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Resisting Medium.

  1. May 14, 2013 #1
    When an object travels through a certain resisting medium the deceleration is proportional to the 4th of the velocity. This, a = -kv^4. Prove v = u(ktu^3 + 1)^1/3 and subsequently x = (1/2ku^2)((ktu^3 + 1)^2/3 - 1).
    v at time 0 = u and x at time 0 = 0.

    Differentiation and integration.

    Attempt at solution:
    I've proved the first part, for v. I keep getting (1/6ku)((ktu^3 + 1)^2/3 - 1) for c though. So, not too far out. It should just be a simple u substitution following on from the derivation of v. Could anyone walk through the second part? I have no clue where I'm making the mistake.
  2. jcsd
  3. May 14, 2013 #2


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    Homework Helper

    To make things easier to understand (as sometimes you might see 'u' to indicate a variable function for velocity)

    ku^3 = A and u=B

    So you have

    v= dx/dt = B(At+1)1/3

    To solve this you can use another substitution or apply the direct formula for

    [tex]\int (ax+b)^n dx[/tex]
  4. May 14, 2013 #3
    I see where I went wrong. Made a stupid blunder and differentiated the u^3 for some reason. Wasn't thinking. Thanks for that.
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