Question

An airplane has to go from a point $A$ to another point $B$, $500\text{km}$ away due ${30}^{\circ }$ East of North. Wind is blowing due North at speed of $20\text{m/s}$. The speed of the airplane is $150\text{m/s}$. Find the direction in which the pilot should head the plane so as to reach the point $B$. Also find the time taken by the plane to go from $A$ to $B$.

• A. $\left({\sin }^{-1}\frac{1}{15}\right)$ East of North, $50\text{min}$
• B. $({30}^{\circ }+{\sin }^{-1}\frac{1}{15})$ East of North, $50\text{min}$
• C. $({60}^{\circ }+{\sin }^{-1}\frac{1}{15})$ East of North, $50\text{min}$
• D. $({30}^{\circ }+{\sin }^{-1}\frac{1}{15})$ East of North, $60\text{min}$

Answer 1

The correct option is B $({30}^{\circ }+{\sin }^{-1}\frac{1}{15})$ East of North, $50\text{min}$


Let $V_w$ and $V_a$ be the velocities of wind and airplane respectively.
If the airplane wants to travel along $AB$ , the component of velocities of wind and airplane perpendicular to line $AB$ must be equal.
$\Rightarrow V_w\cos {60}^{\circ }=V_a\sin \alpha$ (From the figure)
$\Rightarrow \alpha ={\sin }^{-1}\frac{1}{15}$

Therefore, the plane must head $({30}^{\circ }+{\sin }^{-1}\frac{1}{15})$ East of North.

Velocity along line $AB$ $v=V_a\cos \alpha +V_w\sin {60}^{\circ }$
$\Rightarrow v=150\frac{\sqrt{224}}{15}+\frac{20\sqrt{3}}{2}$
$\Rightarrow v=40\sqrt{14}+10\sqrt{3}=167m/s$
Hence, $t=\frac{AB}{167}=\frac{500\times 1000}{167}s=2994s\approx 50\text{min}$