1. ## Two Cyclists

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...

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

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

$\displaystyle 10x = 18 + 6x$

And you can quite simply solve for $\displaystyle x$.

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

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

4. Awesome thanks Those will work...

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

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

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

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

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This techniques with cars moving in opposite directions.

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

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

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

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

. . See?