College Physics 2.40 – Max speed and acceleration of Usain Bolt

Solution:

Part A

There are two parts to the race and must be treated separately since acceleration is not uniform over the race. We will divide the race into $x_1$ (while accelerating) and $x_2$ (with constant speed), where $x_1+x_2=100\:m$.

When the speed is constant, $t=6.69s$, so

$x_2=vt=\left(6.69\:s\right)v=100m-x_1$

When accelerating, $t=3.00s$, so

$x_1=\frac{1}{2}at^2=\frac{1}{2}vt=\left(1.50\:s\right)v$

Plugging $x_1$ into the previous equation, we get the maximum speed:

$100m-\left(1.50s\right)v=\left(6.69s\right)v$

$v=\frac{100m}{6.69s+1.50s}$

$v=12.2\:m/s$

Therefore, his acceleration was

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

$a=\frac{12.2\:m/s}{3.00\:s}$

$a=4.07\:m/s^2$

Part B

Similar to part (a), we can plug in the different values for time and total distance:

$200m-\left(1.50\:s\right)v=\left(16.30\:s\right)v$

$v=\frac{200\:m}{16.30\:s+1.50\:s}$

$v=11.2\:m/s$