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Homework Help: Heading a Soccer Ball

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data
    When a soccer ball is kicked toward a player and the player deflects the ball by “heading” it, the acceleration of the head during the collision can be significant. Figure 2-31 gives the measured acceleration a(t) of a soccer player's head for a bare head and a helmeted head, starting from rest. At time t = 7.0 ms, what is the difference in the speed acquired by the bare head and the speed acquired by the helmeted head?
    *i attached the problem and graph


    2. Relevant equations
    Would you just find the area and subtract?


    3. The attempt at a solution
    I attempted to find the area of the collision with the bare head (.75m/s) and the area with the helmeted head (.26m/s) then subtracted to get .49m/s, but the answer at the back of the book says .56m/s. i haven't had calculus yet, so i'm still trying to understand integrals, any advice would be great!
     

    Attached Files:

  2. jcsd
  3. Aug 29, 2011 #2

    collinsmark

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    Hello ang359,

    Welcome to physics forums!
    Try the above again. I think something went wrong.
    That part sounds good to me.
    You're doing fine. You're on the right track. A (definite) integral is "the area under the curve." And that's what you're doing. :smile:

    [Edit: misinterpreted the graph myself (by a factor of 10) in my original post. Made corrections above.]
     
    Last edited: Aug 29, 2011
  4. Aug 29, 2011 #3

    PeterO

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    The only error I see is you have mis-calculated the "area ... with the bare head"

    Not sure how you did it, but with shapes defined by a series of straight line segments like these I just "count the squares" - or in this case rectangles - then convert

    from the scales [ignoring units] you can find that each rectangle represents 20
    The area under the "helmeted head" totals 13 squares so 260 units. Now considering the scales involved - m/s^2 and milliseconds that easily yields your 0.26 m/s.
     
  5. Aug 31, 2011 #4
    Okay thanks!! i'm not quite sure what i did wrong with the area of the bare head but i'll try it again. it's good to know that i'm somewhat on the right track
     
  6. Aug 31, 2011 #5
    yeah that makes sense, i'll try the calculations again to see if i made a silly mistake. thanks so much!!
     
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