Jack and jill exercise in a 25.0-m-long swimming pool. jack swims nine lengths of the pool in 157.9 seconds (2 minutes and 37.9 seconds), while jill, the faster swimmer, covers ten lengths in the same time interval. find the average velocity and average speed of each swimmer.

solution:

Average velocity

Average velocity is,

Here x is the displacement and t is the time.

At the end of swimming nine length of the pool, Jack reaches the other end of pool, hence the displacement x is the length of pool, that is 25.0 m. The total time taken, t is 157.9s.

Substitute 25.0 m for x and 157.9 s for t to calculate the average velocity. The average velocity of Jack is,

v_{a} =25.0m/157.9s

v_{a} =0.158m/s

At the end of swimming ten length of the pool, Jill reaches where he started, hence the displacement x is zero. The total time taken, t is 157.9s.

Substitute 0.00 m for x and 157.9s for t to calculate the average velocity. The average velocity of Jill is,

v_{a} =0m/157.9s

v_{a}= 0m/s

The average velocity of Jack is 0.158m/s and the average velocity of Jill is 0 m/s.

Average speed

Average speed is,

Here, d is the distance travelled and t is the time.

At the end of swimming nine length of the pool, distance travelled by Jack is nine times the length of pool; that is,

The total time taken, t is 157.9s.

Substitute 225 m for d and 157.9 s for t to calculate the average speed. The average speed of Jack is,

s_{a} =225m/157.9s

1.58m/s

  =1.42m/s

At the end of swimming ten length of the pool, distance travelled by Jill is ten times the length of pool; that is,

The total time taken, t is 157.9s.

Substitute 250 m for d and 157.9s for t to calculate the average speed. The average speed of Jill is,

s_{a} =250m/157.9s

1.58m/s

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

true tan speed = ang vel x radius