Question

The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of m/s in a direction east of north relative to the Earth. It encounters a wind that has a velocity of m/s in a direction of south of west relative to the Earth. What is the velocity of the wind relative to the water?

Short answer

The velocity of the wind relative to that of water is 68 m/s. The resultant vector has the direction of $53.3^{°}$SW.

Step-by-step solution

Definition of velocity

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.

Velocity of the sandal relative to the ship

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:

	X	Y
Vae	-2.89	-3.45
Vew	-1.1	-1.91
Resultant	-3.99	-5.36

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

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

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

Velocity of water

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

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

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

Determine the resultant vector

Hence the resultant vector will be

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

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

The direction of the velocity will be;

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

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