Problema Solution
Airplane a travel 6300 miles at a certain speed, plane b travels 4500 miles at 50 mph faster than plane a in three hr less time. find speed of both planes.
Answer provided by our tutors
let
v1 = the speed of airplane a
d1 = 6300 miles the distance that airplane a traveled
t1 = the time that plane a traveled
v2 = the speed of plane b
d2 = 4500 miles the distance that plane b traveled
t2 = the time that plane b traveled
plane b travels 4500 miles at 50 mph faster than plane a
v2 = v1 + 50
in 3 hr less time
t2 = t1 - 3
since speed = distance/time => time = distance/speed
t1 = d1/v1 = 6300/v1
t2 = d2/v2 = 4500/(v1 + 50)
if we plug these equations in t2 = t1 - 3 we get
4500/(v1 + 50) = 6300/v1 - 3 multiply both sides be v1(v1 + 50)
4500v1 = 6300(v1 + 50) -3v1(v1 + 50)
6300(v1 + 50) -3v1(v1 + 50) = 4500
by solving the equation we find and consider the positive roots since v1>0
v1 = 2099.3 mph
v2 = v1 + 50 = 2149.3 mph
click here to see the step by step solution of the equations
the speed of plane a is 2099.3 mph.
the speed of plane b is 2149.3 mph.