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Grade: 11
        
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.
2 months ago

Answers : (3)

Vikas TU
8478 Points
							
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.
2 months ago
Khimraj
3008 Points
							

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.

one month ago
aswanth nayak
66 Points
							
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
one month ago
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