
Find the time an aeroplane having velocity $ v $ takes to fly around a square of side $ 'a' $ if the wind is blowing at a velocity $ u $ along one side of the square.
$ \left( A \right)\dfrac{{2a}}{{{v^2} - {u^2}}}\left[ {v + \sqrt {{v^2} - {u^2}} } \right] \\
\left( B \right)\dfrac{a}{{{v^2} - {u^2}}}\left[ {v - \sqrt {{v^2} - {u^2}} } \right] \\
\left( C \right)\dfrac{{2a}}{{\sqrt {{v^2} - {u^2}} }} \\
\left( D \right)\dfrac{a}{{\sqrt {{v^2} - {u^2}} }} \\ $
Answer
539.4k+ views
Hint: In order to solve this question, we are going to find the different speeds of the aeroplane through the different sides of the square. Then, by taking the distance and the speed, we can easily calculate the time taken for the aeroplane to fly around the four sides of the square and they are added to find total time.
If $ v $ is the speed of the aeroplane along the side $ 'a' $ of the square, and $ u $ is the speed of the wind blowing along one side of the square, then, the relative speed is:
$ {v_A} = v + u $
The time taken to cover a distance $ 'a' $ with velocity $ v + u $ is
$ {t_{AB}} = \dfrac{a}{{v + u}} $
Complete step by step solution:
Let the velocity of aeroplane while flying through $ AB $ be $ {v_A} $ , then,
$ {v_A} = v + u $
Time taken for it is,
$ {t_{AB}} = \dfrac{a}{{v + u}} $
Now, the velocity of the aeroplane while flying through $ BC $ , be
$ {v_A} = \sqrt {{v^2} - {u^2}} $
The time taken for this trajectory is
$ {t_{BC}} = \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} $
Velocity of aeroplane while flying through $ CD $ ,
Velocity of aeroplane while flying through $ DA $ ,
$ {v_A} = \sqrt {{v^2} - {u^2}} $
The time taken for this trajectory is
$ {t_{DA}} = \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} $
Thus, the total time taken for the aeroplane to complete the trajectory around this square, is equal to the sum of the time taken for all the four sides,
$ \dfrac{a}{{v + u}} + \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} + \dfrac{a}{{v - u}} + \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} $
On solving this expression, we get,
$ \dfrac{{2a}}{{{v^2} - {u^2}}}\left( {v + \sqrt {{v^2} - {u^2}} } \right) $
Hence, option $ \left( A \right)\dfrac{{2a}}{{{v^2} - {u^2}}}\left[ {v + \sqrt {{v^2} - {u^2}} } \right] $ is the correct answer.
Note:
It is important that the air resistance is a very significant factor for the trajectories in the air of any object, even if it is as big as an aeroplane. When the wind is in the direction of motion of the aeroplane, it just adds up to the original speed of the aeroplane. The relative speeds along the different sides are taken very carefully.
If $ v $ is the speed of the aeroplane along the side $ 'a' $ of the square, and $ u $ is the speed of the wind blowing along one side of the square, then, the relative speed is:
$ {v_A} = v + u $
The time taken to cover a distance $ 'a' $ with velocity $ v + u $ is
$ {t_{AB}} = \dfrac{a}{{v + u}} $
Complete step by step solution:
Let the velocity of aeroplane while flying through $ AB $ be $ {v_A} $ , then,
$ {v_A} = v + u $
Time taken for it is,
$ {t_{AB}} = \dfrac{a}{{v + u}} $
Now, the velocity of the aeroplane while flying through $ BC $ , be
$ {v_A} = \sqrt {{v^2} - {u^2}} $
The time taken for this trajectory is
$ {t_{BC}} = \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} $
Velocity of aeroplane while flying through $ CD $ ,
Velocity of aeroplane while flying through $ DA $ ,
$ {v_A} = \sqrt {{v^2} - {u^2}} $
The time taken for this trajectory is
$ {t_{DA}} = \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} $
Thus, the total time taken for the aeroplane to complete the trajectory around this square, is equal to the sum of the time taken for all the four sides,
$ \dfrac{a}{{v + u}} + \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} + \dfrac{a}{{v - u}} + \dfrac{a}{{\sqrt {{v^2} - {u^2}} }} $
On solving this expression, we get,
$ \dfrac{{2a}}{{{v^2} - {u^2}}}\left( {v + \sqrt {{v^2} - {u^2}} } \right) $
Hence, option $ \left( A \right)\dfrac{{2a}}{{{v^2} - {u^2}}}\left[ {v + \sqrt {{v^2} - {u^2}} } \right] $ is the correct answer.
Note:
It is important that the air resistance is a very significant factor for the trajectories in the air of any object, even if it is as big as an aeroplane. When the wind is in the direction of motion of the aeroplane, it just adds up to the original speed of the aeroplane. The relative speeds along the different sides are taken very carefully.
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