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Homework Help: Finding the time elapsed, distance and speed

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A horse race takes place on a long straight track, at the beginning of the race the horse accelerates from rest to some top speed with a (constant) acceleration of 1.13 m/s^2 and then maintains that top speed to the end of the race. The total time taken from the start of the race to its end is 2.05 minutes and the total distance covered is 2.41 km.

    (a) What was the time elapsed during the acceleration phase of the horse's motion?
    (b) What distance did the force cover during the acceleration phase of its motion?
    (c)What was the top speed reached by the horse?

    2. Relevant equations

    3. The attempt at a solution
    a= 1.13 m/s^2
    distance/displacement = 2410 m
    t= 123s
    v1= 0

    V2^2 = V1^2+2ad
    V2^2 = 0+2(1.13)(2410)
    v2 = 73.8 m/s

    a=1.13 m/s^2

    v2 = v1+at
    Therefore, t = v2-v1/a

    t = (73.8-0)/(1.13) = 65.3 s

    t= 65.3s
    d= ?
    v1= 0

    d = v2^2-v1^2/2a
    d= (73.8)-(0)/2(1.13) = 2130 m

    (C) Top speed was found in (A) to be 73.8 m/s

    This was my guess on how to do it...help?
  2. jcsd
  3. Sep 20, 2011 #2


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    Homework Helper

    During the acceleration phase, d is much less than 2410.
    You don't know the distance, so you can't use this formula to find V2.

    You could sketch a velocity graph with unknown t for the acceleration time and unknown V2. The area under the graph is the known distance, so you could write an equation for that with 2 unknowns, then use v2 = at as a second equation.

    Or skip the area and write the total distance as the sum of the distance while under acceleration and the distance while at constant speed. Same two unknowns, same two equations.

    Looks like it will be a quadratic equation, only one solution reasonable.
    If the acceleration phase was very short, then the speed would be close to 2410/123 = 20 m/s. The final V2 will be a little larger than that to compensate for the reduced speed during the acceleration phase. Nowhere near 73, though.
  4. Sep 21, 2011 #3
  5. Sep 22, 2011 #4


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    Homework Helper

    geonik, do you know the two distance equations for accelerated and constant speed motion?
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