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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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An aircraf flies a $400 km/hr$ in still air. A wind of $200 \sqrt 2 km/hr$ is blowing from south. A pilot wishes to travel from A to a point B north east of A .Find the direction he must steer and time of journey if $AB=1000km$

$ a)\;45^{\circ}\; from\; north , 2.61 hrs \\ b)\;60^{\circ}\; from\; north , 1.50\; hrs \\ c)\; 80^{\circ}\; from\; north , 2.15 hrs \\ d)\;75^{\circ}\; from \;north , 1.83 hrs $

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Let $v_w$=Velocity of wind
The aircraft should fly along $AC$ with a velocity of $v_{aw}=400 km/hr$ so that the resultant is along $AB$
Let $v_{aw}$ makes angle $\alpha$ with $AB$
$\large\frac{AC}{\sin 45}=\frac{BC}{\sin \alpha}$
$\sin \alpha=\large\frac{BC}{AC} $$\sin 45$
$\qquad= \large\frac{200 \sqrt 2}{400} \times \frac{1}{\sqrt 2}=\frac{1}{2}$
Therefore $ \alpha=30^{\circ}$
The piolt should steer in a direction $(45 + \alpha)=75^{\circ}$ from North towards East
Also $\large\frac{v_a}{\sin (180-75)}=\frac{400}{\sin 45}$
$v_a=\large\frac{\cos 15}{\sin 45} $$\times 400 $
$\quad= 546.47 km/hr$
Therefore $ t=\large\frac{1000}{546.47}$
$\qquad=1.83\; hr$
Hence d is the correct answer


answered Jul 4, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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