A mosquito is moving with a velocity vector v = 0.5t^2 hat i + 3t hat j + 9 hat k m/s and accelerating in uniform conditions.

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 2s ?

(1) \(tan^{-1}(\frac 23)\) from x-axis

(2) \(tan^-(\frac 23)\) from y-axis

(3) \(tan^{-1}(\frac 52)\) from y-axis

(4) \(tan^{-1}(\frac 52)\) from x-axis

Answer 1

Given :

\(\vec v=0.5t^2\hat i+3t\hat j+9\hat k\)

\(\vec v\)_{at t = 2} = \(2\hat i+6\hat j+9\hat k\)

\(\therefore\) Angle made by direction of motion of mosquito will be,

\(cos^{-1}\frac{2}{11}\) (from x-axis) = \(tan^{-1}\frac{\sqrt{117}}2\)

\(cos^{-1}\frac{6}{11}\) (from y-axis) = \(tan^{-1}\frac{\sqrt{85}}6\)

\(cos^{-1}\frac{9}{11}\) (from z-axis) = \(tan^{-1}\frac{\sqrt{40}}9\)

None of the option is matching.