91影视

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 $胃={30}^{掳}$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,

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

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

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

The direction of the vector is given by,

$\tan 胃=\frac{V_y}{V_x}$

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

${\stackrel{鈫�}{V}}_{wg}=2\hat{i}$

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

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

The velocity of boat with respect to ground is,

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

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

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

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,

$$\begin{gathered} \tan 胃=\frac{V_{bgy}}{Vbgx} \\ =\frac{6.93m/s}{-2m/s} \\ 胃={\tan }^{-1}\left(\frac{6.93m/s}{-2m/s}\right) \\ =-73.9^{掳} \\ 鈮�-{74}^{掳} \end{gathered}$$

Therefore, the direction of the boat鈥檚 velocity relative to the ground is ${74}^{掳}$west of south of ${106}^{掳}$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.

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

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