Question

An aeroplane has to go along straight line from $A$ to $B$ and back along. The relative speed w.r.t. wind is $V$. The wind blows perpendicular to line $AB$ with speed $V_o$. The distance between $A$ & $B$ is $l$. The total time for the round trip is

• A. $\frac{2l}{\sqrt{V^2-V_0^2}}$
• B. $\frac{2Vl}{\sqrt{V^2-V_0^2}}$
• C. $\frac{2V_{0}l}{\sqrt{V^2-V_0^2}}$
• D. $\frac{2l}{\sqrt{V^2+V_0^2}}$

Answer 1

The correct option is A $\frac{2l}{\sqrt{V^2-V_0^2}}$


Direction of wind is perpendicular to the line $AB$ so if we have to reach at point $B$, we have to start moving at some angle (as shown in figure) due to drift because of the wind.
Here,
$V_{aw}=$ Speed of aeroplane w.r.t. wind = $V$
$V_{WG}=$ speed of wind w.r.t ground $=V_o$
$V_{aG}=$ Speed of aeroplane w.r.t. ground

$(V_{aG}{)}^2+(V_{WG}{)}^2=(V_{aw}{)}^2$
$\Rightarrow V_{aG}^2=V^2-V_0^2$
$\Rightarrow V_{aG}=\sqrt{V^2-V_o^2}$
$\therefore$ Time taken by aeroplane to reach from point $A$ to point $B$ is
$t_1=\frac{l}{V_{aG}}=\frac{l}{\sqrt{V^2-V_0^2}}...\left(1\right)$

Again, for point $B$ to point $A$ (return)

$V_{aw}^2=V_{WG}^2+V_{aG}^2$
$V_{aG}=\sqrt{V^2-V_0^2}$
So, time taken to reach from point $B$ to point $A$ is
$t_2=\frac{l}{V_{aG}}=\frac{l}{\sqrt{V^2-V_0^2}}$

Thus, time taken by aeroplane for the overall trip is (from $A$ to $B$ then from $B$ to $A$)
$t=t_1+t_2=\frac{l}{\sqrt{V^2-V_0^2}}+\frac{l}{\sqrt{V^2-V_0^2}}$
$\Rightarrow t=\frac{2l}{\sqrt{V^2-V_0^2}}$