A plane is to fly due to north. The speed of the plane relative to the air is $200 k m / h$ , and the wind is blowing from west to east at $90 k m / h$ .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The plane should head in a direction given by $θ = \sin^{- 1} (0.65)$

The plane should head in a direction given by $θ = \sin^{- 1} (0.45)$

The velocity of plane relative to the ground is $179 k m / h$

The velocity of plane relative to the ground is $159 k m / h$ .

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Since the wind is blowing toward the east, the plane must head west of north as shown in figure. The velocity of the plane relative to the ground ${\vec{v}}_{p g}$ will be the sum of the velocity of the plane relative to the air ${\vec{v}}_{p a}$ and the velocity of the air relative to the ground ${\vec{v}}_{a g}$ .

The velocity of the plane relative to the ground is given by equation:

${\vec{v}}_{p g} = {\vec{v}}_{p a} + {\vec{v}}_{a g}$ . In this case $\vec{v_{p g}} i s d u e n o r t h .$

The sine of the angle $θ$ between the velocity of the plane and north equals the ratio of $v_{a g}$ and $v_{p a}$ .

$\sin θ = \frac{v_{a g}}{v_{p a}} = \frac{90 k m / h}{200 k m / h} = 0.45$

Since $v_{a g}$ and $v_{p g}$ are perpendicular, we can use the Pythagorean theorem to find the magnitude of ${\vec{v}}_{p g}$ .

$v_{p a}^{2} = v_{a g}^{2} + v_{p g}^{2} v_{p g}^{2} = \sqrt{v_{p a}^{2} - v_{a g}^{2}} = \sqrt{{(200 k m / h)}^{2} - {(90 k m / h)}^{2}} = 179 k m / h$