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# After rounding the final turn in the bell lap, two runners emerge ahead of the pack. When Runner A. is 200 ft from the finish line, his speed is 22 ft/sec, a speed that he maintains until he crosses the line. At that instant of time, Runner B, who is 20 ft behind Runner A and running at a speed of 20 ft/sec, begins to sprint. Assuming that Runner B sprints with a constant acceleration, what minimum acceleration will enable him to cross the finish line ahead of Runner A?

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From the text of the exercise we have that

\begin{align*} v_{A}&=22\mathrm{\,\, \frac{ft}{s}}\\ d_{A}&=200\mathrm{\,\, ft}\,\, \text{(distance from the finish)}\\ v_{B}(0)&=20\mathrm{\,\, \frac{ft}{s}}\\ d_{B}&=220\mathrm{\,\, ft} \end{align*}

We also know that

$v_{B}(t)=v_{B}(0)+a\cdot t=20+a\cdot t$

Since the distance runner $A$ runs is given by $\pmb{v_{A}\cdot t}$ (because he is moving at a constant speed), we have that he will reach the finish line in

$t_0=\frac{200}{22}=\frac{100}{11}\mathrm{\,\, s}$

This means that the constant acceleration of the runner $B$ must be large enough so that runner $B$ will reach the finish line in less than $\frac{100}{11}$ seconds. In other words,

$S_{B}\left(\frac{100}{11}\right)=220$

where $S_{B}(t)$ is the distance runner $B$ covers in $t$ seconds. We have the following

$\begin{array}{ccccccc} S_{B}^\prime(t) & = & v_{B}(t) & = & 20 & + & at\\ S_{B}(0)& = & 0 \end{array}\Bigg\}$

Integrating $S_{B}^\prime(t)$ we get

\begin{align*} S_{B}&=\int (20+at)\, dt\\ &=20\int \, dt+\int at\, dt\Rightarrow\pmb{S_{B}(t)=20t+\frac{at^2}{2}+C} \end{align*}

From the initial condition we have

$0=S(0)=0+0+C\Rightarrow\pmb{C=0}$

Thus,

$S_{B}(t)=20t+\frac{at^2}{2}$

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