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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.)

 Jan 9, 2024

Best Answer 

 #1
avatar+36916 
+1

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 

 Jan 9, 2024
 #1
avatar+36916 
+1
Best Answer

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 

ElectricPavlov Jan 9, 2024

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