- #1
aaronstonedd
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Homework Statement
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.
Homework Equations
Velocity of A with respect to B = Velocity of A – Velocity of B
(VAB = VA – VB)
The Attempt at a Solution
Let the actual speed of rain be VR. Let the person's speed be represented by VP.
Using [itex]\widehat{i}[/itex]-[itex]\widehat{j}[/itex]-[itex]\widehat{k}[/itex] notation for vectors, VP = v [itex]\widehat{i}[/itex]
VR = VRsinθ [itex]\widehat{i}[/itex] – VRcosθ [itex]\widehat{j}[/itex]
VRP = VR – VP
VRP = VRsinθ [itex]\widehat{i}[/itex] – VRcosθ [itex]\widehat{j}[/itex] – v [itex]\widehat{i}[/itex] = (VRsinθ – v) [itex]\widehat{i}[/itex] –VRcosθ [itex]\widehat{j}[/itex]
∵ Rain is vertically downwards, ∴ (VRsinθ – v) [itex]\widehat{i}[/itex] = 0
So, VRsinθ = v
Now, let VP be 2v [itex]\widehat{i}[/itex].
∴ VRP = VR – VP
= +VRsinθ [itex]\widehat{i}[/itex] –VRcosθ [itex]\widehat{j}[/itex] –2v [itex]\widehat{i}[/itex]
= (+VRsinθ –2v) [itex]\widehat{i}[/itex] –VRcosθ [itex]\widehat{j}[/itex]
= –v [itex]\widehat{i}[/itex] –vtanθ [itex]\widehat{j}[/itex]
Now [itex]\frac{-v}{-vtan\theta}[/itex] = tan30° = √3̅
∴ tanθ = [itex]\frac{1}{\sqrt{3}}[/itex]
∴ θ = 60°.
VR = [itex]\frac{2v}{\sqrt{3}}[/itex].
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?