# Thread: Running Around a Track

1. ## Running Around a Track

Hello, helpful people. I'm in some need of help.

I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:

"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"

Thanks.

2. ## Re: Running Around a Track

Hello, MyHappyHarmony!

This one is tricky to set up.
I have a rather clunky approach . . .

Brooks and Avery are running laps around the outdoor track, in the same direction.
Brooks completes a lap every 78 seconds, while Avery completes a lap every 91 seconds.
Brooks (the faster runner) has just passed Avery.
How much time will it take for Brooks to overtake Avery again?"

We will use: . $\text{Distance} \:=\:\text{Rate} \times \text{Time} \quad\Rightarrow\quad r \,=\,\frac{d}{t}$
Let $d$ = distance in one lap (say, in feet).

Brooks' rate is $\frac{d}{78}$ ft/sec.

Avery's rate is $\frac{d}{91}$ ft/sec.

In $t$ seconds, Avery runs $\frac{d}{91}t$ feet.

In the same $t$ seconds, Brooks runs $\frac{d}{78}$ feet,
. . which is $d$ feet (one lap) more than Avery's distance.

There is our equation! . . . $\frac{d}{78}t \;=\;\frac{d}{91}t + d$

Divide by $d\!:\;\frac{t}{78} \:=\:\frac{t}{91} + 1$

Got it?

3. ## Re: Running Around a Track

Yes, I do! Thanks. Quick question: Would it be right to solve with a common denominator from there? The big numbers do look quite intimidating.

4. ## Re: Running Around a Track

Originally Posted by MyHappyHarmony
Hello, helpful people. I'm in some need of help.

I'm not the best at setting up equations for word problems, especially those related to time. Therefore, it'd be the greatest help if you could guide me in setting up an equation to solve this problem:

"Brooks and Avery are running laps around the outdoor track, in the same direction. Brooks completes a lap every 78 seconds while Avery needs 91 seconds for every tour of the track. Brooks (the faster runner) has just passed Avery. How much time will it take for Brooks to overtake Avery again?"

Thanks.
Try to construct your solution from this figure

cross posting

5. ## Re: Running Around a Track

I'm sorry. I'm not drawing anything from that figure.

6. ## Re: Running Around a Track

another method
assume track is 1200ft B rate 1200/78=15.38 fps A rate 1200/91= 13.19fps
after B runs 1 lap A is 1029ft ahead of B 91 * 13.19 = 1029 ft
Rate of closure = 15.38-13.19 = 2.2 fps
closure time 1029/2.2 = 467.7 sec
total elapsed time = 467.7 +78= 555.7 sec

7. ## Re: Running Around a Track

Originally Posted by bjhopper
total elapsed time = 467.7 +78= 555.7 sec
typo; 545.7 ; 546 really: LCM(91,78)

These are quite simple: have the faster runner start behind the slower,
the distance behind being the track length; how long does it take to catch up?

8. ## Re: Running Around a Track

Hello Wilmer,
Long 'time no see.How are you?
Soroban' s equation gives elapsed time until runners meet. 546 sec exactly
My answer 555.7 sec.There is no typo error.
Can you show what I did wrong?

Here is another solution
B rate 1/78 laps per sec A rate 1/91 laps per sec
(1/78 -1/91) = rate of closure laps per sec when B is behind A
after 1 lap B is 78/91 =0.857 laps behind A
(1/78 -1/91)t = 0.857
(0.00183)t =0.857 t=468.3sec add 78 for first lap Total elapsed time = 556.3sec

What say you?

9. ## Re: Running Around a Track

Hey BJ. Nice to "type" with you again!!

Originally Posted by bjhopper
My answer 555.7 sec.There is no typo error.
But you show this in your previous post:
"total elapsed time = 467.7 +78= 555.7 sec"

But 467.7 + 78 = 545.7, not 555.7

That's what I was trying to tell you.