Solving a distance, rate, time problem

Question

A long distance runner starts at the beginning of a trail and runs at a rate of
4 miles per hour. Three hours later, a cyclist starts at the beginning of the trail and travels at a rate of
19 miles per hour. What is the amount of time that the cyclist travels before overtaking the runner?

Do not do any rounding.

Answer 1

Speed = Distance/Time or s=d/t. We can solve this equation for distance, d:

s=d/t

st = d

The runner's speed is 4mi/hr, so the distance he travels after 't' hours is given by this equation:

4t = d

The runner starts 3 hours before the cyclist, so his total traveling time will be t+3 (3 hours more than the cyclist's):

4(t+3) = d

The cyclist's speed is 19mi/hr, and the distance he travels after 't' hours is given by this equation:

19t = d

 

Now we need to find at what time, 't' are the two distances, 'd' equal. We can replace the d in the first equation with 19t from the second equation and solve:

 

4(t+3) = 19t

4t+12 = 19t

12 = 15t

12/15 = t

4/5 = t

 

After traveling for 4/5 hr, the cyclist will overtake the runner. We can also write this as 0.8 hours, or multiply by 60 to get the minutes:

0.8hr (60min/hr) = 48 minutes