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Tough Physics integration problem dealing with kinematics.

  1. Sep 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A train traveling at v0 = 22.0 m/s begins to brake by applying a velocity-dependent instantaneous acceleration a(v) = α /(v + u) m/s2, where α = – 23.0 m2/s3, v is the instantaneous velocity of the train, and u = 0.5 m/s. Determine the distance traveled by the train until it comes to a complete stop.

    I know I need to integrate, but I don't know how to get to d(t).


    2. Relevant equations

    (v+u)dv = (α)dt

    3. The attempt at a solution

    -22v-11 = 23.0(dt) is where I'm at. I'm not sure how to integrate both sides. How do I get to v(t), so I can solve d(t)?
     
  2. jcsd
  3. Sep 1, 2012 #2

    TSny

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    Hello, lorenzosounds. You'll need to eliminate time, t, in favor of distance, x. Can you express dt in terms of v and dx?
     
  4. Sep 1, 2012 #3

    Would it be -(1/2v^2+0.5v)=-23.0t ?
     
  5. Sep 1, 2012 #4

    TSny

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    No. If you integrate (v+u)dv = (α)dt, then the left side would be integrated from vo to v. So, vo as well as v would appear on the left side.

    But, then you'll have to solve this for v as a function of t, which will get messy. After that you would have to integrate one more time to find the distance traveled.

    There's an easier way. Instead of trying to get v as a function of t, try to get v as a function of x. Go back to (v+u)dv = (α)dt. Is there a way to express the differential dt in terms of dx? Hint: v = dx/dt.
     
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