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?
To find the time it takes for the second cyclist to catch up with the first, we need to calculate the distance each cyclist travels and set their distances equal. Let 't' be the time in hours that the second cyclist takes to catch up after starting. The first cyclist travels at 6 miles per hour and has a 3-hour head start. Therefore, in those 3 hours, the first cyclist travels 18 miles (6 miles per hour * 3 hours). From the time the second cyclist starts, the first cyclist continues biking at 6 miles per hour, so in 't' hours, the first cyclist will travel an additional 6t miles. The second cyclist travels at 10 miles per hour, so in 't' hours, the second cyclist will travel 10t miles. Setting the distances equal: 18 miles + 6t miles = 10t miles. Solving for 't', we get: 18 = 4t; t = 4.5 hours. Therefore, the second cyclist will catch up with the first cyclist 4.5 hours after starting.
The answer is 18/4=4.5