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Kinematics of a Particle

  1. Jan 13, 2004 #1
    Problem: A car starts from rest and moves along a straight line with an acceleration of a=(3s^-1/3)m/sec^2, where s is in metres. Determine the car's acceleration when t=4sec. ANS: 1.06 m/sec^2

    Alright...I know nothing about integrals...really, nothing. I was never taught anything about integrals even though I've taken calculus courses before.

    Here's what I think I should do.

    Take the equation a = d^2s/dt^2 and INTEGRATE it to find the position (s). But how do you do this.
     
  2. jcsd
  3. Jan 13, 2004 #2
    Forget numbers for a minute.

    If ao is acceleration, vo is initial velocity, xo is initial position, then,

    Acceleration at time t is: a(t) = ao
    Velocity at time t is: v(t) = vo + ao * t
    Position at time t is: x(t) = xo + vo * t + (ao * t^2)/2


    They haven't taught you integration yet. Maybe you've been given these formulae, though, or something similar.
     
  4. Jan 13, 2004 #3
    what's the initial position (xo)?
     
  5. Jan 13, 2004 #4
    You don't need to know to answer the question. It's the distance, d, that's needed ...
    d = x(4) - xo
    = ...
     
  6. Jan 13, 2004 #5
    d = 24s^-1/3

    Sorry, I'm still not understanding.
    I really appreciate your help, though.
     
  7. Jan 14, 2004 #6
    How far you go doesn't depend on where you start from, so take xo to be as simple as possible ... xo = 0.
     
  8. Jan 14, 2004 #7
    The equations are valid only when a is constant
     
  9. Jan 14, 2004 #8
    Thanks for that, himanshu121, I didn't notice the 's' sitting in the expression for a.

    I just assumed that it was just a simple problem that didn't need integration and got side-tracked into seeing if jimmy had seen these equations before.
     
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