Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour

Question

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?

pls. answer this with a complete solution. tnx.

Answer 1

The formula for distance is rate times time: d = r * t

For the second biker to catch up to the first, their distances must be equal::

In equation form:

10 * t = 6 * (t + 3)  

Since the first biker has been riding for three hours longer that value must be added to 't' on the right side.

Multiply out the right side of the equation:

10 * t = 6 * t + 18

Subtract 6 * t from both sides:

4 * t = 18

Divide both sides by 4:

t = 4.5 hours

Check:

First rider: d = 6 * (4.5 + 3) = 45 miles

Second rider: d = 10 * 4.5 = 45 miles

Answer 2

The two displacements for the cyclists are:

s1 = 6(t+3)

s2 = 10t

Where t is the time in hours, what you want to find is:

10t = 6(t+3)

So solve for t:

10t = 6t + 18

4t = 18

t = 4.5 hours