The eye of a hurricane passes over grand bahama island in a direction 60.0â° north of west with a speed of 41.0 km/h. three hours later the course of the hur- ricane suddenly shifts due north, and its speed slows to 25.0 km/h. how far from grand bahama is the hurri- cane 4.50 h after it passes over the island?

initial velocity is given as 41 km/h at 60 degree North of West

After some time the velocity is given as

now we can find the acceleration

now the distance is given by

so the magnitude of distance is

Explanation:

It is given that, the eye of a hurricane passes over grand Bahama island in a direction 60° north of west with a speed of 41.0 km/h. Since, velocity is a vector quantity. It can be written as :

Three hours later the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h,

We have to find how far from grand Bahama is the hurricane 4.50 h after it passes over the island.

.............(1)

Acceleration,

a =( -6.8i+3.5j ) km/h²

Put the value of a in equation (1) and calculating x and y component of displacement as :

and

Magnitude of distance,

s = 196.5 km  

Hence, this is the required solution.

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow