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.
Δ x = v 0 t + 1 2 a t 2 20 m = ( 0 m/s ) t + 1 2 ( 4.20 m/s 2 ) t 2 20 = 2.1 t 2 20 2.1 = 2.1 t 2 2.1 t 2 = 9.5238 t 2 = 9.5238 t 1 = 3.09 s \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*} Δ x 20 m 20 2.1 20 t 2 t 2 t 1 = v 0 t + 2 1 a t 2 = ( 0 m/s ) t + 2 1 ( 4.20 m/s 2 ) t 2 = 2.1 t 2 = 2.1 2.1 t 2 = 9.5238 = 9.5238 = 3.09 s We can compute the velocity of the sprinter at the end of the first 20 meters.
v 2 = v 0 2 + 2 a x v 2 = ( 0 m/s ) 2 + 2 ( 4.20 m/s 2 ) ( 20 m ) v = 2 ( 4.20 ) ( 20 ) m/s v = 12.96 m/s \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*} v 2 v 2 v v = v 0 2 + 2 a x = ( 0 m/s ) 2 + 2 ( 4.20 m/s 2 ) ( 20 m ) = 2 ( 4.20 ) ( 20 ) m/s = 12.96 m/s 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.
Δ x = v t 80 m = ( 12.96 m/s ) t 2 80 m 12.96 m/s = 12.96 m/s t 2 12.96 m/s t 2 = 80 12.96 s t 2 = 6.17 s \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*} Δ x 80 m 12.96 m/s 80 m t 2 t 2 = v t = ( 12.96 m/s ) t 2 = 12.96 m/s 12.96 m/s t 2 = 12.96 80 s = 6.17 s The sprinter’s total time, t T t_{T} t T
t T = t 1 + t 2 t T = 3.09 s + 6.17 s t T = 9.26 s ( Answer ) \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*} t T t T t T = t 1 + t 2 = 3.09 s + 6.17 s = 9.26 s ( Answer )
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