how do u come up w/the answer?

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?

Math assistance needed.

The distance they travel is the same...

dfirst=6*time
dsecond=10*(time-3)

set the two equations equal, solve for time.

To come up with the answer, you can use the concept of relative velocity.

Let's break down the problem. The second cyclist is traveling at a speed of 10 miles per hour, while the first cyclist is traveling at a speed of 6 miles per hour. We need to determine how much time it will take for the second cyclist to catch up with the first cyclist.

Let's assume that the time it takes for the second cyclist to catch up with the first cyclist is 't' hours.

Now, let's consider the distance traveled by each cyclist during this time ‘t’.

The first cyclist has been biking for a total of (t + 3) hours (since the second cyclist started 3 hours later). Therefore, the distance traveled by the first cyclist is 6 * (t + 3) miles.

The second cyclist has been biking for 't' hours. Therefore, the distance traveled by the second cyclist is 10 * t miles.

Since the second cyclist catches up with the first cyclist, the distances traveled by both cyclists must be the same:

6 * (t + 3) = 10 * t

Now, we can solve this equation to find 't':

6t + 18 = 10t

Subtracting 6t from both sides:

18 = 10t - 6t

Simplifying:

18 = 4t

Dividing both sides by 4:

t = 18/4

Therefore, t = 4.5 hours.

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