If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race?
Solution:
Solving for the time it takes to reach the first 20 meters.
\begin{align*} \Delta x & =v_{0}t+\frac{1}{2}at^{2} \\ 20 \ \text{m} & = \left( 0 \ \text{m/s} \right)t + \frac{1}{2}\left( 4.20 \ \text{m/s}^{2} \right)t^{2} \\ 20 & = 2.1 t^{2} \\ \frac{20}{2.1}& = \frac{\cancel{2.1} \ t^{2}}{\cancel{2.1}} \\ t^{2} & = 9.5238 \\ \sqrt{t^{2}} & = \sqrt{9.5238} \\ t_{1} & = 3.09 \ \text{s} \end{align*}
We can compute the velocity of the sprinter at the end of the first 20 meters.
\begin{align*} v^2 & = v_{0}^2 + 2ax \\ v^2 & = \left( 0 \ \text{m/s} \right)^2 + 2\left( 4.20 \ \text{m/s}^2 \right) \left( 20 \ \text{m} \right) \\ v & = \sqrt{2\left( 4.20 \right)\left( 20 \right)} \ \text{m/s} \\ v & = 12.96 \ \text{m/s} \end{align*}
For the remaining 80 meters, the sprinter has a constant velocity of 12.96 m/s. The sprinter’s time to run the last 80 meters can be computed as follows.
\begin{align*} \Delta x & = vt \\ 80 \ \text{m} & = \left( 12.96 \ \text{m/s} \right)\ t_{2} \\ \frac{80 \ \text{m}}{12.96 \ \text{m/s}} & =\frac{\cancel{12.96 \ \text{m/s} } \ \ t_{2}}{\cancel{12.96 \ \text{m/s}}} \\ t_{2} & = \frac{80}{12.96} \ \text{s} \\ t_{2} & = 6.17 \ \text{s} \\ \end{align*}
The sprinter’s total time, t_{T}, to finish the 100-m race is the sum of the two times.
\begin{align*} t_{T} & = t_{1} + t_{2} \\ t_{T} & = 3.09 \ \text{s} + 6.17 \ \text{s} \\ t_{T} & = 9.26 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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