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Drag force deceleration

  1. Sep 30, 2016 #1
    1. The problem statement, all variables and given/known data
    13.51.PNG

    I forgot if the deceleration is -0.003v or same here -0.003v2

    2. Relevant equations
    v = u + at (first acceleration)
    dv/dt = a (deceleratioin) = -0.003v^2

    3. The attempt at a solution
    From first equation,
    342.2 = 0 + 30*t1
    t1= 11.407s

    From 2nd equation,
    dv/v2 = -0.003
    [-1/3.422] - [-1/342.2] = -0.003 (t2-t1)

    ------------------------------------------

    If the decelerate is -0.003v
    dv/v = -0.003
    ln (3.422) - ln (342.2) = -0.003 (t2-t1)

    t2 (total time) = 1544.46 s

    Is the answer making sense?
     
  2. jcsd
  3. Sep 30, 2016 #2
    Updates: should be -0.003 v
     
  4. Oct 1, 2016 #3

    ehild

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    Correct, so what is t2, the total time?
    ------------------------------------------
    In the problem text, the deceleration is a = -0.003 v2, and you did not calculate the time. In case of a= -0.003v, your answer is correct. Use four significant digits.
     
  5. Oct 2, 2016 #4
    thanks for reply. The question was using a = -0.003v as deceleration.
    And my t is 1544.46s, which is 25 min, is it making sense? cuz it is soooo long.
     
  6. Oct 2, 2016 #5

    ehild

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    Yes, it is correct, with that low deceleration, which is linear in velocity. At that high speed, quadratic deceleration would be more realistic
     
  7. Oct 2, 2016 #6
    Okay thanks, I double checked the question, it uses v instead of v2, which you mentioned to be more realistic.
     
  8. Oct 2, 2016 #7

    haruspex

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    On a pedantic note, the question should not give the acceleration as either -0.003v m/s2 or -0.003v2 m/s2.
    Rather, the units should be given as s-1 and m-1 respectively. This is because v will come with its own units. The units specified in the expression for a should only be the units of the -0.003 constant.
     
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