Question

The eye of a hurricane passes over grand bahama island in a direction 60.0° 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?

Answer 1

Step-by-step 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 :

`v_i=41cos(60)i+41sin(60)j`

`v_i=(20.5i+35.5j)\ km/h`

Three hours later the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h,
`v_f=25j\ km/h`

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

`s=ut+(1)/(2)at^2`.............(1)

Acceleration,
`a=(v_f-v_i)/(t)`

`a=(25j-(20.5i+35.5j))/(3)`

`a=((-20.5i+10.5j))/(3)`

a =( -6.8i+3.5j ) km/h²

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

`s_x=v_i_xt+(1)/(2)a_xt^2`

`s_x=20.5* 4.5+(1)/(2)* (-6.8)* 4.5^2`

`s_x=23.4\ km`

and

`s_y=v_i_yt+(1)/(2)a_yt^2`

`s_y=35.5* 4.5+(1)/(2)* (3.5)* 4.5^2`

`s_y=195.18\ km`

Magnitude of distance,
`s=√(s_x+s_y)`

`s=√(23.4^2+195.18^2)`

s = 196.5 km

Hence, this is the required solution.

Answer 2

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

`v_i = -41 cos60\hat i + 41 sin60 \hat j`

`v_i = -20.5 \hat i + 35.5 \hat j`

After some time the velocity is given as

`v_f = 25 \hat j`

now we can find the acceleration

`a = (v_f - v_i)/(t)`

`a = (25\hat j - (-20.5 \hat i + 35.5 \hat j))/(3)`

`a = 6.83 \hat i - 3.5 \hat j`

now the distance is given by

`d = v*t + (1)/(2) at^2`

`d = (-20.5 \hat i + 35.5 \hat j)*4.5 + 0.5*(6.83 \hat i - 1.5 \hat j)* 4.5^2`

`d = -92.25\hat i + 159.75\hat j + 69.15\hat i - 15.19\hat j`

`d = -23.1 \hat i + 144.56\hat j`

so the magnitude of distance is

`d = √(23.1^2 + 144.56^2) = 146.4 km`