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 eho 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?
The faster cyclist "closes the gap" at a rate of 10 - 6 = 4 miles/hour. The initial lead of the slower cyclist is 3*6 = 18 miles.
Divide 18 miles by 4 miles/hour for the answer in hours
5566
To find out how much time will pass before the second cyclist catches up with the first, we need to determine the distance covered by both cyclists.
Let's assume that the time passed after the second cyclist started biking is 't' hours.
The distance traveled by the first cyclist is given by:
Distance = Speed × Time
Distance = 6 miles/hour × (t + 3) hours
Distance = 6(t + 3) miles
The distance traveled by the second cyclist is given by:
Distance = Speed × Time
Distance = 10 miles/hour × t hours
Distance = 10t miles
We want to find the time when the second cyclist catches up with the first. This means that the distance covered by both cyclists is equal:
6(t + 3) = 10t
Now we can solve for 't':
6t + 18 = 10t
18 = 10t - 6t
18 = 4t
t = 18/4
t = 4.5
Therefore, it will take 4.5 hours or 4 hours and 30 minutes for the second cyclist to catch up with the first cyclist from the time the second cyclist started biking.