Question

An aircraft flies at 400 km/h in still air. A wind of $200\sqrt{2}$ km/h is blowing from the south. The pilot wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if AB=1000 km.

Answer 1

Let the time taken by the aircraft to reach point B be $t$

From figure $AM=ABcos{45}^o=1000\times \frac{1}{\sqrt{2}}=500\sqrt{2}$ m

Also $AP=ABsin{45}^o=1000\times \frac{1}{\sqrt{2}}=500\sqrt{2}$ m

S-N direction : $AM=[V_w+V_{sw}sin(45-\alpha \left)\right]t$

$\therefore$ $500\sqrt{2}=[V_w+V_{sw}sin(45-\alpha \left)\right]t$ .............(1)

W-E direction : $AP=\left[V_{sw}cos\right(45-\alpha \left)\right]t$

$\therefore$ $500\sqrt{2}=\left[V_{sw}cos\right(45-\alpha \left)\right]t$ .............(2)

Equating (1) & (2), $V_{sw}cos(45-\alpha )=V_{sw}sin(45-\alpha )+V_w$

$\therefore$ $400\left[cos\right(45-\alpha )-sin(45-\alpha \left)\right]=200\sqrt{2}$

OR $\frac{1}{\sqrt{2}}cos(45-\alpha )-\frac{1}{\sqrt{2}}sin(45-\alpha )=\frac{1}{2}$

OR $sin\left(45\right)cos(45-\alpha )-cos\left(45\right)sin(45-\alpha )=\frac{1}{2}$

$\therefore$ $sin\left(\alpha \right)=\frac{1}{2}$ $⟹\alpha ={30}^o$

From (2), $500\sqrt{2}=400cos\left(15\right)t$

$\therefore$ $500\sqrt{2}=400\times 0.966\times t$ $⟹t=1.83$ hr