1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    mfb

    User Avatar
    2016 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.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Using relative motion (2D) to get rain speed & direction when running
  1. 2D Motion Find speed. (Replies: 1)

Loading...