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Motion Problems

  1. Oct 23, 2007 #1
    1. The problem statement, all variables and given/known data

    If v[tex]^{2}[/tex]=4s-3, find the acceleration when t=1s.

    2. Relevant equations



    3. The attempt at a solution

    I know that acceleration is the derivative of velocity, which is the derivative of distance but how do I get the derivative with both v and s in the equation?
     
  2. jcsd
  3. Oct 23, 2007 #2

    Dick

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    v is velocity, s is distance or time??? There's a problem with v^2=4s-3. The units on both sides don't appear to match.
     
  4. Oct 23, 2007 #3
    S is distance but I'm not sure how to solve this problem.
     
  5. Oct 23, 2007 #4

    HallsofIvy

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    Since, as you say, "acceleration is the derivative of velocity" (with respect to time), you find the acceleration by differentiating wih respect to time.

    Differentiating both sides of v2= 4s- 3 with respect to time: using the chain rule, of course: 2v v'= 4s'. Now v'= a, the acceleration, and s'= v, so that equation is
    2va= 4v and a= 2, a constant!

    Dick, if you assume the "4" and "3" given are dimensonless then, yes, the right side has units of "distance2/time2" while the right side has two parts: one with units of "distance/time" and the other dimensionless- so they can't even be combined. However, if we assume that "4" is really "4 m/s2" and the "-3" is really "-3 m2/s2",then we are alright
     
  6. Oct 23, 2007 #5
    Thank you so much!

    I have another question similar to this and was wondering if I did it right. Here's the question and my attempt:

    s[tex]^{2}[/tex] - 6v[tex]^{2}[/tex] = 10. Find the acceleration in terms of the distance.

    2s (ds/dt) - 12v dv/dt = 0
    2s (ds/dt) - 12(ds/dt)(a) = 0
    2s (ds/dt) = 12(ds/dt)(a)

    a = s/12.

    Am I right?
     
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