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Homework Help: Linear motion: a runner with two accelerations: 1 constant other not

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Problem 3.7 Suppose that motion studies of a runner show that the maximum speed he
    can maintain for a period of about 10 s is 12 m/s. If in a 100-m dash this runner
    accelerates with constant acceleration until he reaches this maximum speed and then
    maintains this speed for the rest of the race, what acceleration will he require if his total
    time is 11 s?

    2. Relevant equations

    a = v2/(22v - 200)

    3. The attempt at a solution

    a = 2.25 m/s2

    I've read through the solution. What I don't understand stand, however, is why v here would be the 12 m/s value. Sure, that's the velocity given, but how and why should I know to use 12 m/s? Why do I plug that into the relevant equation? I could say, sure, because it's the only velocity value given, but why that particular value?
  2. jcsd
  3. Sep 11, 2012 #2
    Acceleration is based on how much velocity needs to be attained. Velocity itself is determined as distance/time. So either you need to plug in the velocity or the date and time to arrive at an acceleration value!
  4. Sep 11, 2012 #3


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    He reaches the maximum velocity in time t1 with acceleration a. He covers the remaining distance in time (t - t1) with maximum velocity.
    Use the time velocity graph to find the total displacement.
    S = 1/2*t1vMax + vmax*(t - t1).
    t and vmax is given. Find t1 ,and hence find acceleration.
  5. Sep 12, 2012 #4


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    In case it's not obvious the title is wrong..

    The runner first accelerates and then continues at constant velocity.
    Last edited: Sep 12, 2012
  6. Sep 12, 2012 #5


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    Well first assume he's trying to win so he will run at max velocity for as long as possible. It's reasonable to assume that will be done at the end of the race.

    So in effect you are being asked to work out how fast he has to accelerate to reach a max velocity. That max velocity is bound to figure in the equations somewhere!

    The equations of motion normally need the starting velocity (normally U) and the final velocity (normally V). In this case U=0. V=12m/s.
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