Question

An aircraft flies at a speed of $800\text{km/h}$ in still air. The wind is blowing at $400\sqrt{2}\text{km/h}$ from south direction and the pilot wishes to travel from point $A$ to point $B$, North-East of $A$. Find the direction with reference to the North, the pilot must steer and time ($\text{hr}$) of his journey if $AB=1200\text{km}$. (Take $\cos {15}^{\circ }=0.96$)

• A. ${30}^{\circ },2.0\text{hr}$
• B. ${75}^{\circ },\frac{25}{32}\sqrt{2}\text{hr}$
  
• C. ${45}^{\circ },\frac{25}{27}\sqrt{2}\text{hr}$
  
• D. ${45}^{\circ },\frac{15}{32}\sqrt{3}\text{hr}$
  

Answer 1

The correct option is B ${75}^{\circ },\frac{25}{32}\sqrt{2}\text{hr}$


Speed of wind w.r.t ground,
$v_w=400\sqrt{2}\text{km/h}$
Speed of aircraft w.r.t wind,
$v_{aw}=800\text{km/h}$
$\Rightarrow$Velocity of aircraft w.r.t ground ${\overrightarrow{v}}_a$ should be along $AB$ or in north-east direction. Thus,${\overrightarrow{v}}_a$ should be the resultant vector of ${\overrightarrow{v}}_w$ and ${\overrightarrow{v}}_{aw}$, which should come along $AB$ or north-east direction.

${\overrightarrow{v}}_a={\overrightarrow{v}}_{aw}+{\overrightarrow{v}}_w$


Let ${\overrightarrow{v}}_{aw}$ makes an angle $\alpha$ with $AB$ as shown in figure.

Applying sine law in triangle $ABC$, we get:

$\frac{AC}{\sin {45}^{\circ }}=\frac{BC}{\sin \alpha }$

$\sin \alpha =\left(\frac{BC}{AC}\right)\times \sin {45}^{\circ }=\left(\frac{400\sqrt{2}}{800}\right)\frac{1}{\sqrt{2}}=\frac{1}{2}$
$\therefore \alpha ={30}^{\circ }$
Therefore, pilot should steer in a direction at an angle of $(45+\alpha )$ as ${75}^{\circ }$ from north towards east.
Again using sine law in triangle;
$\frac{AB}{\sin ({180}^{\circ }-{45}^{\circ }-{30}^{\circ })}=\frac{AC}{\sin {45}^{\circ }}$

$\Rightarrow \frac{{\overrightarrow{v}}_a}{\sin ({180}^{\circ }-{45}^{\circ }-{30}^{\circ })}=\frac{800}{\sin {45}^{\circ }}$

$\Rightarrow {\overrightarrow{v}}_a=\frac{800}{\sin {45}^{\circ }}\times \sin {105}^{\circ }\left(\text{km/h}\right)$
$\because \sin {105}^{\circ }=\sin (90+15{)}^{\circ }=\cos {15}^{\circ }$

$(\overrightarrow{v}{)}_a=\frac{\cos {15}^{\circ }}{\sin {45}^{\circ }}\times 800\text{km/h}$

$\therefore (\overrightarrow{v}{)}_a=\frac{0.96}{\frac{1}{\sqrt{2}}}\times 800\text{km/h}$

${\overrightarrow{v}}_a=768\sqrt{2}\text{km/h}$

$\therefore$ The time of journey from $A$ to $B$ is:

$t=\frac{AB}{\left({\overrightarrow{v}}_a\right)}=\frac{1200}{768\sqrt{2}}=\frac{25}{32}\sqrt{2}\text{hr}$