Chapter 9, Problem 59RE

To determine

To find: The actual speed of the swimmer and the downstream distance if the swimmer ends up from the point directly across the river from the starting point.

Answer to Problem 59RE

The actual speed of the swimmer and the downstream distance if the swimmer ends up from the point directly across the river from the starting point are $5.39\text{miles/hr}$ and $0.4\text{miles}$ respectively.

Explanation of Solution

Given information:

The velocity of the swimmer is $v_1=5\text{miles/hr}$

The velocity of the current is $v_2=2\text{miles/hr}$

Calculation:

Calculate the resultant velocity.

  $\begin{array}{c}v=\sqrt{v_1^2+v_2^2}\\ =\sqrt{{\left(5\text{miles/hr}\right)}^2+{\left(2\text{miles/hr}\right)}^2}\\ =\sqrt{\left(29\text{miles/hr}\right)}\\ =5.39\text{miles/hr}\end{array}$

Thus, the actual velocity of the swimmer is $5.39\text{miles/hr}$ .

Calculate the angle made by the resultant vector.

  $\begin{array}{c}\tan \theta =\frac{v_2}{v_1}\\ \tan \theta =\frac{2\text{miles/hr}}{5\text{miles/hr}}\\ \theta ={\tan }^{-1}\left(\frac{2}{5}\right)\\ \theta =21.8°\end{array}$

In the given figure the swimmer started his swinging from point A and the point B is in the front of that point. The swimmer is expected to reach point B but due to the flow of current of the river he ends up at the point C.

The downstream distance from point B is,

  $\begin{array}{c}\tan \left(21.8\right)=\frac{s}{v_1}\\ s=\frac{2}{5}\\ s=0.4\text{miles}\end{array}$

Therefore, the actual speed of the swimmer and the downstream distance if the swimmer ends up from the point directly across the river from the starting point are $5.39\text{miles/hr}$ and $0.4\text{miles}$ respectively.