Question

An airplane is flying with velocity 50√2 km/hr in north-east direction. Wind is blowing at 25 km/hr from north to south. What is the resultant displacement of airplane in 2 hours?


Solution required WITH PROPER EXPLANATION. ​

Answer 1

Explanation:

please refer the attachment.

Answer 2

Aeroplane travels a resultant displacement of 91.6 kilometres.

Given,

Velocity of the aeroplane in the north-east direction, Va=50√2 km/h

Velocity of the wind from north to south direction, Vw=25 km/h

Time=2 hours.

To find,

the resultant displacement of aeroplane in 2 hours.

Solution:

• The concept here is from the chapter "Motion in a Plane".
• In this question, there are to be vector applications used.
• To find the resultant displacement, resultant velocity is required.
• As the velocity of the aeroplane is along the north-east direction, its velocity will be along z-axis.
• As the velocity of wind is from north to south, its velocity will be along negative y-axis.

Velocity of the aeroplane=50√2 k̂ km/h

Velocity of the wind=-25 ĵ km/h.

The z-axis is always perpendicular to the y-axis.

As the velocity of the wind in along the negative y-axis so the angle between the z-axis and the negative y-axis will be 180°.

The resultant R of two vectors A and B having angle θ between them is calculated as:

$|R|=\sqrt{A^{2}+B^{2}+2ABcosθ }$

The resultant velocity of the aeroplane, Vnet will be:

$|R|=\sqrt{A^{2}+B^{2}+2ABcosθ }\\|Vnet|=\sqrt{Va^{2}+Vw^{2} +2VaVwcos180} \\|Vnet|=\sqrt{(50\sqrt{2}) ^{2} +25^{2} +2X50\sqrt{2} X25X(-1)}\\|Vnet|=\sqrt{5000+625-2500\sqrt{2} }\\|Vnet|=\sqrt{5625-2500\sqrt{2} }\\|Vnet|=\sqrt{5600-2500X1.4}\\|Vnet|=\sqrt{5600-3500}\\|Vnet|=\sqrt{2100}\\|Vnet|=45.8 km/h.$

The displacement in 2 hours will be:

$Velocity=\frac{Displacement}{Time} \\45.8=\frac{Displacement}{2}\\Displacement=45.8X2\\Displacement=91.6 km.$

The resultant displacement is 91.6 km.

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