How Do You Calculate Acceleration in Asafa Powell's Record-Breaking Dash?

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Asafa Powell's record-breaking 100m dash time of 9.77 seconds can be analyzed by dividing the race into two segments: the first 3 seconds with constant acceleration and the remaining 6.77 seconds at maximum speed. The distance covered during the acceleration phase is calculated using the formula for average velocity, leading to the final velocity being determined as 12.092 m/s. By applying the acceleration formula, the acceleration during the initial 3 seconds is found to be approximately 4.03 m/s². This approach effectively uses kinematic equations to solve for unknowns in the context of Powell's sprint. The calculations demonstrate the relationship between time, distance, and acceleration in sprinting performance.
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Homework Statement


On June 14th, 2005, Asafa Powell of Jamaica set a world's record for the 100-m dash with a time t=9.77s. Assuming he reached his maximum speed in 3.00s, and then maintained that speed until the finish, estimate his acceleration during the first 3.00s.


Homework Equations


vx=v0x+axt
x-x0=v0xt+1/2axt2
vx2=v0x2+2ax(x-x0)


The Attempt at a Solution


I broke the run into two segments, one of 3s with constant acceleration, and the other of 6.77s with 0 acceleration. The distance traveled in the first segment is 100m - the second, and the same is true of the second segment in relation to the first. Beyond that, I've been fumbling around with this problem for two hours and I'm still no closer to solving it than when I started.
 
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The problem is that there are 2 unknowns. You don't know what his acceleration is (obviously) and you don't know what his final velocity is.

It was a good idea to break the problem into two parts. Try using the third equation to write the final velocity in terms of acceleration.
 
Got it! I can't believe I spent so long on it when the solution was so simple!

There was another equation that I omitted because I didn't think I'd be using it:
\Delta x=\frac{1}{2}(v_{0x}+v_{x})t


So here is how I solved it:

I found \Delta x_{1} in terms of the final velocity with the formula above:

\Delta x_{1}=\frac{1}{2}(v_{0x}+v_{fx})t

\Delta x_{1}=\frac{1}{2}(0+v_{fx})*3 sec

\Delta x_{1}=1.5 sec*v_{fx1}

And since there is no acceleration in the second segment:

\Delta x_{2}=v_{x2}t

\Delta x_{2}=6.77 sec*v_{fx1}


\Delta x_{total}=\Delta x_{1}+\Delta x_{2}

100 meters=1.5 sec*v_{fx1}+6.77 sec*v_{fx1}

100 meters=8.27 sec*v_{fx1}

v_{fx1}=12.092 m/s



a=\frac{\Delta v}{\Delta t}

a_{x1}=12.092 m/s \div 3 sec

a_{x1}=4.03 m/s^2


The equations aren't coming out right on the preview. The first equation if for delta x, 1 = delta x 1, 2 = delta x 2, and a = delta v /delta t
 
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