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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=4s. 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

    NateTG

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    If you were trying to find the position, then that would be correct. Since you're trying fo find the acceleration, you can just use the formula.
     
  4. Jan 13, 2004 #3
    what formula?
     
  5. Jan 13, 2004 #4

    NateTG

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    My bad, I thought s was in seconds.

    I don't see an easy way to deal with that one if you don't know how to integrate.
     
  6. Jan 13, 2004 #5
    How do you integrate?
     
  7. Jan 14, 2004 #6
    Are you from the U of S? Because i was working on that exact problem before I came on here, very strange. Anyways, as was said it can't really be done without integrating, which if you are from the U of S, they haven't taught us yet. But we have learnt antiderivatives, which should help you. Start with the relationship ads=vdv (which you can get by eliminating the dt term in a=dv/dt and v=ds/dt). Solve for a to get a=vdv/ds and substitute this into the equation given in the question. Now get the v and dv on the same side, as well as the s and ds terms. It should look something like this:
    vdv=3s^(-1/3)ds. Integrate both sides, which basically means to take the antiderivatives. This leaves v and s: v^2=9s^(2/3). Solve for v, and than substitute ds/dt for v. Once again get the s and ds on the same side, and dt on the other side, and integrate again (antiderivative). You now have s as a function of t! From here you can either substitute in t=4 and find s, than put this s value into the original equation to get a, or you could find the second time derivative of s to get an expression for a as a function of t, than put in 4 for t. Both will give you the same answer. Hope this helped, although I'm sure it's confusing to follow.
     
  8. Jan 14, 2004 #7
    I believe U have also posted the same Question in maths section So look at my reply which is similar to that of eddo
     
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