91Ó°ÊÓ

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 $\mathrm{θ}={30}^{°}$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,

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

Where $\stackrel{→}{\mathbf{V}}\mathit{P}\mathit{A}$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 $\stackrel{→}{V}$is,

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

The direction of the vector is given by,

$\tan \mathrm{θ}=\frac{V_y}{V_x}$

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

${\stackrel{→}{V}}_{wg}=2\hat{i}$

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

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

The velocity of boat with respect to ground is,

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

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

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

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,

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

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 $\mathbf{29}\mathbf{}\mathit{s}$.