Get the answer to "The eye of a hurricane passes over Grand..."

High School
Physics
Force, Motion and Energy
Acceleration Due to Gravity

DofotheroU

Answered question

2020-10-18

The eye of a hurricane passes over Grand Bahama Island. It is moving in a dirction 60.0 degrees north of west with a speed of 41.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h. How far from Grand Bahama is the hurricane 4.50 h after it passesover the island?

Answer & Explanation

Lacey-May Snyder

Skilled2020-10-19Added 88 answers

Jeffrey Jordon

Expert2021-10-06Added 2605 answers

Step 1

The path taken by the hurricane is shown in the figure below.

The speed of the hurricane in the east direction is $v_{A B} = 41.0 k m h^{- 1}$ for time t=3 h. Therefore the distance moved by the hurricane in the east direction is $A B = 41.0 k m h^{- 1} \times 3 h = 123 k m$

The speed of the hurricane in the north direction is $v_{B C} = 25.0 k m h^{- 1}$ for time t=1.5 . Therefore, the distance moved in the north direction is $B C = 25.0 k m h^{- 1} \times 1.5 h = 37.5 k m$

Step 2

Therefore the distance of the haurrican from Grand Bahama is

$A C = \sqrt{(A B)^{2} + (B C)^{2}}$

$= \sqrt{(123 k m)^{2} + (37.5 k m)^{2}}$

=128.589 km

=129 km

Which has three significant figures.

And direction is

$θ = \tan^{- 1} (\frac{B C}{A B})$

$= \tan^{- 1} (\frac{37.5 k m}{123 k m})$

$= {16.955}^{\circ}$

$= {17.0}^{\circ}$

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