# Homework Help: Relative Motion Question Involving Rain and Moving Car

1. Apr 1, 2009

### ff_yy

1. The problem statement, all variables and given/known data
A car travels due east with a speed of 45.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth. The traces of the rain on the side windows of the car make an angle of 45.0° with the vertical. Find the velocity of the rain with respect to the following reference frames.
(a) the car

(b) the Earth

2. Relevant equations
n/a

3. The attempt at a solution
I managed to find the answer of part a) which is 17.7 m/s using simple trigonometry although I didn't really understand why the sign was positive.
I found the velocity of the car to be 12.5 m/s and with my diagram got:
sin45=12.5/Vr (Vr= velocity of rain relative to the car)
Vr = 12.7 m/s

For part b, I thought that
velocity of rain relative to car = velocity of car relative to earth - velocity of rain relative to earth
So, 17.7 = Vre -12.5
So I found the velocity of rain relative to the earth to be 30.2 m/s.

I can't remember where I got that reasoning for part b from, but is that the only thing that's wrong?

2. Apr 2, 2009

### Staff: Mentor

The sign just depends on your coordinate system; all they want is the magnitude of the velocity.
OK.
?? Typo?

velocity of rain relative to car + velocity of car relative to earth = velocity of rain relative to earth
No. Realize that the equation above is a vector equation. Apply it to each component separately.

3. Apr 3, 2009

### ff_yy

So, when you say apply the vector equation to each component separately, can I split say the vector of velocity of rain relative to car into horizontal and vertical components?

And that would mean that
horizontally: Vre= -12.5 +12.5 = 0
vertically: vre =12.5 + 0 = 12.5

So answer is just 12.5 m/s (down)

If that's not right, then I'm not sure what you mean...

4. Apr 3, 2009

You got it.