# Homework Help: Raindrops and a car problem

1. Jul 19, 2012

### EBBAzores

1. The problem statement, all variables and given/known data
A car travels to east with 50 km/h. It's rainning vertically in relation to the Earth. The raindrops on the lateral windows of the car make a 60 degree angle with the vertical. Determine the velocity of the raindrops in relation to:
a) The Earth
b) The car

2. Relevant equations

I really don't know what to do here xD

3. The attempt at a solution

I really don't know what to do here xD

2. Jul 19, 2012

### TSny

Hi. This is a relative velocity problem. I'll try to get you started.

Take toward the right on a sheet of paper to represent east and toward the top of the paper as representing vertically upward from the earth.

Draw a vector representing the velocity vector of the car relative to the earth:$\vec{V}$C/E
Draw a vector representing the velocity of the rain relative to the earth: $\vec{V}$R/E
Draw a vector representing the velocity of the rain relative to the car: $\vec{V}$R/C

Which two of these vectors when added together will equal the third vector?

3. Jul 19, 2012

### EBBAzores

VC/E plus VR/E? That's equal to the 3rd vector

4. Jul 19, 2012

### SammyS

Staff Emeritus
No.

How are $\displaystyle \vec{v}_{C/E}\,,\ \vec{v}_{R/E}\,,\text{ and }\vec{v}_{R/C}$ related?

5. Jul 19, 2012

### EBBAzores

Vr/c = Vc/e + Vr/e

so by that logic Vr/c = 50/sin(60)

and of course Vr/e = Vr/c cos(60) = (50cos(60))/sin(60)

and thats the answer by this logic right?

6. Jul 19, 2012

### TSny

This is not the correct relation between the vectors. Did you draw the three vectors? If you add Vc/e + Vr/e graphically, you will see that the result does not point in the direction that you would expect for vr/c.

Can you see the correct relationship?

When you construct the correct velocity diagram, see if you still get these answers.

7. Jul 19, 2012

### EBBAzores

So from what I draw I understood that the actual velocity diagram is given by Vc/e = Vr/e + Vr/c but besides that the results that I obtained before were right

8. Jul 19, 2012

### TSny

Let's see.

Vr/e = velocity of rain with respect to the earth which is a vector pointing straight down.

Vr/c = velocity of rain with respect to the car which is a vector pointing down and to the left (i.e., down and toward the west).

If I place these vectors "head-to-tail" to add them, I get a resultant vector that points down and to the left (but more steeply downward than Vr/c).

This doesn't agree with Vc/e which is the velocity of the car with respect to the earth and points horizontally toward the right.

9. Jul 19, 2012

### EBBAzores

So the Vr/e + Vc/e = Vr/c

10. Jul 19, 2012

### TSny

There's a general rule for setting up correct relative velocity equations. Suppose you have three objects a, b, and c which move relatively to one another. A correct equation would be

Va/c = Va/b + Vb/c

Another correct relation would be

Vb/a = Vb/c + Vc/a

The general rule is that whatever subscript appears first on the left side of the equation is also the first subscript on the right side of the equation. The last subscript on the left side is also the last subscript on the right side. Whatever subscript does not appear on the left side will appear on the right as "sandwiched between" the first and last subscripts.

Thus in the first equation above, "a" appears first on both sides of the equation and "c" appears last on both sides. "b" is sandwiched between "a" and "c" on the right side.

See if this can help you set up a correct equation for the velocities in your problem.

11. Jul 19, 2012

### TSny

Another rule that is often useful is Va/b = -Vb/a. (You might not need this rule to solve the particular problem you are working on now, but it might come in handy in other problems.)

Example, if a car is moving toward the east at 50 mph relative to the earth, then the earth would be moving at 50 mph toward the west relative to the car: Vc/e = -Ve/c

12. Jul 19, 2012

### EBBAzores

Vr/e=Vr/c + Vc/e....

Seriously I'm feeling so stupid right I have an exam in 3h hours and I'm feeling like a don't know a thing!

13. Jul 19, 2012

### EBBAzores

14. Jul 19, 2012

### TSny

That's the correct equation. Draw a velocity diagram representing this equation and you'll be able to use trig to get the answer. (The answers for the magnitudes of the velocities will agree with what you got before.)