Question

An aeroplane flies along a straight line from A to B air speed V and back again with the same air speed. If the distance between A and B is l and a steady wind blows perpendicular to AB with speed u,the total time taken for round trip is
A) \[2l/\sqrt {{V^2} - {u^{^2}}} \]
B) $2l/\sqrt {{V^2} + {u^2}} $
C) $l/\sqrt {{V^2} - {u^2}} $
D) $3l/\sqrt {{V^2} - {u^2}} $

Answer 1

Hint: In order to solve this first we have to resolve the velocity vector into its components to balance air’s speed. Now the velocity in the direction of movement of aeroplane will be the second velocity vector. Now the distance and speed is known so time can be found. Similarly a second time will also be found and hence by simply adding them we can get total time.


Complete step by step answer:
According to the question:
The velocity of aeroplane is V so,
The components of velocity vector V will be $V\cos \theta $and $V\sin \theta $
$V\sin \theta $will be a vertical component while $V\cos \theta $will be a horizontal component.
Since it is given that u(air speed) is perpendicular to AB so $V\sin \theta $will balance u

$V\sin \theta = u$
On further solving,
$\sin \theta = u/V$
Since ,
$\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $
So,
$\cos \theta = \sqrt {{V^2} - {u^2}} /V$
Since the horizontal component of velocity is ,$V\cos \theta $
So the total time taken from A to B will be:
${T_{A \to B}} = l/V\cos \theta $ ……..(1)
Putting value of $\cos \theta $in equation (1) we get
${T_{A \to B}} = l/\sqrt {{V^2} - {u^2}} $
Since time from A to B is time independent so time from B to A will also be
${T_{B \to A}} = l/\sqrt {{V^2} - {u^2}} $
So the total time taken will be,
${T_{Total}} = {T_{A \to B}} + {T_{B \to A}}$
${T_{Total}} = 2l/\sqrt {{V^2} - {u^2}} $
So the answer will be option A

Note: While solving these types of problems we should have the knowledge of the basic trigonometry and the knowledge of little kinematics. Also time for a round trip is asked hence we should multiply time for one way trip by two.
An alternative method is that we can also use the triangle rule for finding $\cos \theta $ from $\sin \theta $ .