Problema Solution
Joe leaves school going 14mph on his bike.
Three hours later Sam leaves going 54mph.
How far does Joe travel before Sam catches up to him?
Answer provided by our tutors
let
t1 = the time of Joe's travel
v1 = 14 mph the speed of Joe
v2 = 54 mph the speed of Sam
t2 = the time of Sam's travel
Since Sam leaves the school 3 hours later we have
t2 = t1 - 3
the distance of Joe's travel = the distance of Sam's travel
t1*v1 = t2*v2
we are using the formula for speed = distance/time that is v = d / t => d = v*t
14*t1 = 54*t2
by solving the system of equations
t2 = t1 - 3
14*t1 = 54*t2
we find
t1 = 81/20 = 4.05 hours
t2 = 20/21 = 1.05 hours
we need to calculate the distance that Joe travels
t1*v1 = 4.05*14 = 56.7 miles
Joe will make 56.7 miles trip before Sam catches him up.