Question

A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the local ocean current is 1.50 m/s in a direction 40.0º north of east and changes the ship's intended motion. What is the velocity of the ship relative to the Earth?

Answer 1

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

$\overrightarrow{v}_{s,w}=6.5 \widehat{j}$

velocity of water with respect to earth = 1.5 m/s at 40° north of east

$\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)$

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

$\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}$

$\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j} \right )$

$\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}$

The magnitude of the velocity of ship relative to earth is $\sqrt{1.15^{2}+5.54^{2}}$ = 5.66 m/s