Problema Solution
a plane traveled 580 miles to jacksonville and back. the trip there was with the wind. it took 5 hours. the trip back was into the wind. the trip back took 10 hours. find the speed of the plane im still air and the speed of the wind
Answer provided by our tutors
let
p = the plane's speed in still air
w = the speed of the wind
d = 580 miles to jacksonville and back
t1 = 10 hours into the wind
t2 = 5 hour with the wind
the speed of the plane traveling against the wind is: p - w
the speed of the plane traveling with the wind is: p + w
since speed = distance/time => distance = speed*time
traveling into the wind
d/2 = (p - w)t1
(p - w)10 = 580/2
(p - w)10 = 290 divide both sides by 10
p - w = 29
traveling against the wind
d/2 = (p + w)t2
(p + w)t2 = d/2
(p + w)5 = 580/2
(p + w)5 = 290 divide both sides by 5
p + w = 58
by solving the system of equations
p - w = 29
p + w = 58
we find
p = 43.5 mph
w = 14.5 mph
click here to see the step by step solution of the system of equations
the speed of the plane in still air is 43.5 mph.
the speed of the wind is 14.5 mph.
let
p = the plane's speed in still air
w = the speed of the wind
d = 580 miles to jacksonville and back
t1 = 10 hours into the wind
t2 = 5 hour with the wind
the speed of the plane traveling against the wind is: p - w
the speed of the plane traveling with the wind is: p + w
since speed = distance/time => distance = speed*time
traveling against the wind
d/2 = (p - w)t1
(p - w)10 = 580/2
(p - w)10 = 290 divide both sides by 10
p - w = 29
traveling with the wind
d/2 = (p + w)t2
(p + w)t2 = d/2
(p + w)5 = 580/2
(p + w)5 = 290 divide both sides by 5
p + w = 58
by solving the system of equations
p - w = 29
p + w = 58
we find
p = 43.5 mph
w = 14.5 mph
click here to see the step by step solution of the system of equations
the speed of the plane in still air is 43.5 mph.
the speed of the wind is 14.5 mph.