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Karnataka Board PUCPUC Science Class 11

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.

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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. Find the direction in which the pilot should head the plane to reach the point B.   

Sum
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Solution

Given:
Distance between points A and B = 500 km
B from A is 30˚ east of north.
Speed of wind due north, vw = 20 m/s
Airspeed of the plane, va = 150 m/s
Let \[\vec{R}\]  be the resultant direction of the plane to reach point B. 

Using the sine formula in ∆ACB, we get:

\[\frac{20}{\sin A} = \frac{150}{\sin30^\circ}\]

\[ \Rightarrow \sin A = \frac{20}{150}\sin30^\circ\]

\[ = \frac{20}{150} \times \frac{1}{2} = \frac{1}{15}\]

\[ \Rightarrow A = \sin^{- 1} \left( \frac{1}{15} \right)\]

Direction of the aeroplane is \[\sin^{- 1} \left( \frac{1}{15} \right)\]  east of line AB.

 
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Chapter 3: Rest and Motion: Kinematics - Exercise [Page 54]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 3 Rest and Motion: Kinematics
Exercise | Q 49.1 | Page 54

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