1. Aug 29, 2011

### ang359

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:

• ###### soccer ball problem.jpg
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2. Aug 29, 2011

### collinsmark

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.

[Edit: misinterpreted the graph myself (by a factor of 10) in my original post. Made corrections above.]

Last edited: Aug 29, 2011
3. Aug 29, 2011

### PeterO

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.

4. Aug 31, 2011

### ang359

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

5. Aug 31, 2011

### ang359

yeah that makes sense, i'll try the calculations again to see if i made a silly mistake. thanks so much!!