Q15DQ - University Physics with Modern Physics

Question

You are on the west bank of a river that is flowing north with a speed of $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{m}\mathbf{/}\mathbf{s}$ . Your swimming speed relative to the water is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{m}\mathbf{/}\mathbf{s}$, and the river is 60 m wide. What is your path relative to the earth that allows you to cross the river in the shortest time? Explain your reasoning.

Step-by-Step Solution

Answer

The distance from the earth along which the person cross the river is 76.77 m . So, the swimmer should swim across the river with their heads straight in order to do so in the smallest amount of time.

Step 1: Identification of given data:

The given data can be listed below.

 

The speed of river flowing north is ${\mathrm{v}}_{\mathrm{MW}}=1.2 \mathrm{m}/\mathrm{s}$.

Speed of the person is ${\mathrm{v}}_{\mathrm{WE}}=1.5 \mathrm{m}/\mathrm{s}$ .

Width of the river is ${\mathrm{d}}_{\mathrm{y}}=60 \mathrm{m}$ .

Step 2: Concept/Significance of relative speed.

The relative speed of two objects traveling in the same direction at different speeds is equal to the difference between their speeds. Their relative speed is equal to the sum of their speeds when they are going in opposing directions.

Step 3: Determination of path relative to the earth that allows to cross the river in the shortest time.

The free body diagram is given below as,

                               

The relative velocity of the man with respect to water is given by,

 ${\overrightarrow{\mathrm{v}}}_{\mathrm{MW}}={\overrightarrow{\mathrm{v}}}_{\mathrm{ME}}-{\overrightarrow{\mathrm{v}}}_{\mathrm{WE}}$

Here, ${\overrightarrow{\mathrm{v}}}_{\mathrm{ME}}$ is the velocity of the man relative to earth, ${\overrightarrow{\mathrm{v}}}_{\mathrm{WE}}$ is the velocity of water relative to earth.

Rearrange the above equation for velocity of man is given by,

${\overrightarrow{\mathrm{v}}}_{\mathrm{ME}}={\overrightarrow{\mathrm{v}}}_{\mathrm{WE}}-{\overrightarrow{\mathrm{v}}}_{\mathrm{MW}}$ 

The magnitude of velocity of man relative to earth is given by,

${\mathrm{v}}_{\mathrm{ME}}=\sqrt{{\mathrm{v}}_{\mathrm{WE}}^2+{\mathrm{v}}_{\mathrm{MW}}^2}$ 

 

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{ME}}=\sqrt{{\left(1.5 \mathrm{m}/\mathrm{s}\right)}^2+{\left(1.2 \mathrm{m}/\mathrm{s}\right)}^2} \\ =1.92 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

The direction of velocity is given by,

$\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{v}}_{\mathrm{WE}}}{{\mathrm{v}}_{\mathrm{MW}}}\right)$ 

Substitute all the values in the above,

$$\begin{gathered} \mathrm{\theta }={\tan }^{-1}\left(\frac{1.2}{1.5}\right) \\ =38.6° \end{gathered}$$ 

 

From the free body diagram the distance of path travelled by man with respect to water is given by,

$$\begin{gathered} \mathrm{cos\theta }=\frac{{\mathrm{d}}_{\mathrm{y}}}{\mathrm{d}} \\ \mathrm{d}=\frac{{\mathrm{d}}_{\mathrm{y}}}{\mathrm{cos\theta }} \end{gathered}$$ 

Substitute all the values in the above,

$$\begin{gathered} \mathrm{d}=\frac{60 \mathrm{m}}{\cos 38.6°} \\ =\frac{60 \mathrm{m}}{0.7815} \\ =76.77 \mathrm{m} \end{gathered}$$ 

 

Thus, the distance from the earth along which the person cross the river is $76.77 \mathrm{m}$ . So, the swimmer should swim across the river with their heads straight in order to do so in the smallest amount of time.