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JEE Main 2021 (16 Mar, Shift 2) · Motion in a Plane (2-D Kinematics)

Motion in a Plane (2-D Kinematics) question from JEE Main 2021 (16 Mar, Shift 2)

Q6JEE Main 2021 · 16 MAR · SHIFT 2Medium

Motion in a Plane (2-D Kinematics)

A mosquito is moving with a velocity $\vec{v} = 0.5t^{2}\hat{i} + 3t\hat{j} + 9\hat{k}$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?

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Answer: (a)

At $t = 2$ s: $\vec v = 0.5(4)\hat i + 3(2)\hat j + 9\hat k = 2\hat i + 6\hat j + 9\hat k$
Angle $\phi$ from y-axis: $\tan\phi = \frac{\sqrt{v_x^2 + v_z^2}}{v_y} = \frac{\sqrt{4 + 81}}{6} = \frac{\sqrt{85}}{6}$
So the answer is (a) $\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)$ from y-axis.

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