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Homework Help: Using relative motion (2D) to get rain speed & direction when running

  1. May 7, 2013 #1
    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 [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 = VRVP
    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 = VRVP
    = +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?
  2. jcsd
  3. May 7, 2013 #2


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    2017 Award

    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.
  4. May 8, 2013 #3
    Two mistakes

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

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