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Circular motion -- Car on a circular track

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A car has an initial speed v0= 20m/s. It increases its speed along the circular track at s=0, at=(0.9s) m/(s2), where s is in meters. Determine the time needed for the car to travel 35m. The radius of curvature of the track is 40m.

    2. Relevant equations
    an = v^2/p
    a = dv/dt


    3. The attempt at a solution
    I really don't know how to approach the problem. I know that I want to define an integral that evaluates a change in s on one side with a change in t on the other.

    I started by setting a = dv/dt = dv/ds * ds/dt = v * dv/ds so a * ds = v*dv. By doing this integral from 0 to 35m I found the final velocity to be 26.5 m/s. I do not know where to proceed from here.

    Or can i solve that integral without inputting a value for s to get vf = sqrt ( 0.9s^2+400 ) ? Then using that I can do v = ds/dt so dt = (1/v) ds --> t = Integral [ 1/sqrt(0.9s^2+400) ds?
     
  2. jcsd
  3. May 16, 2016 #2

    haruspex

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    Strange question... if I understand it correctly the curvature is irrelevant. Is this perhaps just one part of a longer question?
    I get a larger value for the final speed. Please post your working.
    The approach you describe at the end looks right, but make it easy on yourself by using symbols for the acceleration function etc., not plugging in numbers: ##\int (ks^2+v_0^2)^{-\frac 12}.ds##.
    Can you think of a substitution for s in the integral that will get rid of the square root?
     
  4. May 16, 2016 #3
    It isn't part of a longer problem, might just be irrelevant information. I know you can integrate that using a trig sub with tan(theta) but it looks a little messy. I was hoping there was another approach.
     
  5. May 16, 2016 #4

    haruspex

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    Tan or sinh. If the integral is correct (and I think it is) then there will not be an easier way.
     
  6. May 16, 2016 #5
    can you explain the sinh substitution method.
     
  7. May 16, 2016 #6
    Being lazy, I'd write it as a quadratic equation: (a/2 X t^2) + 20t - 35 =0. Solve that and choose the value of t that solves the problem.
     
  8. May 16, 2016 #7
    I thought the quadratic form could only be used when you are assuming constant acceleration, where in this case its not, its a(s). In fact, when I use a(s) = a(35) and solve for the values of t, neither answers are correct.
     
  9. May 17, 2016 #8

    haruspex

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    Just as tan2+1=sec2, sinh2+1=cosh2.

    Quite so.
     
  10. May 17, 2016 #9
    The quadratic equation should be 0.45t^2 + 20T -35 = 0. Solve using the quadratic formula. I get 1.686 seconds.

    By habit I use the simplest method I can. I've been doing this since before pocket calculators, so the simpler the better.
     
  11. May 17, 2016 #10

    haruspex

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    I find a better strategy is to use the simplest valid method. As Nikstykal pointed out in post #7, your quadratic method is invalid here. The acceleration is not constant.
     
  12. May 17, 2016 #11
    Being lazy I read the acceleration as 0.9 m/s^2. Oops!
     
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