A plane flies from Penthaven to Jackson and then back to Penthaven. When there is no wind, the round trip takes $3$ hours and $20$ minutes, but when there is a wind blowing from Penthaven to Jackson at $70$ miles per hour, the trip takes $3$ hours and $50$ minutes. How many miles is the distance from Penthaven to Jackson?
(Assume that the plane flies at a constant speed, and that the turnaround time is negligible.)
d = distance s = plane speed w = wind speed = 70
d/s = 1.667 hr ( distance/rate = time..... this is for one way) this shows s = d/1.667
with and without the wind
d/(s+w) + d/(s-w) = 3.833 hr re-arrange
d(s+w) + d(s-w) = 3.833 (s^2 - w^2) now sub in some values
d ( s+70) + d(s-70)
2 ds = 3.833 s^2 - 18781.7 Sub in the definition for s (from above)
2d ( d/1.667) = 3.833 d^2/ 2.778 - 18781.7 Continue to simplify and solve for 'd'
1.38 d^2 - 2d^2/1.667 - 18781.7 = 0
.1802 d^2 - 18781.7 = 0
d = 323 miles
d = distance s = plane speed w = wind speed = 70
d/s = 1.667 hr ( distance/rate = time..... this is for one way) this shows s = d/1.667
with and without the wind
d/(s+w) + d/(s-w) = 3.833 hr re-arrange
d(s+w) + d(s-w) = 3.833 (s^2 - w^2) now sub in some values
d ( s+70) + d(s-70)
2 ds = 3.833 s^2 - 18781.7 Sub in the definition for s (from above)
2d ( d/1.667) = 3.833 d^2/ 2.778 - 18781.7 Continue to simplify and solve for 'd'
1.38 d^2 - 2d^2/1.667 - 18781.7 = 0
.1802 d^2 - 18781.7 = 0
d = 323 miles