(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration.
(b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
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 \Delta x_1 (while accelerating) and \Delta x_2 (with constant speed), where \Delta x_1 + \Delta x_2 = 100 \ \text{m} .
For \Delta x_1:
During the accelerating period, we are going to use the formula \Delta x=v_0t+\frac{1}{2}at^2, since we know that \displaystyle a=\frac{\Delta v}{t}=\frac{v_{max}-v_0}{t}=\frac{v_{max}}{t}; and t=3.00 \ \text{s}.
\begin{align*}
\Delta x & =v_0t+\frac{1}{2}at^2 \\
\Delta x_1 & =0+\frac{1}{2}at^2 \\
\Delta x_1 & =\frac{1}{2}at^2 \\
\Delta x_1 & =\frac{1}{2}\left(\frac{v_{max}}{t}\right)t^2 \\
\Delta x_1 & =\frac{1}{2}\left(v_{max}\right)t \\
\Delta x _1&=\frac{1}{2}\left(v_{max}\right)\left(3.00\:\text{s}\right) \\
\Delta x _1&=1.5v_{max}
\end{align*}When the speed is constant, t=6.69 \ \text{s}, so
\begin{align*}
\Delta x_2 & = v_{max}t \\
\Delta x_2 & = v_{max}\left(6.69\:\text{s}\right) \\
\Delta x_2 & =6.69v_{max}
\end{align*}Plugging-in the two equations in the equation \Delta x_1 + \Delta x_2 = 100 \ \text{m} .
\begin{align*}
\Delta x_1 + \Delta x_2 & = 100 \ \text{m} \\
1.5v_{max} + 6.69v_{max} & =100 \ \text{m} \\
8.19\:v_{max} & =100 \\
v_{max} & =\frac{100}{8.19} \\
v_{max} & =12.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Therefore, his acceleration can be computed using the formula
a=\frac{v_{max}}{t}Plugging in the given values
\begin{align*}
a & =\frac{v_{max}}{t} \\
a & = \frac{12.2\:\text{m/s}}{3.00\:\text{s}} \\
a & = 4.07\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
Similar to part (a), we can plug in the different values for time and total distance:
\begin{align*}
\Delta x_1+ \Delta x_2 & =200 \\
1.5\:v_{max}+\left(19.30-3.00\right)v_{max} & =200 \\
1.5\:v_{max}+16.30v_{max} & =200 \\
17.80v_{max} & =200 \\
v_{max} & =\frac{200}{17.80} \\
v_{max} & = 11.2\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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