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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Plane

A jet plane takes off with a velocity of 300 km/s at a bearing of 60$^{\circ}$. A wind with a velocity of 80 km/s NW blows it off course. What is the resultant velocity of the jet plane (magnitude and the bearing)?

1 Answer

Answer: $290 \;km/s, 44.5^{\circ}$
The horizontal and vertical components of the velocity of the plane are: $300\cos 30^{\circ}$ and $300 \sin 30^{\circ}$
The horizontal and vertical components of the velocity of the wind are: $-80\cos 45^{\circ}$ and $80\sin 45^{\circ}$
Therefore the resultant velocity of the airplane is:
Horizontal component: $300\cos 30^{\circ}$ - $80\cos 45^{\circ} = 203.2$
Vertical component: $300 \sin 30^{\circ}$ + $80\sin 45^{\circ} = 206.6$
$R = 203.2 \hat{i} + 206.6 \hat{j}$
$\Rightarrow$ Magnitude of velocity $ = \sqrt{ 203.2^2 + 206.6^2} = 290\;km/s$
$\Rightarrow$ New Direction/Bearing of the airplane $ =\tan^{-} \large\frac{206.6}{203.3}$$ =44.5^{\circ}$
answered Aug 11, 2014 by balaji.thirumalai
 

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