Question:
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?
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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The direction of motion is the direction of the velocity vector at \(t = 2\) s; find its angle with the y-axis from its components.
Updated On: Nov 4, 2026
- \(\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)\) from y-axis
- \(\tan^{-1}\left(\frac{5}{2}\right)\) from y-axis
- \(\tan^{-1}\left(\frac{2}{3}\right)\) from x-axis
- \(\tan^{-1}\left(\frac{5}{2}\right)\) from x-axis
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The Correct Option is A
Solution and Explanation
At \(t = 2\ \mathrm{s}\): \(\vec{v} = 0.5(4)\hat{i} + 3(2)\hat{j} + 9\hat{k} = 2\hat{i} + 6\hat{j} + 9\hat{k}\).
\(|\vec{v}| = \sqrt{4+36+81} = 11\ \mathrm{m/s}\).
Angle \(\alpha\) with the y-axis: \(\cos\alpha = \frac{6}{11}\), so \(\sin\alpha = \frac{\sqrt{121-36}}{11} = \frac{\sqrt{85}}{11}\).
\(\tan\alpha = \frac{\sqrt{85}}{6}\), i.e. the direction makes \(\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)\) with the y-axis.
Answer: (1) \(\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)\) from y-axis
\(|\vec{v}| = \sqrt{4+36+81} = 11\ \mathrm{m/s}\).
Angle \(\alpha\) with the y-axis: \(\cos\alpha = \frac{6}{11}\), so \(\sin\alpha = \frac{\sqrt{121-36}}{11} = \frac{\sqrt{85}}{11}\).
\(\tan\alpha = \frac{\sqrt{85}}{6}\), i.e. the direction makes \(\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)\) with the y-axis.
Answer: (1) \(\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)\) from y-axis
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