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The faster of two runners on a circular race course can complete a lap in 30 seconds. The slower runner can complete a lap in 40 seconds. The runners start from the same spot at the same time. How long will it take the faster runner to gain a full lap on the slower runner?

I can easily do this problem in my head and come up with 120 seconds. I just can't figure out how to write it down and solve it using the distance = rate x time formula. I'm embarrassed to ask but can someone help?

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the laps per second of the first runner is $\dfrac{1}{30}$

the laps per second of the second runner is $\dfrac{1}{40}$

In $x$ second, the first runner will be one lap ahead of the second when

\begin{align} \dfrac{x}{30} - \dfrac{x}{40} &= 1 \\ 4x - 3x &= 120 \\ x &= 120 \end{align}

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  • $\begingroup$ I see someone beat me to it. Thanks guys. $\endgroup$ – Brian488 Jun 19 '17 at 0:38
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Let $x$ be the number of laps completed. When the faster racer has achieved one more lap, you have: $$30(x+1)=40x$$ Gives you $x=3$. So $(30*4)$ or $(40*3)$ gives you 120.

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If $d$ is the length of a lap, then $$ d=v_1t_1=v_2t_2 \qquad \Rightarrow \qquad \frac{v_1}{v_2}=\frac{t_2}{t_1}=\frac{4}{3}. $$ At a common time $t$, the fast runner will have completed $n+1$ laps to the slowest runner $n$ laps so $$ t=\frac{(n+1)d}{v_1}=\frac{nd}{v_2}\qquad\Rightarrow\qquad \frac{n+1}{n}=\frac{v_1}{v_2}=\frac{4}{3} $$ from which $n=3$. Since it takes $40$ seconds for the slow runner to complete one lap, it will take $t=3\times 40=120$seconds for fast runner to overtake the slow runner.

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How does this look?

Let x = the amount of time for the faster runner to gain a lap on the slower runner

The faster runner runs at a rate of $ \frac{1}{30}$ of a lap per second and the slower runner at $ \frac{1}{40}$ of a lap per second. So $ \frac{1}{30}x - \frac{1}{40}x$ = 1 should give me 120 seconds. Does this look good?

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