Question

The eye of a hurricane passes over Grand Bahama Island in a direction 60.0° north of west with a speed of 43.5 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 23.5 km/h. How far from Grand Bahama is the hurricane 4.95 h after it passes over the island?

Answer 1

D = 170.6Km

Explanation:

First of all, we set the reference (origin) at Grand Bahama.

Nw, from the first displacement of 3h we calculate the distance:

D1 = V1*t = 43.5 * 3 = 130.5 Km

The coordinates of this new location is given by:

r1 = ( -D1*cos(60°); D1*sin(60°)) = (-62.5; 158.835) Km

For the second displacement, the duration was of 1.95 hours (4.95 -3), so the distance traveled was:

D2 = V2*t = 23.5 * 1.95 = 45.825 Km

The coordinates of this new location is given by:

r2 = r1 + ( 0; D2) = (-62.25; 158.835) Km

Now we just need to calculate the magnitude of that vector to know the distance to Grand Bahama:

`Dt = \sqrt{D_(2x)^(2)+D_(2y)^(2)}=170.6 Km`