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Simple velocity function

  1. Jan 17, 2008 #1
    The position of a car that is being slowed at a uniform rate is given by:
    s(t) = 25t - 5.25t^2
    where s is the displacement of the car from the point at which braking began and t is the
    time in seconds after braking began. Find the velocity function for the car, including
    its domain.

    The answer is v(t) = 25 -11t 0 < or equal to t < or equal to 25/11

    I assume I need to find the value of t first, then take a derivative, but I'm not sure
    how to get t by itself
    0 = 25t - 5.25t^2
    5.25t^2 = 25t
    and then I don't know what to do.
     
  2. jcsd
  3. Jan 17, 2008 #2
    I may be wrong here, someone correct me if I am.

    I think since s(t) is the position of the car, simply differentiating the equation will give you the v(t), velocity of the car with respect to time.

    Thus, v(t) = 25-10.5t

    I'm not sure how the answer is v(t) = 25-11t
    Maybe the original problem was written incorrectly and should have been s(t)=25t-5.5t^2?

    Hopefully I helped...
     
  4. Jan 17, 2008 #3
    Ohh, thanks!
    I think maybe he might have rounded? That's probably it, but I don't know why.

    so if v(t) = 25 - 11t
    then when the velocity is 0, the car has come to a stop so
    0 = 25 - 11t
    11t = 25
    so t = 25/11
     
  5. Jan 17, 2008 #4
    yup. no problem.
     
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