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Time when car stopped given initial velocity and acceleration

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    At time t = 0, a car has a velocity of 16 m/s. It slows down w/an acceleration given by -0.50t in m/s^2 for t in seconds. It stops at t =?

    2. Relevant equations



    3. The attempt at a solution

    v = 16 - 0.5t
    16/.5 = t
    16 x 2 = t
    32 = t

    Did I set this up wrong? My answer is incorrect. It even works intuitively, if speed is reduced by .5 every second, starting at 16, it will take 32 subtractions of .5 to equal 0.
     
  2. jcsd
  3. Aug 24, 2011 #2

    lightgrav

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    nothing wrong here. is there a part 2 of the question that you didn't answer?
     
  4. Aug 24, 2011 #3
    No, and even if there was, the system allows partial submissions.
     
  5. Aug 24, 2011 #4

    lightgrav

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    they goofed up. it happens.
     
  6. Aug 24, 2011 #5
    Okay. Just to make sure they aren't 2 for 2..
    "A car moving with an initial velocity of 25 m/s north has a constant acceleration of 3 m/s2 south. After 6 seconds its velocity will be:"

    I say 11 m/s south, but this is no option.

    x = 25t - 3t^2
    v = 25 - 6t
    v(6) = 25 - 36
    v = -11 (or 11 south, as the option format it.)
     
  7. Aug 24, 2011 #6

    lewando

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    Could this mean the acceleration is not constant? -0.50t m/s2 is different from -0.50 m/s2.
     
  8. Aug 24, 2011 #7

    lightgrav

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    maybe they do have changing acceleration (didn't notice the "t", sorry)

    for problem 2 they want _velocity_ , not distance ...
     
  9. Aug 24, 2011 #8
    I'll try

    How is what I found distance?
     
  10. Aug 24, 2011 #9

    lightgrav

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    I don't know . . . you're ignoring units! where did the 6 t come from ? a = 3 m/s^2
     
  11. Aug 24, 2011 #10
    Guy, y'all were right. Acceleration was not constant. I defined acceleration with a = -.5t + 16 and integrated for velocity function, for a result of 8 seconds when velocity is zero which is correct.

    I differentiated, but I see the error. My position function should have 1.5 as the second coefficient, and the derivative should be back at 3.

    So, they didn't goof up at all.
     
    Last edited: Aug 24, 2011
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