Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line measure as the velocity of the raindrops? (Assume the car is moving to the right and that the + x-axis is to the right. Enter the magnitude in m/s and the direction in degrees counterclockwise from the + x-axis.)

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

vDE = 7.5 m/s

vPE = 20.2 m/s

Required: vDP

Assume that

vDE to be in direction of - j

vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

vDP = vDE - vPE (I)

Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √ (( - 7.5) ² + ( - 20.2) ²) m/s = 21.7454 m/s

θ = Arctan ( - 7.5 / - 20.2) = 20.3693°

θ is in 3rd quadrant so add 180°

θ = 20.3693° + 180° = 200.3693°