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Kinematics: 3 unknowns, find Time.

  1. Sep 21, 2009 #1
    SOLVED.

    1. The problem statement, all variables and given/known data
    angle = 10 degrees
    delta X = 20
    delta Y = -1.5

    car goes off a 10 degree ramp and lands at the delta X/delta Y.
    what is T?


    2. Relevant equations
    delta X = VxoT
    Vy = Vyo + AyT
    delta Y = VyoT + (1/2)AyT^2
    Vy^2 = Vyo^2 + 2Ay(delta Ymax)


    3. The attempt at a solution
    none
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 21, 2009 #2

    rl.bhat

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    Homework Helper

    Read the rules of PF.
    Don't expect answer from us. Show your attempts.
    You know the relevant equations.
     
  4. Sep 21, 2009 #3
    Thanks for the help >:(

    I found Vxo and Vyo in terms of V using sohcahtoa, then I plugged it into and delta X = VxoT and found out T (t = 20.3/v). I plugged Vyo and T into my third eqn. above and got V=1.

    There's my attempt.
     
  5. Sep 21, 2009 #4

    rl.bhat

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    Your T value is correct.
    Now in the third equation, substitute the values of T. Y and Ay = g. In this Y and g are negative. Solve for v. Then you can find T.
     
  6. Sep 21, 2009 #5
    figured it out. did this earlier but I didn't see the answer for some reason?

    AP physics test tomorrow wish me luck o_O i'll need it!
     
  7. Sep 21, 2009 #6

    rl.bhat

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    Can you show me the calculations, so that I can check it. I am getting v = 9.624 m/s.
    Any way good luck to you.
     
  8. Sep 21, 2009 #7
    i got V = 20.04 m/s (correct answer from teacher is 20.05)

    calculations:

    Vxo = Vcos10
    Vyo = Vsin10

    20 = (Vcos10)t
    t = (20.3/v)

    -1.5 = (Vsin10)(20.3/v) + (1/2)(-9.8)(20.3/v)^2
    -1.5 = (V's cancel out, so do 20.3 times sin10)
    -1.5 = (3.52) - 4.9(20.3/v)^2
    5.02 = 4.9(20.3/v)^2
    1.026 = (20.3/v)^2
    root everything
    1.013 = 20.3/v
    1.013v = 20.3
    v = 20.039 m/s
     
  9. Sep 22, 2009 #8

    rl.bhat

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    Good. I think I missed vsin10.
     
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