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?

they both go distance d

first one travels time = t+3
second one travels time = (t)
d = 6 (t+3) = 10(t)
6 t + 18 = 10 t
4 t = 18
t = 9/2 = 4.5 hr

To solve this problem, we can set up an equation by considering the distance traveled by both cyclists.

Let's assume that the time passed when the second cyclist catches up with the first is 't' hours after the second cyclist starts biking.

Since the first cyclist traveled for 3 hours before the second cyclist started, the distance traveled by the first cyclist would be (3 + t) * 6 miles (since the first cyclist is traveling at a rate of 6 miles per hour).

The distance traveled by the second cyclist would be t * 10 miles (since the second cyclist is traveling at a rate of 10 miles per hour).

To find the time when the second cyclist catches up with the first, we can set up the equation:

(3 + t) * 6 = t * 10

Now, let's solve for 't':

18 + 6t = 10t

18 = 10t - 6t

18 = 4t

t = 18 / 4

t = 4.5

Therefore, it will take 4.5 hours for the second cyclist to catch up with the first from the time the second cyclist started biking.