Two Cyclists

**Caity** · March 10th 2011, 08:12 AM

This is a review problem

Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? I know the answer is 4.5 hours but need to figure an easy way to explain it...

**Quacky** · March 10th 2011, 08:24 AM

In three hours, the first cyclist will have travelled $18$ m.
The first is travelling at a speed of $6m/h$ . After $x$ hours, they will have travelled ' $6x$ ' metres.
The second is travelling at a speed of $10m/h$ . After $x$ hours, they will have travelled $10x$ metres.

Remember that the first has an $18$ metre headstart, and you can form the following equation:

$10x = 18 + 6x$

And you can quite simply solve for $x$ .

**Wilmer** · March 10th 2011, 08:24 AM

HINT:
distance = speed * time (hope you knew that!)

1: distance = 6(t)
2: distance = 10(t - 3)

**Caity** · March 10th 2011, 08:35 AM

Awesome thanks

Those will work...

**Soroban** · March 10th 2011, 02:10 PM

Hello, Caity!

I have a back-door approach to these "two cars" problems.

Two cyclists start biking from a trail's start 3 hours apart.
The second cyclist travels at 10 mph and starts 3 hours after the first cyclist
who is traveling at 6 mph.
How much time will pass before the second cyclist catches up with the first
from the time the second cyclist started biking?

I know the answer is 4.5 hours but need to figure an easy way to explain it.

Cyclist $\,A$ has a $\,3$ -hour headstart.
He is: $3 \times 6 \,=\,18$ miles ahead.

Then cyclist $\,B$ starts after him at $\,10$ mph.
His speed is: $10-6 \,=\,4$ mph faster than $\,A$ 's speed.

Imagine this . . .
Cyclist $\,A$ stops . . . and cyclist $\,B$ approaches him at $\,4$ mph.
How long will it take $\,B$ to make up the $\,18$ miles?

It will take: . $\dfrac{18}{4} \,=\,4.5$ hours.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This techniques with cars moving in opposite directions.

. . Two cars start from the same place and at the same time.
. . Car $\,A$ drives east at $\,45$ mph.
. . Car $\,B$ drives west at $\,35$ mph.
. . How long will it take for them to be $\,180$ miles apart?

Their combine speed is: $45 + 35\,=\,80$ mph.

It is as if car $\,A$ is stationary and car $\,B$ drives away at $\,80$ mph.
When will they be $\,180$ miles apart?

It will take: . $\dfrac{180}{80} \,=\,2.25$ hours.

. . See?

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