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?
First here's how to work it your head with no algebra!
When the second one starts, the first one is 18 miles down the trail, since he
has been going for 3 hours at 6 miles per hour. The second one's approach rate
is 10-6 or 4 miles per hour. So in 4 hours he'll have covered 4×4 or 16 of the
18 miles between them, so to go the remaining 2 miles between them, approaching
by 4 miles per hour, it will take him an additional half hour, so the answer
is 4 1/2 hours.
But your teacher will probably not accept that method, although you should be
able to think it out that way. Here's the way your teacher probably expects
you to do it.
Make this chart:
Distance Rate Time
First bicyclist
Second bicyclist
Let the answer be t hours. Then fill in t for the second
bicyclist's time.
Distance Rate Time
First bicyclist
Second bicyclist t
The first bicyclist travels 3 hours longer than the second,
so fill in his time as t+3
Distance Rate Time
First bicyclist t+3
Second bicyclist t
Fill in their rates as 6 and 10 respectively
Distance Rate Time
First bicyclist 6 t+3
Second bicyclist 10 t
Now use Distance = Rate × Time to fill in the
distances.
Distance Rate Time
First bicyclist 6(t+3) 6 t+3
Second bicyclist 10t 10 t
Now since they traveled the SAME distance, their
distances are equal, so the equation is:
6(t+3) = 10t
6t + 18 = 10t
6t - 10t = -18
-4t = -18
t = (-18)/(-4)
t = 18/4
t = 9/2
t = 4 1/2 hours.
Edwin
