At a particular location, a hurricane's eye is moving with speed 40.8 km/h and in the direction 31.0° north of west. The hurricane keeps going at this velocity for 2.90 h, then changes direction, and heads due north. When its direction changes, it slows down to 22.6 km/h. It travels at this new velocity for 1.50 h. (Let î represent east and ĵ represent north.) Required:
a. Write the hurricane's initial velocity (in km/h).
b. Write the hurricane's final velocity (in km/h) in unit-vector notation.
c. Write the hurricane's displacement (in km) during the first 3.50 hours.

(a) u = (-34.972, 21.014) km/h

(b) v = (22.6 ĵ) km/h

(c) D = (-122.40, 73.55) km

Explanation:

Given;

initial velocity of the hurricane's eye, u = 40.8 km/h

direction of the hurricane, θ = 31.0° North west

final velocity of the hurricane, v = 22.6 km/h

direction of the final velocity, θ = 90°

(a)

The direction in counterclockwise rotation from East (the x axis) = 180 - 31°

θ' = 149°

x - component = 40.8 km/h x Cos 149° = -34.972 km/h

y - component = 40.8 km/h x sin 149° = 21.014 km/h

u = (-34.972, 21.014) km/h

(b)

θ = 90°

x - component (î)  = 22.6 km/h x Cos 90° = 0 î

y - component (ĵ) = 22.6 km/h x sin 90° = 22.6 ĵ

v = (0 î, 22.6 ĵ) km/h

v = (22.6 ĵ) km/h

(c)

displacement = speed x time

D = 3.5(-34.972, 21.014)

D = (-122.40, 73.55)km

1 is a answer of a question