Algebra help with distance, two runners on circular track. How far before they meet.

Question

Two runners race around a circular track, with a 300 ft radius, in opposite directions, from the same starting point. Tom averages 8 mph and Dick averages 12 mph. How far has Dick run before he meets Tom?

Answer 1

You need to set $\text{time} = t$ and decide if you are going to solve it in term so minutes or hours. (Hint: minutes)

You need to find out $$\begin{align} v_{tom} &= \text{Tom's speed in feet per minute}\\ v_{dick} &= \text{Dick's speed in feet per minute}\end{align}$$ For these you need to know there are $5280$ feet in a mile. And that there are $60$ minutes in an hour.

You need to know the length of the track in feet. Call it $C$. You need the formula $$\text{Circumference} = 2 \times \pi \times \text{radius}$$ The radius is $300$ feet.

Now you need to find a formula using time, $t$, as a variable to solve. (Hint: Tom and dick each run for $t$ minutes at their speeds $v_{tom}$ and $v_{dick}$ and in that time their combined distance is $C$. So what is the formula?)

Is that enough to get you started?

Answer 2

The track length is $\quad300\times 2\pi\approx 1885'.\quad $ The two runners are converging at $$\quad12+8=20\text{ mph}= \frac{20\times 5280}{3600}=\frac{88}{3}\text{ fps}.$$

This means they will meet in $$\frac{3}{88}\cdot 1885 \approx 64.26 \text{ seconds}$$

Dick is running at $$\frac{12}{20}\cdot \frac{88}{3}=17.6 \text{ fps}$$

The distance Dick has run when he meets Tom is $17.6\times 64.26\approx 1131'$

We can compare this to the easy way where Dick's distance is $$\frac{12}{20}\cdot 1885=1131'$$