Catchup and Pass Example Problem

We know that Sarah is running at a speed of $800\frac{\text{feet}}{\text{minute}}$ and Joon is running at a speed of $\mathrm{1,200}\frac{\text{feet}}{\text{minute}}$.

The amount of time Joon runs is equal to the amount of time Sarah runs. We can let this time be t.

	Rate	Time	Distance
Joon	$1,200\frac{\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}}{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}$	t minutes
Sarah	$800\frac{\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}}{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}}$	t minutes

Now that we have a rate and time for Joon and Sarah, we can determine their distances using the distance formula:

$\begin{array}{l}\Rightarrow {\text{distance}}_{\mathrm{J}\mathrm{o}\mathrm{o}\mathrm{n}}=\mathrm{1,200}\frac{\text{feet}}{\text{minute}}\times \text{t minutes}=\mathrm{1,200}\mathrm{t}\text{ feet}\\ \Rightarrow {\text{distance}}_{\mathrm{S}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{h}}=800\frac{\text{feet}}{\text{minute}}\times \text{t minutes}=800\mathrm{t}\text{ feet}\end{array}$

	Rate	Time	Distance
Joon	$\mathrm{1,200}\frac{\text{feet}}{\text{minute}}$	t minutes	1,200t feet
Sarah	$800\frac{\text{feet}}{\text{minute}}$	t minutes	800t feet

In a normal catch-up problem we would set the two distances equal and determine t. However, in this problem Joon actually runs 120 more feet than Sarah because Joon starts 80 feet behind Sarah and then passes her by 40 feet. Therefore:

$\begin{array}{l}\Rightarrow {\text{distance}}_{\mathrm{J}\mathrm{o}\mathrm{o}\mathrm{n}}={\text{distance}}_{\mathrm{S}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{h}}+120\text{ feet}\\ \Rightarrow \mathrm{1,200}\mathrm{t}=800\mathrm{t}+120\\ \Rightarrow 400\mathrm{t}=120\\ \Rightarrow \mathrm{t}=\frac{120}{400}\text{=}\frac{3}{10}\text{minute}\end{array}$

We now must convert this time from minutes to seconds:

$\Rightarrow \text{Time in Seconds}=\left(\frac{\frac{3}{10}\text{ minute}}{1}\right)\times \left(\frac{\text{60 seconds}}{\text{1 minute}}\right)$

$\Rightarrow \mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}\text{ in Seconds}=\frac{\frac{3}{10}\cancel{\text{minute}}\times \text{ 60 seconds}}{1\cancel{\text{minute}}}=18\text{ seconds}$

$\begin{array}{}\end{array}$ The total time it takes Joon to catch up and pass Sarah is 18 seconds.

Alternate Solution:

We are given that:

“If Joon is currently 80 feet behind Sarah, how many seconds will it take Joon to catch up to Sarah and run 40 feet ahead of her?”

Since Joon is 80 feet behind Sarah and must catch up to Sarah and pass her by 40 feet, the change in distance is 120 feet. Of course, the change in rate is 1200 ft/min – 800 ft/min = 400 ft/min. Thus:

$\Rightarrow \text{Time}=\frac{\text{Δ Distance}}{\text{Δ Rate}}$

$\Rightarrow \text{Time}=\frac{\mathrm{120 }\mathrm{f}\mathrm{t}}{\mathrm{400 }\mathrm{f}\mathrm{t}/\mathrm{m}\mathrm{i}\mathrm{n}}=\frac{12}{40}\mathrm{m}\mathrm{i}\mathrm{n}=\frac{3}{10}\mathrm{m}\mathrm{i}\mathrm{n}$

We convert 3/10 min to 18 seconds.