Question

As it passes over Grand Bahama Island, the
eye of a hurricane is moving in a direction 40
north of west with a speed of 79 km/h. Three
hours later, it shifts due north, and its speed
slows to 15 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

QUESTION:

As it passes over Grand Bahama Island, the eye of a hurricane is moving in a direction 40 north of west with a speed of $79\mathrm{km / h}$ . Three hours later, it shifts due north, and its speed slows to 15 km/h. How far from Grand Bahama is the eye 4.50 h after it passes over the island? Answer in units of km

SOLUTION

Let's draw a scketch:

ANSWER

The angle ABC is equal to $180 - 40 = 140{}^{\circ}$ . AC is the displacement of the hurricane's eye.

According to the law of cosines

$$\mathrm {A C} ^ {2} = \mathrm {A B} ^ {2} + \mathrm {B C} ^ {2} - 2 \mathrm {A C} \cdot \mathrm {B C} \cdot \cos (1 4 0 {}^ {\circ})$$ $$\mathrm {A B} = \mathrm {v} _ {1} \mathrm {t} _ {1}$$ $$\mathrm {B C} = \mathrm {v} _ {2} \mathrm {t} _ {2}$$ $$\mathrm {v} _ {1} = 7 9 \mathrm {k m / h}$$ $$\mathrm {v} _ {2} = 1 5 \mathrm {k m / h}$$ $$t _ {1} = 3 \mathrm {h}$$ $$t _ {2} = 4. 5 - 3 = 1. 5 \mathrm {h}$$ $$\mathrm {A C} = \sqrt {\left(\mathrm {v} _ {1} \mathrm {t} _ {1}\right) ^ {2} + \left(\mathrm {v} _ {2} \mathrm {t} _ {2}\right) ^ {2} - 2 \cdot \mathrm {v} _ {1} \mathrm {t} _ {1} \mathrm {v} _ {2} \mathrm {t} _ {2} \cos (1 4 0 {}^ {\circ})}$$ $$\mathrm {A C} = \sqrt {6 4 8 4 5 . 1}$$ $$\mathrm {A C} = 2 5 4. 6 5 \mathrm {k m}$$

ANSWER

254.65 km