Question

A woman who can row a boat at 6.4 km/hin still water faces a long, straight river with a width of 6.4 km and a current of 3.2 km/h.$\hat{\mathbf{i}}$ Letpoint directly across the river and$\hat{\mathbf{j}}$point directly downstream. If she rows in a straight line to a point directly opposite her starting position, (a) at what angle to$\hat{\mathbf{i}}$must she point the boat and (b) how long will she take? (c) How long will she take if, instead, she rowsdownthe river and then back to her starting point? (d) How long if she rows 3.2 kmupthe river and then back to her starting point? (e) At what angle to$\hat{\mathbf{i}}$should she point the boat if she wants to cross the river in the shortest possible time? (f) How long is that shortest time?

Short answer

a) The angle made by boat with $\hat{i}$if she rows in a straight line to a point directly opposite her starting point is $-30°$.

b) The time taken by boat if she rows in a straight line to a point directly opposite her starting point is69 min.

c) The time taken by boat if she rows 3.2 km down the river and then back to her starting point is 80 min

d) The time taken by boat if she rows 3.2 km up the river and then back to her starting point is 80 min

e) The boat should be pointed at $0°$angle to $\hat{i}$if she wants to cross the river in the shortest possible time.

f) The shortest time is 60 min long.

Step-by-step solution

The given data

1. The velocity of water with respect to ground,$V_{wg}=3.2\mathrm{km}/\mathrm{h}\mathrm{towards}\mathrm{east}$
2. The velocity of boat with respect to water, $V_{bw}=6.4\mathrm{km}/\mathrm{h}$.
3. The width of the river,$x=6.4\mathrm{km}$.
4. The distance rowed by the girl,$d=3.2\mathrm{km}$.

Understanding the concept of the relative motion

The observer or measurer of the velocity particle is determined by the frame of reference. The coordinate system of a physical object is attached to the reference of the frame.

Formulae:

The upstream velocity of a boat in the river,$V_{up}=V_1-V_2$ (i)

The downstream velocity of a boat in the river, $V_{down}=V_1+V_2$ (ii)

The time of travel or time taken by the boat, $t=\frac{x}{v}$ (iii)

a) Calculation of the angle made by the boat if she rows the boat to go to the opposite of the river

If the boat points as shown in the above diagram, then only will it row in a straight line and reach the opposite point of the starting point

To reach a point directly opposite means the velocity of the boat relative to ground must be equal to,

$\overrightarrow{V_{bg}}=\overrightarrow{V_{bw}}+\overrightarrow{V_{wg}}$

Thus, all the perpendicular components must cancel in the vector sum to get,

$\overrightarrow{V_{bg}}=\overrightarrow{V_{bg}}\hat{i}$

This gives us,

$V_{bw}\sin \theta =\left(-3.2\mathrm{km}/\mathrm{h}\right)$

The angle made by it withis given by solving the above equation as follows:

$$\begin{gathered} \theta ={\sin }^{-1}\left(-\frac{3.2}{6.4}\right) \\ =-30° \end{gathered}$$

Hence, the value of the angle made by the boat is$-30°$.

b) Calculation of the time taken by the boat to row to the opposite point

From above diagram, we can write that,

$$\begin{gathered} V_{bg}=V_{bw}\cos \theta \\ =6.4\mathrm{km}/\mathrm{h}\cos \left(30°\right) \\ =5.5\mathrm{km}/\mathrm{h} \end{gathered}$$

So, the time required to travel distance is given using equation (iii) as:

$$\begin{gathered} t=\frac{64\mathrm{km}}{5.5\mathrm{km}/\mathrm{h}} \\ =1.15\mathrm{h}\times \frac{60\min }{1\mathrm{h}} \\ =69\min \end{gathered}$$

Hence, the value of the time is 69 min.

c) Calculation of the time taken by the boat to travel 3.2 km down and up the river

The time taken by boat if she rows down the river and then back to her starting point is given using equations (i), (ii) and (iii) as follows:

$$\begin{gathered} t=\frac{3.2\mathrm{km}}{6.4\mathrm{km}/\mathrm{h}+3.2\mathrm{km}/\mathrm{h}}+\frac{3.2\mathrm{km}}{\left(6.4-3.2\right)\mathrm{km}/\mathrm{h}} \\ =1.33\mathrm{h}\times \frac{60\min }{1\mathrm{h}} \\ =80\min \end{gathered}$$

Hence, the value of the time taken is $80\min$.

d) Calculation of the time taken by the boat to travel 3.2 km up and down the river

The time taken by boat if she rows 3.2 km up the river and then back to her starting point is given using equations (i), (ii) and (iii) as follows:

$$\begin{gathered} t=\frac{3.2\mathrm{km}}{6.4\mathrm{km}/\mathrm{h}+3.2\mathrm{km}/\mathrm{h}}+\frac{3.2\mathrm{km}}{\left(6.4-3.2\right)\mathrm{km}/\mathrm{h}} \\ =1.33\mathrm{h}\times \frac{60\min }{1\mathrm{h}} \\ =80\min \end{gathered}$$

Hence, the value of the time taken is 80 min.

e) Calculation of the angle made by the boat while crossing the boat in shortest time

If she wants to cross the river in the shortest possible time then the boat should row in a straight line along $\hat{i}$, so the angle made by boat with$\hat{i}is0°$

f) Calculation of the shortest time

The shortest time is given using equation (iii) as follows:

$$\begin{gathered} t=\frac{6.4\mathrm{km}/\mathrm{h}}{64\mathrm{km}} \\ =1\mathrm{h}\times \frac{60\min }{1\mathrm{h}} \\ =60\min \end{gathered}$$

Hence, the value of the time is $60\min$.