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An aeroplane fly along 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 at angle ‘theta’ with line AB. Then calculate total time of the trip.
Hiii Speed of wind = UHorizontal component of U will oppse the aeroplane in travelling from A to B.Speed while going from A to B = V – U cosQSpeed while coming from Bto A = V + U cosQTotal 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 is1/ V +vAnd the time in the upwind half is1/ V-vSo, 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^2Combining like terms,2 V / V^2 -v^2If V = 1, and v = .1, when AB = 1,2 / 1 - .012 / .99Total time is 2.02 unitsTo put this into a real-world scenario:Instead of AB being 1,AB= 100 milesV=100 mphv=10 mphDownwind time: 100/110Upwind time: 100/90Total time: (100/110) + (100/90)(9000/9900) + (11000/9900)20,000/9900Total time is 2.02 hoursThis is not the 2.00 hours found by doubling the distance and assuming the winds cancel out.
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 is1/ V +vAnd the time in the upwind half is1/ V-vSo, 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^2Combining like terms,2 V / V^2 -v^2If V = 1, and v = .1, when AB = 1,2 / 1 - .012 / .99Total time is 2.02 unitsTo put this into a real-world scenario:Instead of AB being 1,AB= 100 milesV=100 mphv=10 mphDownwind time: 100/110Upwind time: 100/90Total time: (100/110) + (100/90)(9000/9900) + (11000/9900)20,000/9900Total time is 2.02 hoursThis is not the 2.00 hours found by doubling the distance and assuming the winds cancel out. Hope this helps you regards
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