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Homework Help: Kinematics Question Please Help!

  1. Dec 28, 2014 #1
    Okay so I have tried for hours to figure this out and can't. Please show me how to get the answers; what do I already have, what do I need, and what kinematics formula(s) do I use? Thanky ou so much!
    A jogger accelerates constantly to a velocity of +2.3m/s in 6.5s. After jogging 11m, the jogger stops. What was the initial velocity of the jogger?
    I can use any rearrangement of the 5 kinematics formulas (each one only has 4 variables) to solve this question.
    I do not understand if the final velocity is 0m/s or 2.3m/s. And does the 11m come after he accelerates or is the 11m the total distance that was traveled?
  2. jcsd
  3. Dec 28, 2014 #2

    Stephen Tashi

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    I can't interpret the problem either. Did you quote it exactly?
  4. Dec 28, 2014 #3


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    Staff: Mentor

    Hi Ksenia. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    We could recast this data into a legitimate kinematics question. But I would like to see the original wording just the same.

    Do you know what answer they are looking for?
    Last edited by a moderator: May 7, 2017
  5. Dec 28, 2014 #4
    Yes I worded it just like it was in the problem. And no there are no answers as this came from my correspondence module
  6. Dec 28, 2014 #5
    No, the answer isn't given and the wording is exactly as in my booklet.
    Last edited by a moderator: May 7, 2017
  7. Dec 28, 2014 #6


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    Why not, as a start, work out how far the jogger would travel if they started from rest and accelerated as described.
  8. Dec 28, 2014 #7

    Stephen Tashi

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    My guess is that "After jogging 11m, the jogger stops" is a badly written sentence that should have been worded to convey the idea that during the time the jogger is accelerating he travels 11 meters. So he "stops accelerating" not "stops moving".

    If you following the template for homework, you will state the 5 kinematic equations. My guess is that your are expected to work with [itex] v = v_0 + (a)(t) [/itex] and [itex] d = d0 + v_0 t + (1/2)(a)(t^2) [/itex] where we take [itex] d_0 = 0 [/itex] as the distance where the jogger began accelerating.
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