91Ó°ÊÓ

1. The velocity of water with respect to the ground $\left(V_{\mathrm{wg}}\right)=1.1m/s$towards the east.
2. The velocity of the boat with respect to water,$\left({\mathrm{V}}_{\mathrm{bw}}\right)=4\mathrm{m}/\mathrm{s}$

The velocity of a particle Pas measured by an observer in frame Ais different than the velocity measured in frame B when two frames of reference Aand Bare moving relative to each other at a constant velocity. The relation between two measured velocities is given as,

${\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{PA}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{P}\mathbf{B}}\mathbf{+}{\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{B}\mathbf{A}}$

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

Using the relative motion concept, we can find the magnitude and direction of one of the vectors.

Formulae:

The magnitude of the resultant vector $\stackrel{→}{V}$is given by,$V=\sqrt{V_x^2+V_y^2}$ (i)

The direction of the vector is given by,

$\mathrm{θ}={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$ (ii)

The velocity of the boat is 4m/s. So, if the time taken by the boat to reach the opposite end is t, then we can say that the distance covered by the boat is given by:

$r_{bg}=4t$

In the same time t, boat would travel 200 m across the river and (85.1.1t)m of distance upstream. This is because water is moving with 1.1m/s and would drag the boat with the same speed. In time t, it would drag the boat by 1.1t downstream. So, to counter this, we have to add (1.1t)m of distance along upstream. So, the magnitude of the position can be given using equation (i) as:

$$\begin{gathered} r_{bg}=\sqrt{{r_{bw}}^2+{r_{wg}}^2} \\ 4t=\sqrt{{\left(200\right)}^2+{\left(82+1.1t\right)}^2} \\ 16t={\left(200\right)}^2+{\left(82+1.1t\right)}^2 \\ -14.8t^2+180.4t+46724=0 \end{gathered}$$

After solving this quadratic equation, we got the time taken by the boat to cross the river and land in the clearing as: t=62.6 s

The direction in which the boat must be pointed in order to travel in a straight line and land in the clearing on the north bank is given using equation (ii) as follows: (from the figure)

$$\begin{gathered} \tan \mathrm{θ}=\frac{82+1.1t}{200} \\ =\frac{82+1.1\left(62.6\right)}{200} \\ =\frac{151}{200} \\ \mathrm{θ}={\tan }^{-1}\left(\frac{151}{200}\right) \\ =37.05° \\ ≈37° \end{gathered}$$

Hence, the value of the direction of the boat is .

From part (a) calculations, we can get that the value of time taken by the boat is $62.6s$.