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Velocity v(t) = b(t-T)^2, find position function x(t)

  1. Nov 4, 2006 #1
    If a car's x-position at time t = 0 is x(0) = 0 and it has an x-velocity of
    vx(t) = b(t-T)^2, where b and T are constants, which function best describes x(t)?

    a. x(t) = 2b(t - T)
    b. x(t) = 3b(t - T)3
    c. x(t) = (1/3)b(t - T)3
    d. x(t) = (1/2)b(t - T)
    e. x(t) = (1/3)b[(t - T)3 + T3]
    f. None of the above

    I was thinking d, but not sure.
  2. jcsd
  3. Nov 4, 2006 #2
    doesn't look like it to me.

    v(t) = dx(t)/dt

    what is the derivative of option d?
  4. Nov 4, 2006 #3
    ok when I did the derivative of vx(t)=b(t-T)^2, i got "a".

    or do they want the antiderivative of that equation?

    i'm not sure how to do the derivative of option d though because I'm not sure what to do about those constants. Wouldn't the derivative of b and T be 0 since they are constant?
  5. Nov 4, 2006 #4
    the derivative of a constant is zero. what the above equation i posted says is that the velocity (v) is the time derivative of position (x). if you differentiate the initial equation (velocity) you get the acceleration. that's not what you're after.
  6. Nov 4, 2006 #5
    ooo ok ok...I need to get vx(t) in terms of x(t)

    so i should take the antiderivative of the equation like use the reverse kinematic chain...so let me see here...

    oooo ok ok I think I got this...I'm thinking the answer will be c...because if you take the derivative of c it would be (1/3)3b(t-T)^2...the 3's cancel out and you are left with b(t-T)^2 which is vx(t).

    Does this sound right?
  7. Nov 4, 2006 #6
    yes, looks correct to me.
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