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Homework Help: 1D Kinematics question, need help!

  1. May 25, 2009 #1
    A sprinter running a 100 meter race starts at rest, accelerates at constant acceleration
    with magnitude A for 2 seconds, and then runs at constant speed until the end.

    a) Find the position (relative to the start position) and speed of the runner at the end of
    the 2 seconds in terms of A. ANS: 2A , Vx=2A

    b) Assume that the runner takes a total of 10 seconds to run the 100 meters. Find the
    value of the acceleration A. You can leave your answer in terms of a fraction but clearly
    indicate the units. ANS: 100/18

    .....struggling with b) !!!!!!!!!!!!!!!!
  2. jcsd
  3. May 25, 2009 #2


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    Gold Member

    You need to show us some work before we can help.
  4. May 26, 2009 #3
    Relevant equations
    all kinematic eq'n with x, v, a in it

    What I know so far:

    I got that x(2)=2A and v(2)=2A

    when it comes to b) i have tried various inputs...
    Which way do I approach this??? Do approach it from x=100 and t=10, do I incorporate x=2A and v=2A.

    The closest I've gotten is using x=100-2A, t=8, v=2A...
  5. May 26, 2009 #4
    Consider the 100m broken down into two parts
    - the part with constant acceleration from part a). You were asked to get the displacement of this in terms of A in part a). This will help you with that.
    - and consider the second part where velocity is constant. what formulae can you use?
  6. May 26, 2009 #5
    according to your post, it can be broken down into two times

    t=2 and t=8
    at t=2 , what I know is x=2A, v=2A, a=A

    at t=10, what I know is x=100, v=10 (because v=d/t=100/10, is this relavant?)

    there are two eq'ns that come to mind

    x(t)= Xi + Vit + 0.5at^2

    Vf^2= Vi^2+2a(Xf - Xi)
  7. May 26, 2009 #6
    For the second part of the race, the eight seconds at constant velocity, just think displacement = velocity X time.
    Your v=d/t = 100/10 isn't correct because velocity wasn't constant for the entire 100m.
    For the first two seconds, acceleration was constant, and velocity was increasing (at a constant rate= acceleration). For the remaining eight seconds then, velocity is constant, so you can use v = d/t here.
    Your two eqns are correct and should then help you find a value for A.

    You're eventually looking for the slope of the velocity/time graph (slope here = acc.) over the first two seconds. What do you need to find this?
  8. May 26, 2009 #7
    still struggling
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