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Q65E

Expert-verifiedFound in: Page 125

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**An athlete crosses a ** **-m-wide river by swimming perpendicular to the water current at a speed of ** **m/s relative to the water. He reaches the opposite side at a distance ** ** m downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground?**

The total time taken to cross the river is $50$s. The velocity of the water will be $0.8$ m/s. The velocity of the swimmer w.r.t his friend at rest on the ground is $9.8$m/s.

**Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.**

The velocity of the swimmer relative to the water is V_{sw }= $0.5$m/s.

He is swimming at a velocity perpendicular to the current. We know the distance covered by him in the positive direction

The time is taken by him to cross the river:

$\begin{array}{rcl}V& =& \frac{d}{t}\\ 0.5& =& \frac{25}{t}\\ t& =& 50s\end{array}$

The total time taken to cross the river is $50$ s.

The velocity of the water w.r.t earth:

$\begin{array}{rcl}{V}_{we}& =& \frac{d}{t}\\ & =& \frac{40}{50}\\ & =& 0.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The velocity of the water will be $0.8$m/s

The velocity will be calculated by the following equation:

$\begin{array}{rcl}t& =& 50s\\ D& =& \sqrt{{25}^{2}+{40}^{2}}\\ D& =& 47.2m\end{array}$

Using the above equation, we can calculate the speed:

$\begin{array}{rcl}V& =& \frac{d}{t}\\ & =& \frac{47.2}{50}\\ & =& 0.94\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The velocity of the swimmer w.r.t his friend at rest on the ground is $0.94$ m/s.

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