A Tricky Word Problem, Assistance please?

Mia

New member
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

Moderator
Staff member
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

Elite Member
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.

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

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

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

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

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

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

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

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

Staff member
All done!!!

Denis

Senior Member
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!