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Homework Help: Football player kinematics in one dimension

  1. Jun 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of 3 seconds. Then during a negligible amount of time, he changes the magnitude of his acceleration to a value of 1.1 m/secondssquared. With this acceleration, he continues in the same direction for another 2 seconds until he reaches a speed of 6.4 m/sec. What's the value of his acceleration (assumed to be constant) during the initial 3 sec period?

    2. Relevant equations

    kinematics

    3. The attempt at a solution

    I tried drawing a graph but to no avail. How would I do it with just equations?
     
  2. jcsd
  3. Jun 14, 2007 #2

    Dick

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    You'll have to pick out an equation to use first. As you are given times, velocities and accelerations - something relating the three of those could be handy. Can you find one?
     
  4. Jun 14, 2007 #3
    v=vinital+at
    but i dont understand what 1.1 m/s sqrd is for....is that the acceleration during the 3sec-5sec time period? i dont know what to plug in
     
  5. Jun 14, 2007 #4

    Dick

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    Good equation choice and yes, that's what the 1.1m/sec^2 is. So first concentrate on the 3-5 sec period. You know the final velocity (at 5 sec) and you know the acceleration. Can you determine the velocity at 3 sec? BTW write the equation as v=vinitial+a*(t-tinitial) where tinitial is 3 sec for this time period.
     
    Last edited: Jun 14, 2007
  6. Jun 14, 2007 #5
    Ok, so I got that stuff plugged in but the acceleration isn't there...what would that be?
     
  7. Jun 14, 2007 #6
    Ohhh ok, so 1.1 is the magnitude of the acceleration so that means it IS the acceleration from 3-5 seconds right?
     
  8. Jun 14, 2007 #7

    Dick

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    Right. You may find it easier to work with your equation in the form (change of v)=a*(change of t).
     
  9. Jun 15, 2007 #8
    Ok, so i got 1.4 m/s^2 but i dont know if that's the right answer? can anyone do it and see what they get/help me? =)
     
  10. Jun 15, 2007 #9

    Dick

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    Yes, that's right.
     
  11. Jun 15, 2007 #10
    Yay, thank you so much for your help =)
     
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