Question

A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current is 1.52 m/s in a direction 40° north of east. What is the velocity of the ship relative to the earth?.

Answer 1

Answer:

The velocity of the ship relative to the earth V = 9.05 $\frac{m}{s}$

Explanation:

The local ocean current is  = 1.52 m/s

Direction $\theta$ = 40°

Velocity component in X - direction $V_{x}$ = 1.52 $\cos 40$°

$V_{x}$ = 1.164 $\frac{m}{s}$

Velocity component in Y - direction $V_{y}$ = 8 + 1.52 $\sin 40$°

$V_{y}$ = 8.97 $\frac{m}{s}$

The velocity of the ship relative to the earth

$V = \sqrt{V_{x}^{2} + V_{y} ^{2} }$

Put the values of $V_{x}$ and $V_{y}$ we get,

⇒ $V = \sqrt{1.164^{2} + 8.97 ^{2} }$

⇒ V = 9.05 $\frac{m}{s}$

This is the velocity of the ship relative to the earth.