A 200-m -wide river has a uniform flow speed of1.1 m/sthrough a jungle and toward the east. An explorer wishes to leave a small clearing on the south bank and cross the river in a powerboat that moves at a constant speed of4.0 m/swith respect to the water. There is a clearing on the north bank82 mupstream from a point directly opposite the clearing on the south bank. (a) In what direction must the boat be pointed in order to travel in a straight line and land in the clearing on the north bank? (b) How long will the boat take to cross the river and land in the clearing?

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,

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

Here${\overrightarrow{\mathbf{V}}}_{\mathbf{BA}}$ 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 $\overrightarrow{V}$is given by,$V=\sqrt{V_x^2+V_y^2}$ (i)

The direction of the vector is given by,

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

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

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 $62.6s$.