Question

The eye of a hurricane passes over Grand Bahama Island in a direction 60.0° north of west with a speed of 43.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 23.0 km/h. How far from Grand Bahama is the hurricane 4.80 h after it passes over the island?

Answer 1

The hurricane will be 166.15km away from the island.

Step-by-step explanation:

We can solve this problem with trigonometry and the formulas of constant velocity motion.

we need to find the displacement on the X and Y axis.

the speed on the X axis on the first trame of movemont is defined as:

`Vx=V*cos(ang)`

we will take the reference to the left of the X axis as positive, so:

`Vx=43*cos(60)\\Vx=21.5 km/h`

So the displacement on X is:

`dx=Vx*t\\dx=21.5km/h*3h\\dx=64.5km`

we will do the same for the Y axis:

`Vy=V*sin(ang)\\Vy=43sin(60)\\Vy=37.24km/h`

the Y displacement is:

`dy=Vy*t\\dy=37.24km/h*3h\\dy=111.72km`

Then on the second trame of the movement we only have displacement on the Y axis.

`dy2=Vy2*t\\dy2=23km/h*1.8h\\dy2=41.4km`

now we need the total displacement on the X and Y axis:

`Y=dy+dy2\\Y=153.12km\\X=64.5km`

The magnitud of the total displacement will be:

`D=√((X)^2+(Y)^2)\\D=√((64.5)^2+(153.12^2)\\ \\D=166.15km`