 min and 34.5 seconds are 2*60sec +34.5 sec= 154.5 s

Jill covers 10*25.0m = 250 m in 154.5 s.

Average Speed = $\frac{Distance traveled}{Time of travel}$

Average Velocity = $\frac{Displacement}{Time}$

Thus,

the speed of Jack is : 225.0 meters /154.5 s =(225/154.5) m/s = 1.46 m/s

the speed of Jill is : 250.0 meters /154.5 s =(250/154.5) m/s = 1.62 m/s

Assuming Jack and Jill depart from point 0 towards the positive direction,

each even number of lengths, means Displacement = 0, and each odd number of lengths means Displacement = 25 m

So, average Velocity of Jill is 0,

average velocity of Jack is 25/ 154.5 = 0.16 m/s

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