Question 226005
Torque left home on his bicycle at 9:00 A.M., traveling at an average rate of 12 mph.
 At noon, Torque's brother set out after him on a motorcycle, following the same route, averaging 39 mph.
 How long had Torque been riding when his brother caught up?
:
Let t = time Torque had been riding
then the brother starts, 3hrs later, therefore
(t-3) = travel time of the brother
:
A good fact to remember in these "catch up" problems, is when this occurs,
they both will have traveled the same distance
:
write a distance equation from this fact; (dist = speed * time)
:
bro dist = T's dist
39(t - 3) = 12t
39t - 117 = 12t
39t - 12t = +117
27t = 117
t = {{{117/27}}}
t = 4{{{1/3}}} hrs (4 hr 20 min) is the riding time of T
;
:
Check the solution by finding the dist traveled by each
(bro traveled 4.33 - 3 = 1.33 hr)
4.33 * 12 ~ 52 mi
1.33 * 39 ~ 52 mi