Subfield: Algebra


Question: In the diagram, a straight, flat road joins $A$ to $B$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d6fc47bd20a6486a92ceg-1.jpg?height=78&width=575&top_left_y=2059&top_left_x=840)

Karuna runs from $A$ to $B$, turns around instantly, and runs back to $A$. Karuna runs at $6 \mathrm{~m} / \mathrm{s}$. Starting at the same time as Karuna, Jorge runs from $B$ to $A$, turns around instantly, and runs back to $B$. Jorge runs from $B$ to $A$ at $5 \mathrm{~m} / \mathrm{s}$ and from $A$ to $B$ at $7.5 \mathrm{~m} / \mathrm{s}$. The distance from $A$ to $B$ is $297 \mathrm{~m}$ and each runner takes exactly $99 \mathrm{~s}$ to run their route. Determine the two values of $t$ for which Karuna and Jorge are at the same place on the road after running for $t$ seconds.

Solution:
Suppose that Karuna and Jorge meet for the first time after $t_{1}$ seconds and for the second time after $t_{2}$ seconds.

When they meet for the first time, Karuna has run partway from $A$ to $B$ and Jorge has run partway from $B$ to $A$.

![](https://cdn.mathpix.com/cropped/2023_12_21_77d89371ac5d12b13590g-1.jpg?height=170&width=588&top_left_y=392&top_left_x=866)

At this instant, the sum of the distances that they have run equals the total distance from $A$ to $B$.

Since Karuna runs at $6 \mathrm{~m} / \mathrm{s}$ for these $t_{1}$ seconds, she has run $6 t_{1} \mathrm{~m}$.

Since Jorge runs at $5 \mathrm{~m} / \mathrm{s}$ for these $t_{1}$ seconds, he has run $5 t_{1} \mathrm{~m}$.

Therefore, $6 t_{1}+5 t_{1}=297$ and so $11 t_{1}=297$ or $t_{1}=27$.

When they meet for the second time, Karuna has run from $A$ to $B$ and is running back to $A$ and Jorge has run from $B$ to $A$ and is running back to $B$. This is because Jorge gets to $A$ halfway through his run before Karuna gets back to $A$ at the end of her run.

![](https://cdn.mathpix.com/cropped/2023_12_21_77d89371ac5d12b13590g-1.jpg?height=236&width=588&top_left_y=1026&top_left_x=866)

Since they each finish running after 99 seconds, then each has $99-t_{2}$ seconds left to run. At this instant, the sum of the distances that they have left to run equals the total distance from $A$ to $B$.

Since Karuna runs at $6 \mathrm{~m} / \mathrm{s}$ for these $\left(99-t_{2}\right)$ seconds, she has to run $6\left(99-t_{2}\right) \mathrm{m}$.

Since Jorge runs at $7.5 \mathrm{~m} / \mathrm{s}$ for these $\left(99-t_{2}\right)$ seconds, he has to run $7.5\left(99-t_{2}\right) \mathrm{m}$.

Therefore, $6\left(99-t_{2}\right)+7.5\left(99-t_{2}\right)=297$ and so $13.5\left(99-t_{2}\right)=297$ or $99-t_{2}=22$ and so $t_{2}=77$.

Alternatively, to calculate the value of $t_{2}$, we note that when Karuna and Jorge meet for the second time, they have each run the distance from $A$ to $B$ one full time and are on their return trips.

This means that they have each run the full distance from $A$ to $B$ once and the distances that they have run on their return trip add up to another full distance from $A$ to $B$, for a total distance of $3 \cdot 297 \mathrm{~m}=891 \mathrm{~m}$.

Karuna has run at $6 \mathrm{~m} / \mathrm{s}$ for $t_{2}$ seconds, for a total distance of $6 t_{2} \mathrm{~m}$.

Jorge ran the first $297 \mathrm{~m}$ at $5 \mathrm{~m} / \mathrm{s}$, which took $\frac{297}{5} \mathrm{~s}$ and ran the remaining $\left(t_{2}-\frac{297}{5}\right)$ seconds at $7.5 \mathrm{~m} / \mathrm{s}$, for a total distance of $\left(297+7.5\left(t_{2}-\frac{297}{5}\right)\right) \mathrm{m}$.

Therefore,

$$
\begin{aligned}
6 t_{2}+297+7.5\left(t_{2}-\frac{297}{5}\right) & =891 \\
13.5 t_{2} & =891-297+7.5 \cdot \frac{297}{5} \\
13.5 t_{2} & =1039.5 \\
t_{2} & =77
\end{aligned}
$$

Therefore, Karuna and Jorge meet after 27 seconds and after 77 seconds.

Final answer: 27, 77