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Homework Help: Problem understanding simple constant acceeleration problem

  1. Dec 17, 2014 #1

    i have an exam and i was trying solving old exams and i came upon this problem

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

    Due to air resistance, a 60- kg parachutist experiences an acceleration of 3.0 m/s2 while
    falling in the air. If she falls 80 m in 12 seconds what is her final speed (in m/s)?
    2. Relevant equations
    i know its a constant acceleration problem so

    3. The attempt at a solution
    D is Y2-Y1 =(-80-0)=-80
    by using 2nd equation vi=24.7
    and then using 1st equation vf should be -11.3
    so final speed should be 11.3
    but when i checked the answer i found it different

    here is the answer that i found
    v=24.7=25 m/s
    final answer is 25
  2. jcsd
  3. Dec 17, 2014 #2


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    Welcome to Physics Forums.

    I'm not sure if your answer is 11.3, or 24.7, but I agree with the answer of 24.7 m/s

    Since acceleration is constant, you can find the average speed by taking half of the sum of the initial and final speeds.
    In 12 seconds at 3m/s/s, the initial speed would be 36 m/s less than the final speed, so you get [itex]V_{avg}=\frac{1}{2}(V_f+(V_f-36))=V_f-18[/itex]
    But the average speed is also the total distance divided by the total time so you get [itex]V_{avg}=V_f-18=\frac{80}{12}[/itex]
    and thus [itex]V_f≈24.7[/itex]
  4. Dec 17, 2014 #3
    Don't you think it's a little weird that the initial velocity was upward at 11.3 m/s? Do you think that whoever made up the problem didn't realize this?

  5. Dec 17, 2014 #4


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    Homework Helper

    Oh wow, I didn't realize this either! Well, that doesn't make much sense... :confused:
  6. Dec 17, 2014 #5


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    Gold Member

    Was looking at it and thought I needed to take a nap. Tried pulling the rip cord at t=0, putting the chute into an updraft. Gave up.
  7. Dec 17, 2014 #6
    I think that whoever made up the problem was taking a nap.

  8. Dec 17, 2014 #7
    this was a short problem from last year exam (it's the official answer according to physics department)

    so i am not doing something wrong? Untitled.png
  9. Dec 17, 2014 #8
    I can't see how they got this answer. For one thing, the solution to their equation is -24.7, not + 24.7. The equation itself doesn't make sense to me.

    Here's how I would have done it (with downward velocities, displacements, and accelerations treated as positive):
    [tex]80 = v_i(12)+\frac{3}{2}(12)^2[/tex]
    v_i = -11.33 m/s

    v_f = v_i + 3(12) = 24.7 m/s

    Nathanael's method in post #2 is equally acceptable (maybe better), and gives the same answer.

    Last edited: Dec 17, 2014
  10. Dec 17, 2014 #9
    thank you all for clarification

    i will ask my dr about it
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