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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
Dear Student, 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)) Regards
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