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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 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.

Respuesta :

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

fichoh

The velocity is the ratio of the displacement to the time spent while distance is the ratio of the speed to the time, Hence, the velocity of Jack and Jill can be calculated thus :

  • Velocity = (Displacement / time)

  • Speed = (distance / time)

Jack :

  • Total Distance covered = 25 × 9 = 225 meters

  • Displacement = 25 meters

  • Time taken = 2 min 34.5 seconds = (120 + 34.5) = 154.5 seconds

Speed = (225 / 154.5) = 1.456 m/s

Velocity = (25 / 154.5) = 0.162 m/s

Jill :

  • Total Distance covered = 25 × 10 = 250 meters

  • Displacement = 0 meters

  • Time taken = 2 min 34.5 seconds = (120 + 34.5) = 154.5 seconds

Speed = (250 / 154.5) = 1.618 m/s

Velocity = 0/154.5 = 0 m/s

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