Problema Solution
Two cyclists 90 miles apart, start riding towards each other at the same time. Once cycles twice as fast as the other. If they meet 2 hours later, at what average speed is each cyclist traveling?
Answer provided by our tutors
let
v1 = the speed of the first cyclist
v2 = the speed of the second cyclist
d = 90 miles the total distance
t = 2 hours the time of the travel
once cycles twice as fast as the other
v1 = 2*v2
wince avg.speed = distance/time => distance = time*avg.speed
they meet t=2 hours later
t*v1 + t*v2 = d
2*v1 + 2*v2 = 90
by solving the system of equations
v1 = 2*v2
2*v1 + 2*v2 = 90
we find
v1 = 30 mph
v2 = 15 mph
click here to see the step by step solution of the system of equations
the average speed of the first cyclist is 30 mph.
the average speed of the second cyclist is 15 mph.