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?
( This question is very hard, please help me, please help with the working. thanks )
OR
let the time taken from when the second rider leaves till they meet be t hours
10t = 6t + 18 , (their distances are the same)
4t = 18
t = 4.5
When the second cyclist starts, they are 18 miles apart. (3 hours times the speed of the first biker).
After that, the distance between the bikers shrinks at a rate of 10 - 6 = 4 miles/hr.
The distance between bikers becomes zero after 18/4 = 4.5 hours. That is when the late-starting biker [passes the other.
thanks a lot !!
To solve this problem, we can use the formula:
Time = Distance / Speed
Let's assume that the time passed when the second cyclist catches up with the first cyclist is represented by "t" (in hours).
By the time the second cyclist starts biking, the first cyclist has already been biking for 3 hours. Therefore, the first cyclist has already covered a distance of:
Distance1 = Speed1 * Time1
= 6 * 3
= 18 miles
Now, we need to find the time it takes for the second cyclist to cover the same distance as the first cyclist.
Distance2 = Speed2 * Time2
Distance2 = 10 * t
We know that Distance1 = Distance2, so we can equate the two equations:
18 = 10 * t
To solve for t, we can divide both sides by 10:
t = 18 / 10
t = 1.8 hours
Therefore, it will take approximately 1.8 hours (or 1 hour and 48 minutes) for the second cyclist to catch up with the first cyclist from the time when the second cyclist started biking.