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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.

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.

Grade:11

4 Answers

Vikas TU
14149 Points
4 years ago
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.
Khimraj
3007 Points
4 years ago

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.

aswanth nayak
100 Points
4 years ago
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
aswanth nayak
100 Points
4 years ago
Dear Student,
 
 
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))
 
Regards

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