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Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 155.5 s (2 min and 35.5 s) , whereas Jill, the faster swimmer, covers 10 lengths in the same time interval. Find the average velocity and average speed of each swimmer.

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9 votes

Answer:

Jack:

Average velocity:
\vec v_(avg, Ja) = +0.161\,(m)/(s), Average speed:
v_(avg,Ja) = 1.447\,(m)/(s)

Jill:

Average velocity:
\vec v_(avg, Ji) = 0\,(m)/(s), Average speed:
v_(avg,Ja) = 1.608\,(m)/(s)

Step-by-step explanation:

We assume that each swimmer returns to the point of the departure each two lengths of the pool.

The average velocity is defined by the following definition:


\vec v_(avg) = (1)/(\Delta t) \cdot \Sigma_(i=1)^(n) \vec r_(i) (1)

Where:


\Delta t - Total time, measured in seconds.


\vec r_(i) - i-th Relative vector position, measured in meters.


n - Number of lengths done by the swimmer.

And the average speed is represented by the following formula:


v_(avg) = (1)/(\Delta t) \cdot \|\vec r_(i)\| (2)

Where
\|\vec r_(i)\| is the norm of the i-th relative vector position, measured in meters.

Now, we proceed to calculate both values for both swimmers:

Jack - Average velocity (
\Delta t = 155.5\,s,
\vec r_(1) = \vec r_(3) = \vec r_(5) = \vec r_(7) = \vec r_(9) = +25\,m,
\vec r_(2) = \vec r_(4) = \vec r_(6) = \vec r_(8) = -25\,m)


\vec v_(avg,Ja) = (5\cdot (25\,m)+4\cdot (-25\,m))/(155.5\,s)


\vec v_(avg, Ja) = +0.161\,(m)/(s)

Jack - Average speed (
\Delta t = 155.5\,s,
\|\vec r_(i)\| = 25\,m,
n = 9)


v_(avg,Ja) = (9\cdot (25\,m))/(155.5\,s)


v_(avg,Ja) = 1.447\,(m)/(s)

Jill - Average velocity (
\Delta t = 155.5\,s,
\vec r_(1) = \vec r_(3) = \vec r_(5) = \vec r_(7) = \vec r_(9) = +25\,m,
\vec r_(2) = \vec r_(4) = \vec r_(6) = \vec r_(8) = \vec r_(10) = -25\,m)


\vec v_(avg,Ji) = (5\cdot (25\,m)+5\cdot (-25\,m))/(155.5\,s)


\vec v_(avg, Ji) = 0\,(m)/(s)

Jill - Average speed (
\Delta t = 155.5\,s,
\|\vec r_(i)\| = 25\,m,
n = 10)


v_(avg,Ji) = (10\cdot (25\,m))/(155.5\,s)


v_(avg,Ja) = 1.608\,(m)/(s)

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