Question

An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

Answer 1

Explanation:

Step 1:

Given :

$\mathrm{u}=2 \mathrm{Om} / \mathrm{s}$

$v=150 \mathrm{m} / \mathrm{s}$

Wind travels at a Speed = $u$

Aeroplane travels at a Speed  = $V$

To arrive at the point $B$, The pilot should go along $\mathrm{AB}$.  

Step 2:

Let us expect that $\theta$ be the point at which the plane heads towards east of line $\mathrm{AB}$.  

so point between two speeds is $\theta+30^{\circ}$

$\tan \theta=\frac{u \sin (\theta+30)}{v+u \cos (\theta+30)}$

$\frac{\sin \theta}{\cos \theta}=\frac{u \sin (\theta+30)}{v+u \cos (\theta+30)}$

By cross multiplying we get

$\sin \theta[v+u \cos (\theta+30)]=\cos \theta u \sin (\theta+30)$

$\left.\sin \theta^{\circ}\left[v+u \cos \theta \cos 30^{\circ}-u \sin \theta \sin 30\right)\right]=u \cos \theta^{\circ} \sin \theta^{\circ} \cos 30^{\circ}+u \cos \theta^{\circ} \cos \theta^{\circ} \sin 30$

As we know that,

$\text { since } \cos (A+B)=\cos A \cos B-\sin A \sin B$

$\sin (A+B)=\sin A \cos B+\cos A \sin B$

$v \sin \theta^{\circ} \mathrm{u} \cos \theta^{\circ} \sin \theta^{\circ} \frac{\sqrt{3}}{2}-\mathrm{u} \sin ^{2} \theta \frac{1}{2}=u \cos \theta \sin \theta \frac{\sqrt{3}}{2}+u \cos ^{2} \theta \frac{1}{2}$

Step 3:

on solving the above equation,

we get

$v \sin \theta^{\circ}-\frac{u \sin ^{2} \theta}{2}=\frac{u \cos ^{2} \theta}{2}$

$v \sin \theta^{\circ}=\frac{\mathrm{u}}{2}\left[\cos ^{2} \theta+\sin ^{2} \theta\right]$

$\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$v \sin \theta^{\circ}=\frac{\mathrm{u}}{2}$

$v \sin \theta^{\circ}=\frac{\mathbf{u}}{2 \mathrm{v}}$

$v \sin \theta^{\circ}=\frac{20}{2 \times 150}$

$v \sin \theta^{\circ}=\frac{20}{300}$

$v \sin \theta^{\circ}=\frac{1}{15}$

$\theta^{\circ}=\sin ^{-1} \frac{1}{15}$

$\theta^{\circ}=\sin ^{-1} 0.066$

∴ so as to arrive at point $B$, the plane should move with an edge of $\sin ^{-1}[1 / 15]$east of line $A B$.