1. ## I need help please!!

Can I get some help with this problem? I don't know how to set it up to solve it. On the answer sheet it says the correct answer is 4 1/2 hours but I can't figure it out. Thanks

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?

2. Hi cnotez

Let:
Both the cyclist start at point O. After 3 hours, first cyclist reaches point A. Then after travelling for t hours, both of them meet at point B.

Using s = vt, you can find OA then you can set up two equations (when they meet) and solve for t

3. Originally Posted by cnotez
Can I get some help with this problem? I don't know how to set it up to solve it. On the answer sheet it says the correct answer is 4 1/2 hours but I can't figure it out. Thanks

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?
Asked and replied to here: http://www.mathhelpforum.com/math-he...-problems.html

4. If you break this question down, it really isn't that bad!

Firstly, we need to recognize that both cyclists will be at the same distance when they're equal with each other. Or, in other words, the distance traveled by Cyclist 1 will equal the distance traveled by Cyclist 2 when the second cyclist catches up (this is what we're attempting to find).

By using the distance formula:
$d=s*t$

Distance traveled by cyclist 2:
$d=10*t$
$=10t$

Distance traveled by cyclist 1:
$d=6*(t+3)$
$=6t+18$

The reason I've added a 3 in the bracket of the second equation is because Cyclist 1 has an additional three hours of riding time, relative to the second cyclist. This is outlined in the question:
The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist
As I earlier stated, the distance traveled by both cyclists must be equal when Cyclist 2 has caught up with Cyclist 1.

Therefore:
$d(of cyclist 1)=d(of cyclist 2)$
$6t+18=10t$

Solving simultaneously to find the time when this occurs:
$18=4t$
$t=18/4$
$t=4.5$

Hence, the time both cyclists are level is 4.5 hours.

If you struggled to follow this explanation, I'm more than happy to further explain the concepts involved

5. Originally Posted by The Second Solution