91Ó°ÊÓ

Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.

On the ship, you will see that the body is in the y-direction only. The details are given in the sketch. W represents the water, a represents the wind.

Hence the velocity can be calculated by the following equation.

$V_{aw}=V_{ae}+V_{ew}$

The velocity of the sandal relative to the ship is -17.1 m/s

The component table will be as below:

Here we need to find the x component and the y component.

$$\begin{gathered} Xcomponent \\ {V}_{aex}=V_i\cos \mathrm{θ} \\ {V}_{aex}=\left(-4.5\right)\cos {50}^{°} \\ {V}_{aex}=-2.89 \end{gathered}$$

$$\begin{gathered} Ycomponent \\ {V}_{aey}=V_i\sin \mathrm{θ} \\ {V}_{aey}=\left(-4.5\right)\sin {50}^{°} \\ {V}_{aey}=-3.45 \end{gathered}$$

Hence, the velocity of water w.r.t earth:

$$\begin{gathered} V_{wex}=V_{we}\cos \mathrm{θ} \\ {V}_{wex}=\left(2.2\right)\cos {30}^{°} \\ {V}_{wex}=1.10 \\ {V}_{ewx}=-1.10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

$$\begin{gathered} Ycomponent \\ {V}_{wey}=V_{we}\sin \mathrm{θ} \\ {V}_{wey}=\left(2.2\right)\sin {30}^{°} \\ {V}_{wey}=1.91 \\ {V}_{ewy}=-1.91\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the resultant vector will be

$$\begin{gathered} \stackrel{â‡\P Ä}{R}=\sqrt{{\left(X\right)}^2+{\left(Y\right)}^2} \\ \stackrel{â‡\P Ä}{R}=\sqrt{{\left(-3.99\right)}^2+{\left(-5.36\right)}^2} \\ \stackrel{â‡\P Ä}{R}=6.68\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the velocity of the wind relative to that of water is 6.68 m/s.

The direction of the velocity will be;

$\begin{array}{rcl}\tan \mathrm{θ}& =& \frac{Y}{X}\\ \tan \mathrm{θ}& =& \frac{5.36}{3.99}\\ \mathrm{θ}& =& 53.3\end{array}$

The resultant vector has the direction of $53.3^{°}SW$.