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Find constant when velocity is proportional to time

  1. Jul 1, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    I am having difficulty understanding the question statement . I do not understand how can body come to rest when velocity is directly proportional to time .As time progresses , velocity increases .

    Could someone help me understand the problem


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  2. jcsd
  3. Jul 1, 2017 #2


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    Instead of saying "before coming to rest", I think the problem should say "after starting from rest".
    Edit: But then, it can't have non-zero initial kinetic energy. So assume K is the KE of the body after travelling the distance d.
    Last edited: Jul 1, 2017
  4. Jul 1, 2017 #3
    Are you getting one of the options ?
  5. Jul 1, 2017 #4


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    See I edited the post. I haven't tried it yet.
  6. Jul 1, 2017 #5
    @Chestermiller ,@gneill could you help me understand the question .
  7. Jul 1, 2017 #6


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    Staff: Mentor

    The concept of direct proportionality does not specify the sign of the relationship (the constant of proportionality can be either positive or negative).

    I'm not impressed by the author's way of presenting the problem. I think it's intended to be an exercise in unit analysis, but they should have made it more clear that "V = Ct" was simply meant to demonstrate what direct proportionality means, not a formula that specifies the actual equation of motion for the given scenario.

    I suggest that you approach the problem in terms of unit analysis. Eliminate the answers that cannot be correct on the basis of the units that would result if direct proportionality (e.g. V = Ct) holds.
  8. Jul 1, 2017 #7


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    Using these assumptions, I am getting an answer from the options. Also, dimensional analysis, as suggested by gneill, gives the same answer.
  9. Jul 1, 2017 #8
    Yes . Your assumption does give the answer which in turn matches with dimensional analysis .

    Thanks .
  10. Jul 1, 2017 #9


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    You're welcome!
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