91Ó°ÊÓ

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,$d=3.2\mathrm{km}$.

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)

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,

$\stackrel{→}{V_{bg}}=\stackrel{→}{V_{bw}}+\stackrel{→}{V_{wg}}$

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

$\stackrel{→}{V_{bg}}=\stackrel{→}{V_{bg}}\hat{i}$

This gives us,

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

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

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

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

From above diagram, we can write that,

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

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

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

Hence, the value of the time is 69 min.

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:

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

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

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:

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

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

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$\hat{i}is0°$

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

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

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