Question

As it passes over Grand Bahama Island, the eye of a hurricane is moving in a direction 30◦ north of west with a speed of 51 km/h. Three hours later, it shifts due north, and its speed slows to 34 km/h. How far from Grand Bahama is the eye 4.50 h after it passes over the island? Answer in units of km.

Answer 1

Answer:183.52 km

Explanation:

Given

First hurricane is moving in a direction $30^{\circ}$

speed of hurricane =51 km/h

After 3 hr it changes its direction changes to North and speed decreasing to 34 km/h

Position vector of Hurricane after 3 h

$r_1=51\times 3(-cos30\hat{i}+sin30\hat{j})$

After 1.5 hr position vector of hurricane w.r.t previous one

$r_{21}=34\times 1.5(\hat{j})$

thus $r_2=r_{21}+r_1$

$r_2=-132\hat{i}+127.5\hat{j}$

So total distance$=|r_2|=\sqrt{132^2+127.5^2}$

$=\sqrt{33,680.25}=183.52 km$