# Using relative motion (2D) to get rain speed & direction when running

1. May 7, 2013

### aaronstonedd

1. The problem statement, all variables and given/known data
A person moving towards east with a speed 'v' observes the rain to be falling vertically downwards. When he doubles his speed, the rain appears to come at 30° angle with the vertical. Find the actual speed and direction of rain with vertical.

2. Relevant equations
Velocity of A with respect to B = Velocity of A – Velocity of B
(VAB = VAVB)

3. The attempt at a solution
Let the actual speed of rain be VR. Let the person's speed be represented by VP.
Using $\widehat{i}$-$\widehat{j}$-$\widehat{k}$ notation for vectors, VP = v $\widehat{i}$

VR = VRsinθ $\widehat{i}$ – VRcosθ $\widehat{j}$

VRP = VRVP
VRP = VRsinθ $\widehat{i}$ – VRcosθ $\widehat{j}$ – v $\widehat{i}$ = (VRsinθ – v) $\widehat{i}$ –VRcosθ $\widehat{j}$

∵ Rain is vertically downwards, ∴ (VRsinθ – v) $\widehat{i}$ = 0

So, VRsinθ = v

Now, let VP be 2v $\widehat{i}$.

VRP = VRVP
= +VRsinθ $\widehat{i}$ –VRcosθ $\widehat{j}$ –2v $\widehat{i}$
= (+VRsinθ –2v) $\widehat{i}$ –VRcosθ $\widehat{j}$
= –v $\widehat{i}$ –vtanθ $\widehat{j}$

Now $\frac{-v}{-vtan\theta}$ = tan30° = √3̅

∴ tanθ = $\frac{1}{\sqrt{3}}$

∴ θ = 60°.

VR = $\frac{2v}{\sqrt{3}}$.

Both these answers are wrong, and I don't know how or why. I'm in Class 11. The correct answers should be θ = 30° and ∴ VR = 2v. What am I missing here?

2. May 7, 2013

### Staff: Mentor

tan 30° is not the square root of 3.

It is confusing that you use the same symbol for vectors and scalar quantities.

3. May 8, 2013

### aaronstonedd

Two mistakes

Sorry, I made another mistake. Though you're right, sin30° = $\frac{1}{\sqrt{3}}$, I made another mistake: -VRcosθ $\widehat{j}$ ≠ -vtanθ $\widehat{j}$, but, = -vcotθ $\widehat{j}$.

So still, θ = 60°. Why is it so? (I know my answer's wrong, but I don't know why.)