An aeroplane flyalong a straight path AB and return back again, the distance between A and B is L. The aeroplane maintains the consrant speed V w.r.t wind, there is a steady wind with speed U and atangle ‘theta’ with line AB. Then calculate total time of the trip.

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.