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-Dominant- [34]
3 years ago
9

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 hurrica

ne 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.
Physics
1 answer:
Furkat [3]3 years ago
7 0

Answer:

(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

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