Question

A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 8.5 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:

Given

Velocity of ship relative to water is $8.5 \m/s$

$V_{sw}=V_{s}-V_{w}=8.5\hat{j}----1$

Velocity of water with respect to earth

$V_{we}=V_{w}-V{e}=1.5(\cos (40)\hat{i}+\sin (40)\hat{j})----2$

adding 1 and 2

$V_{s}-V_{e}=1.5\cos (40)\hat{i}+8.5\hat{j}+1.5\sin (40)\hat{j}$

$V_{s}-V_{e}=1.5\cos (40)\hat{i}+9.46\hat{j}$

$|V_{se}|=\sqrt{(1.149)^2+(9.46)^2}$

$|V_{se}|=\sqrt{90.8118}$

$|V_{se}|=9.53\ m/s$