Q:

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.

Accepted Solution

A:
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.


[tex]\displaystyle{ Average\ Speed= \frac{Distance\ traveled}{Time \ of \ travel}\\\\ [/tex]



[tex]\displaystyle{ Average \ Velocity = \frac{Displacement}{time} [/tex]



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

Answer: 

Average Speed of Jack : 1.46 m/s

Average Speed of Jill : 1.62 m/s

Average Velocity of Jack : 0

Average Velocity of Jill : 0.16 m/s