Hiii 
Speed of wind = U
Horizontal component of U will oppse the aeroplane in travelling from A to B.
Speed while going from A to B = V – U cosQ
Speed while coming from Bto A = V + U cosQ
Total time = L(1/(V – U cosQ) + 1/V + U cosQ))
Hope this helps.
						

							
Does “AB=1” mean “the distance from A to B is 1?”
If so, the time (distance divided by speed) in the downwind half is
1/ V +v
And the time in the upwind half is
1/ V-v
So, the total time is (1/V+v) + (1/V-v)
Working toward a common denominator,
{(V-v) + (V+v)} / (V-v) (V+v)
Then,
2 V / V^2 + V v - V v -v^2
Combining like terms,
2 V / V^2 -v^2
If V = 1, and v = .1, when AB = 1,
2 / 1 - .01
2 / .99
Total time is 2.02 units
To put this into a real-world scenario:
Instead of AB being 1,
AB= 100 miles
V=100 mph
v=10 mph
Downwind time: 100/110
Upwind time: 100/90
Total time: (100/110) + (100/90)
(9000/9900) + (11000/9900)
20,000/9900
Total time is 2.02 hours
This is not the 2.00 hours found by doubling the distance and assuming the winds cancel out.
						

							
Dear Student,
 
 
 
Does “AB=1” mean “the distance from A to B is 1?”
If so, the time (distance divided by speed) in the downwind half is
1/ V +v
And the time in the upwind half is
1/ V-v
So, the total time is (1/V+v) + (1/V-v)
Working toward a common denominator,
{(V-v) + (V+v)} / (V-v) (V+v)
Then,
2 V / V^2 + V v - V v -v^2
Combining like terms,
2 V / V^2 -v^2
If V = 1, and v = .1, when AB = 1,
2 / 1 - .01
2 / .99
Total time is 2.02 units
To put this into a real-world scenario:
Instead of AB being 1,
AB= 100 miles
V=100 mph
v=10 mph
Downwind time: 100/110
Upwind time: 100/90
Total time: (100/110) + (100/90)
(9000/9900) + (11000/9900)
20,000/9900
Total time is 2.02 hours
This is not the 2.00 hours found by doubling the distance and assuming the winds cancel out.
 
 
Hope this helps you
 
 
regards