A Tricky Word Problem, Assistance please?

Mia

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Mar 31, 2011
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So this is the problem that has got me and most of my peers stumped.

Johnny and Ernesto were participating in a 30 mile time trial on their bikes. Every 30 seconds a new rider would leave the finish line. At 9:05 am Ernesto starts riding at the speed of 440yards per minute. At 9:20am, Johnny starts riding but is going 20% faster than Ernesto. At what time will Johnny catch Ernest? How many miles will they each have ridden at that time?
I haven't figured out much. Could someone please help me and explain all of the steps it took to get that answer? Thank you!
 

tkhunny

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If you have figured out anything, you have not shared it with us.

1) Name Stuff. Write Down clear and useful definitions. I'll get you started.

Distance = Rate * Time

D = The Distance both must travel to be in the same place. Note: 0 miles < D < 30 miles
T = Time Johnny Travels
E = Time Ernesto Travels = T + 15 min
S = Johnny's Speed = 440 ypm
U = Ernesto's Speed = 1.2*S

The question asks for T, E, and D. Let's see what you get.
 

soroban

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Hello, Mia!

For these "catch-up" problems, I have a back-door approach.


Johnny and Ernesto were participating in a 30-mile time trial on their bikes. .Irrelevant!
Every 30 seconds a new rider would leave the finish line. .Is this necessary?

At 9:05 am Ernesto starts riding at the speed of 440 yards per minute.
At 9:20 am, Johnny starts riding but is going 20% faster than Ernesto.

At what time will Johnny catch Ernesto?
How many miles will they each have ridden at that time?

I must assume that they start at the same place.

\(\displaystyle \text{Ernesto has a 15 minute headstart.}\)
\(\displaystyle \text{In that time he has gone: }\,15\times 440 \:=\:6600\text{ yards.}\)

\(\displaystyle \text{Then Johnny starts at a speed which is: }20\% \times 440 \:=\:88\text{ yd/min faster.}\)

\(\displaystyle \text{Johnny is gaining on Ernesto at the rate of 88 yd/min.}\)

\(\displaystyle \text{It is }as\:i\!f\text{ Ernesto has } stopped\text{ and Johnny approaches him at 88 yd/min.}\)

\(\displaystyle \text{How long does it take Johnny to cover the 6600 yards?}\)

. . \(\displaystyle \text{It takes him: }\:\frac{6600}{88} \,=\,\boxed{75\text{ minutes}}\)


\(\displaystyle \text{Ernesto had already traveled 6600 yars.}\)
\(\displaystyle \text{In the next 75 minutes, he travels: }\:75\times 440 \,=\,33,\!000\text{ yards.}\)

\(\displaystyle \text{Therefore, Ernesto's total distance is: }\:6,\!600 + 33,\!000 \:=\:\boxed{39,600\text{ yards}}\)
. . \(\displaystyle \text{(And, of course, Johnny's total distance is the same.)}\)

 

Subhotosh Khan

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All done!!!
 

Denis

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Soroban's alternative is nice, but if the intent is for you to LEARN by using the
regualr formula speed = distance / time (learn to skate before playing hockey!), then:
Code:
Ernesto:.....@440.........[d]...............> x hours

Johnny:......@528.........[d]...............> x-1/4 hours
Since speed = distance / time, then distance [d] = speed * time; so:
d = 440x (Ernesto)
d = 528(x - 1/4) (Johnny) ; and you have:
440x = 528(x - 1/4)
Solve for x.
This is probably what your teacher expects; for YOU to do at least some work :idea:

I suggest you don't hand in Soroban's, but complete above instead; you'll have
egg on your face if teacher asks you to redo using regular formula!
 
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