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Homework Help: Kinematics motion

  1. Dec 6, 2013 #1
    After 35mins of runnin, at the 9km point in a 10km race, you find yourself 100m behind the leader and moving at the same speed.
    What should your acceleration be if you're to catch up by the finish line? Assume that the leader maintains constant speed.
    Last edited: Dec 6, 2013
  2. jcsd
  3. Dec 6, 2013 #2


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    Please show your own attempt.
  4. Dec 6, 2013 #3
    vf = 9000m/35min = 257m min^-1
    vf = 9100m/35min = 260 min^-1

    10,000-9100m = 900m
    900m/ 260m min^-1 = 3.46min

    Runner requires 3.46min to tie the leader at the finish line.

    xf - xi = 0.5(vf + vi)t
    1000m = 0.5(vf+257m min^-1)3.46min
    vf = 321m min^-1

    vf doesn't tally with the answer sheet
  5. Dec 6, 2013 #4
    Can someone help me? I've spent 4 hrs on this question already. I don't want to get into the unhealthy habit of not sleeping for days until I figure out the answer.
  6. Dec 6, 2013 #5


    Staff: Mentor

    The numbers that are significant in this problem are the 1000 m that the slower runner needs to cover, and the 900 m that the faster runner has left.

    Don't use the same variable for both runners:
    vf can't possibly be equal to two different numbers.
  7. Dec 6, 2013 #6


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    I agree. You have to get to the finish in less than 3.46min.

    Than I would look at one of the SUVAT equations to work out the minimum constant acceleration required. http://en.wikipedia.org/wiki/Equations_of_motion

    Perhaps this one and solve for a...

    s = ut + 0.5at2

    You know..

    s = 900m
    u = 257m/min or 4.28m/S
    t = 3.46min or 207.6 seconds
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