# Calculating Distance - Issue

I am having difficulty with the following problem.

A man covers a distance on a scooter. Had he moved 3Kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40 min more. What is the distance (Ans=40)?

Here is what i came up with

$s = (v+3)(t - 40/60)$

$s = (v-2)(t+ 40/60)$

3 variables, two equations? Any suggestions what else I am missing or an easier way to solve this ?

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what about $s=v t$? – Julian Jun 30 '12 at 14:12

You have $s=v t$. So since $0=-\frac{4}{6}(v+3)+3t=\frac{4}{6}(v-2)-2t$ it follows that $t=\frac{10}{3}$, $v=12$, $s=40$

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How did you get 4/6 ? – Rajeshwar Jun 30 '12 at 14:36
Could you explain your answer in a little detail – Rajeshwar Jun 30 '12 at 14:36
+1: @Rajeshwar, see my answer for more details. – Cameron Buie Jun 30 '12 at 15:06

Recall also that he actually traveled the distance $s$ at velocity $v$ in time $t$, so we have $s=vt$ as our first equation. Rewriting the other two equations slightly, by distributing (FOIL) and by simplifying the fraction $\frac{40}{60}=\frac{2}{3}$, we have $$\left\{\begin{array}{l}s=vt\\s=vt-\frac{2}{3}v+3t-2\\s=vt+\frac{2}{3}v-2t-\frac{4}{3}\end{array}\right.$$

We readily see (by subtracting the first equation from the other two) that $$\left\{\begin{array}{l}0=-\frac{2}{3}v+3t-2\\0=\frac{2}{3}v-2t-\frac{4}{3}\end{array}\right.$$ and adding those two equations together, we find that $0=t-\frac{10}{3}$, so $t=\frac{10}{3}$. Plugging that back into the first equation in the second system yields $0=-\frac{2}{3}v+10-2$, so $\frac{2}{3}v=8$, so $v=12$. Finally, since $s=vt$, we have $s=12\cdot\frac{10}{3}=40$.

I only posted to expand on Julian's answer! He likely assumed that you could then solve the system of three linear equations in three variables, and so glossed over the details. Feel free to upvote my answer, if it's helpful, but you should arguably accept his over mine.

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Thank you for specifying the detail. – Rajeshwar Jun 30 '12 at 15:48