Please help me solve this!

[missellison2007]

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?

 

[galactus]

The second cyclist will catch up to the first when their distances are the same.

Remember, d=rt

The first travels a distance of 6t.

The second starts 3 hours later and travels 10(t-3)

Equate and solve for t.

 

[soroban]

Hello, missellison2007!

Here's a back-door approach to this problem . . .

Two cyclists start biking from a trail's start 3 hours apart.
The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph.
How much time will pass before the 2nd cyclist catches up with the 1st
from the time the 2nd cyclist started biking?

Cyclist \(\displaystyle A\) travels at 6 mph and has a 3-hour headstart.
. . He is 18 miles ahead of cyclist \(\displaystyle B\).
How long does it take cyclist \(\displaystyle B\) to catch up?

Cyclist \(\displaystyle B\) travels \(\displaystyle 10 - 6 \:=\:4\) mph faster than cyclist \(\displaystyle A.\)

Think . . . It is as if cyclist \(\displaystyle A\) has stopped and cyclist \(\displaystyle B\) approaches him at 4 mph.


\(\displaystyle \text{To cover 18 miles, it take cyclist }B\!:\;\text{18 miles} \div \text{4 mph} \:=\:\boxed{4\tfrac{1}{2}\text{ hours}}\)