Question

A ship sets sail from Rotterdam, The Netherlands, heading due north at m/s relative to the water. The local ocean current is 1.50m/s in a direction${\mathbf{40}}^{\mathbf{o}}$north of east. What is the velocity of the ship relative to the Earth?

Short answer

The velocity of the ship with respect to earth is 8.04 m/s.

Step-by-step solution

Definition of velocity

The directional speed of an item in motion is an indication of its rate of change in position, as seen from a specific point of view and measured by a specific time standard.

Stating given data

The ship is heading to the north with a velocity with respect to water of 7m/s .

The local ocean current is 1.50m/s which is relative to the earth .

Velocity of ship relative to earth

Velocity of the ship with respect to earth

$$\begin{gathered} V_{se}=V_{sw}+V_{we} \\ \overrightarrow{R}=V_1+V_2 \end{gathered}$$

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

role="math" localid="1668668321262" $$\begin{gathered} V_x=V_i\cos \theta \\ {V}_x=(1.5)\cos 40 \\ {V}_x=1.15 \\ Ycomponent \\ {V}_y=V_i\sin \theta \\ {V}_y=(1.5)\sin 40 \\ {V}_y=0.964 \end{gathered}$$

Hence the resultant vector for the x component is

$$\begin{gathered} V_1=0+1.15 \\ =1.15m/s \end{gathered}$$

Hence the resultant vector for the y component is

Resultant velocity of ship relative to earth

The resultant velocity is

$$\begin{gathered} \overrightarrow{R}=\sqrt{{\left(\sum X\right)}^2+{\left(\sum Y\right)}^2} \\ \overrightarrow{R}=\sqrt{{(1.15)}^2+{(7.96)}^2} \\ =8.04m/s \end{gathered}$$

The velocity of the ship with respect to the earth is 8.04m/s.