A $\mathbf{200}\mathbf{}\mathit{m}$wide river flows due east at a uniform speed of $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. A boat with a speed of $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$relative to the water leaves the south bank pointed in a direction ${\mathbf{30}}^{\mathbf{°}}$west of north. What are the (a) magnitude and (b) direction of the boat’s velocity relative to the ground? (c)How long does the boat take to cross the river?

1. The velocity of water with respect to the ground$\left(V_{wg}\right)$is $2m/s.$.

2. The velocity of the boat with respect to water$\left(V_{bw}\right)$is $8m/s$at $\theta ={30}^{°}$west of north

When two frames of reference A and B are moving relative to each other at a constant velocity, the velocity of a particle P as measured by an observer in frame A usually differs from that measured from frame B. The two measured velocities are related by,

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

Where $\overrightarrow{\mathbf{V}}\mathit{P}\mathit{A}$is the velocity of B with respect to A. Using the relative motion concept, find the magnitude and direction of one of the vectors.

Formulae:

The magnitude of the vector $\overrightarrow{V}$is,

$V=\sqrt{V_x^2+V_y^2}$

The direction of the vector is given by,

$\tan \theta =\frac{V_y}{V_x}$

The velocity of water with respect to ground can be written in the vector notation as

${\overrightarrow{V}}_{wg}=2\hat{i}$

The velocity of the boat with respect to water can be written in the vector notation as,

$$\begin{gathered} \overrightarrow{V}\theta i_{bw}=W_{bw}\sin \theta \hat{j}+{}_{bw}\cos \\ =-\left(8m/s\right)\sin \left({30}^{°}\right)\hat{i}+\left(8m/s\right)\cos \left({30}^{°}\right)\hat{j} \\ =\left(-4m/s\right)\hat{i}+\left(6.93m/s\right)\hat{j} \end{gathered}$$

The velocity of boat with respect to ground is,

$\begin{array}{rcl}{\overrightarrow{V}}_{bg}& =& {\overrightarrow{V}}_{bw}+{\overrightarrow{V}}_{wg}\\ & =& \left(-4\hat{i}+6.93\hat{j}\right) \text{m/s}+\left(2\hat{i}\right) \text{m/s}\\ & =& \left(-2\hat{i}+6.93\hat{j}\right) \text{m/s}\\ & & \end{array}$

The magnitude of velocity of boat with respect to ground is,

$$\begin{gathered} V_{bg}=\sqrt{{V_{bgx}}^2+{V_{bgy}}^2} \\ =\sqrt{{\left(-2m/s\right)}^2+{\left(6.93m/s\right)}^2} \\ =7.21m/s \\ \approx 7.2m/s \end{gathered}$$

Therefore, the magnitude of velocity of boat with respect to ground is$7.2m/s$.

The direction of velocity of boat with respect to ground is given by,

$$\begin{gathered} \tan \theta =\frac{V_{bgy}}{Vbgx} \\ =\frac{6.93m/s}{-2m/s} \\ \theta ={\tan }^{-1}\left(\frac{6.93m/s}{-2m/s}\right) \\ =-73.9^{°} \\ \approx -{74}^{°} \end{gathered}$$

Therefore, the direction of the boat’s velocity relative to the ground is ${74}^{°}$west of south of ${106}^{°}$from x-axis.

The time taken by boat to cross the river is calculated by dividing the distance traveled by the y component of the velocity.

$$\begin{gathered} ∆t=yV{/}_{bgy} \\ =\frac{200m}{\left(7.2m/s\right)\left(\sin {106}^{°}\right)} \\ =29s \end{gathered}$$

Therefore, the time taken by the boat to cross the river is $\mathbf{29}\mathbf{}\mathit{s}$.