Roger can run one mile in 14minutes. jeff can run one mile in 12minutes. if jeff gives roger a 1minute head start, how long will it take before jeff catches up to roger? how far will each have run?

v1 = 1/14 miles per min the average speed of Roger

v2 = 1/12 miles per min the average speed of Jeff

t = the time in minutes that Jeff runs before catches up to Roger

t + 1 = the time in minutes that Roger runs before Jess catches up with him

since the both run the same distance and distance = speed*time we have

v1 * (t + 1) = v2*t

(1/14) (t + 1) = (1/12) * t

by solving we find

t = 6 min

the distance each of them runs is v2*t = (1/12) * 6 = 1/2 mile