An aeroplane is flying with the velocity of V1 = 800 kmph relative to the air towards south. A wind with velocity of V2 = 15 ms–1 is blowing from West to East. What is the velocity of the aeroplane with respect to the earth.

V12=800×518=20009=222.2m/s(in south V¯12 is -ve)∴V¯12=−222.2j^;V¯2=15i^V¯12=V¯1−V¯2=−222.2j^−15i^V¯1=V¯12+V¯2=−222.2j^+15i^V¯1=(222.2)2+(15)2=222.7 ms−1Aeroplane is moving towards south-east direction.

An aeroplane is flying with the velocity of V1 = 800 kmph relative to the air towards south. A wind with velocity of V2 = 15 ms–1 is blowing from West to East. What is the velocity of the aeroplane with respect to the earth.

V12=800×518=20009=222.2m/s(in south V¯12 is -ve)∴V¯12=−222.2j^;V¯2=15i^V¯12=V¯1−V¯2=−222.2j^−15i^V¯1=V¯12+V¯2=−222.2j^+15i^V¯1=(222.2)2+(15)2=222.7 ms−1Aeroplane is moving towards south-east direction.

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An aeroplane is flying with the velocity of V₁= 800 kmph relative to the air towards south. A wind with velocity of V₂= 15 ms^–1 is blowing from West to East. What is the velocity of the aeroplane with respect to the earth.

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150 ms^–1

40 ms^–1

82 ms^–1

222.7 ms^–1

answer is A.

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Detailed Solution

$V_{12} = 800 \times \frac{5}{18} = \frac{2000}{9} = 222.2 m / s$
(in south ${\bar{V}}_{12}$ is -ve)
$\begin{array}{l} ∴ {\bar{V}}_{12} = - 222.2 \hat{j}; {\bar{V}}_{2} = 15 \hat{i} \\ {\bar{V}}_{12} = {\bar{V}}_{1} - {\bar{V}}_{2} = - 222.2 \hat{j} - 15 \hat{i} \\ {\bar{V}}_{1} = {\bar{V}}_{12} + {\bar{V}}_{2} = - 222.2 \hat{j} + 15 \hat{i} \\ |{\bar{V}}_{1}| = \sqrt{(222.2)^{2} + (15)^{2}} = 222.7 {ms}^{- 1} \end{array}$