91Ó°ÊÓ

• Speed of the boat in still water,$v_{xb}=6.40\mathrm{km}/\mathrm{h}$.
• Speed of water current,$v_{yr}=3.20\mathrm{km}/\mathrm{h}$.
• Width of the river, $d=6.40\mathrm{km}$.

When there is a relative motion between the two frames of reference Aand Bat a constant velocity, the velocity of a particle P is different for each observer in frames A and B. The two measured velocities are related by

${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{PA}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{PB}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{BA}}$

Here${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{B}\mathbf{A}}$ is the velocity of B with respect to A.

We can find the time by using the velocity equation. We can consider the velocities of boat and river as resultant velocity components and then we can find resultant velocity. We use trigonometric functions to find angles.

Formula:

The velocity of a body in motion,$v=\frac{d}{t}$ (i)

Assuming that the velocityof the boat is along x and the velocityof the river is along y, we can find the directionof the boat using tan(y/x). Hence, the direction is given as:

$$\begin{gathered} \tan \mathrm{θ}=-\frac{3.20}{6.40} \\ \mathrm{θ}={\tan }^{-1}\left(-\frac{3.20}{6.40}\right) \\ =-26.56° \end{gathered}$$

So, the boat should be $26.56°$with an x axis to reach the opposite point.

From equation (i), we get the time taken to cross the river by the boat. Substituting the values of the given quantities in equation.

$$\begin{gathered} t=\frac{\mathrm{d}}{{\mathrm{V}}_{\mathrm{ix}}} \\ =\frac{6.4\mathrm{km}}{\left(6.4\mathrm{km}/\mathrm{h}\right)\left(\cos \left(26.56°\right)\right)} \\ =1.12\mathrm{h} \end{gathered}$$

Hence, the time taken by the boat to cross the river is $1.12\mathrm{h}$.

Resultant velocity of the boat for downstream,

$$\begin{gathered} v_d=v_{xb}+v_{yr} \\ =6.4\mathrm{km}/\mathrm{h}+3.2\mathrm{km}/\mathrm{h} \\ =9.6\mathrm{km}/\mathrm{h} \end{gathered}$$

Resultant velocity for upstream,

$$\begin{gathered} v_u=v_{xb}+v_{yr} \\ =6.4\mathrm{km}/\mathrm{h}-3.2\mathrm{km}/\mathrm{h} \\ =3.2\mathrm{km}/\mathrm{h} \end{gathered}$$

Total time taken by boat to row down and row back to starting point can be found by using equation (i). Substitute the values of the quantities in equation (i), we get,

$$\begin{gathered} t=\frac{d}{v_d}+\frac{d}{v_u} \\ =\left(\frac{3.2\mathrm{km}}{9.6\mathrm{km}/\mathrm{h}}\right)+\left(\frac{3.2\mathrm{km}}{3.2\mathrm{km}/\mathrm{h}}\right) \\ =1.33\mathrm{h} \end{gathered}$$

Hence, the total time taken for going both upstream and downstream is $1.33\mathrm{h}$.

Using same equation above, we get the total time taken for going up and returning back. Therefore, substitute the value of the quantities in the equation (i).

$$\begin{gathered} t=\left(\frac{3.2\mathrm{km}}{3.2\mathrm{km}/\mathrm{h}}\right)+\left(\frac{3.2\mathrm{km}}{9.6\mathrm{km}/\mathrm{h}}\right) \\ =1.33\mathrm{h} \end{gathered}$$

Hence, the total time taken for going up and returning back is $1.33\mathrm{h}$.

For the shortesttime the distanceshould be shortest, so the angle should be 0.

Hence, using equation (i) and$\mathrm{θ}=0°$we get

$$\begin{gathered} t=\frac{6.4\mathrm{km}}{6.4\mathrm{km}/\mathrm{h}·\cos \left(0°\right)} \\ =1.0\mathrm{h} \end{gathered}$$

Hence, the shortest time taken is $1.0\mathrm{h}$.