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Basic kinematics - calculating time

  1. Oct 6, 2012 #1
    A boy in a wagon starts from a rest position and then goes down a straight ramp, gaining
    speed with uniform acceleration as he travels. In 6.00 s, he is travelling at 4.2 m/s.

    a) What is his velocity at 10.0 s?

    a = v2 – v1 / Δt
    a = 4.2 m/s [down] – 0 / 6.0 s
    a = 0.70 m/s^2 [down]

    v2 = v1+a Δt
    v2 = 0 + (0.70 m/s2) x 10.0 s
    v2 = 7.0 m/s [down]

    His velocity at 10.0 s is 7.0 m/s [down]


    b) How long will it take him to reach a speed of 28.0 m/s?

    This is where I am having trouble. I am unsure of the formula for calculating time. My book mentions Δt = v2 – v1 / Δt, but I believe this to be incorrect, so I tried Δt = v2 – v1 / a, but still could not get the answer they were looking for which is 40 seconds. Please advise.
     
  2. jcsd
  3. Oct 6, 2012 #2

    Doc Al

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    Staff: Mentor

    That's the correct formula. (Just a rearrangement of the one you were using.) Why can't you get the answer? What did you plug in?
     
  4. Oct 6, 2012 #3

    sophiecentaur

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    You are using the right equation but I wonder whether you are reading the question as they are hoping you should. I think they mean total time to reach 28m/s and not the extra time.
     
  5. Oct 6, 2012 #4
    I am not sure what velocity to use, I was using 7.0 m/s for v2 and 0 for v1 and dividing it by 0.70 for acceleration.
     
  6. Oct 6, 2012 #5

    sophiecentaur

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    For part b, they are dividing 28 (total change in speed from zero) by 0.7 (the accn. you calculated). I think you should read the question again!
     
  7. Oct 6, 2012 #6

    Doc Al

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    Staff: Mentor

    As sophiecentaur suggests, you need to reread the question. You were given the speed of 28.0 m/s for part b.

    (7.0 m/s was the speed you found after 10 seconds.)
     
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