Question

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?

Short answer

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.

Step-by-step solution

Definition of velocity

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

Time is taken by the swimmer

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.

Velocity of the water with respect to earth

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

Velocity of the swimmer w.r.t friend at rest on the ground

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.