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Homework Help: Motion and Forces : Dynamics

  1. Sep 22, 2006 #1
    The 100-m dash can be run by the best sprinters in 10.0s. A 60-kg sprinter accelerates uniformly for the first 30m to reach her top speed, which she maintains for the remaining 70m. (a) what is the horizontal component of force exerted on her feet by the ground during the acceleration? (b) what is the speed of the sprinter over the last 70m of the race (i.e , her top speed)

    (a) = 170N
    i jux do not get question a, can u guyz tell me wt formula i should use so that i can solve a????
  2. jcsd
  3. Sep 22, 2006 #2


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    With constant acceleration, a, an object can increase its speed from 0 to vf in t seconds according to vf= at (That's just the definition: 'acceleration = change is speed divided by t, a= v/t, multiplied by t). During that time, the object will have moved distance x= (a/2)t2.

    You are told that an object (the runner) has constant acceleration while moving a distance 30m. Okay, 30= (a/2)t2 but you don't yet know a or t. If you had been told the speed at the end of those 30 m, you could use v= at to get a second equation to solve for those two "unknowns", but you aren't so you now have two equation in three "unknowns". I assume that this sprinter completes the entire 100 m in the 10.0 s given.
    Running at constant speed v, a runner goes a distance x= vt in time t. Letting t' be the time that the runner takes for the final 70 m, we have 70= vt'. Since the total time for the 100 m is t+ t'= 10.0 s, t'= 10- t and that equation is 70= (10-t)v.

    You know have:
    (a/2)t2= 30
    v= at
    (10-t)v= 70
    three equations in the three "unknowns", a, v, t.
    For (A), once you know a, you can use F= ma (here, m= 60 kg.)
    Part (B) asks for v.
  4. Sep 22, 2006 #3
    thz there, but i really dun get how can u solve for the unknown there; i cant even solve for one.... which one should i solve first????
  5. Sep 23, 2006 #4
    You solve the three equations simultaneously. For example you can substitute the fact that v = at into the last equation to get it in terms of a and t only. Then you substitute in that at^2 = 60 into the new equation and you can solve for at, which is your value for v. Then you do some more substitution to solve for the other variables.
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