Question

A particle moves in x-y plane. The position vector of particle at any time t is $\overrightarrow{r} = \{(2t)\widehat{i} + (2t^{2})\widehat{j}\}$m. The rate of change of q at time t = 2s. (where q is the angle which its velocity vector makes with positive x-axis ) is

• A. $\frac{2}{17}$rad/s
• B. $\frac{1}{14}$rad/s
• C. $\frac{4}{7}$rad/s
• D. $\frac{6}{5}$rad/s

Answer 1

Answer: (A)

$\frac{2}{17}$rad/s

Answer 2

x = 2t ̃ V_x = $\frac{dx}{dt}$= 2

y = 2t² ̃ v_y = $\frac{dy}{dt}$= 4t

\ tan q = $\frac{v_{y}}{v_{x}}$= $\frac{4t}{2}$= 2t Differentiating with respect to time we get,

(sec²q) $\frac{d\theta}{dt}$ = 2

or (1 + tan²q) $\frac{d\theta}{dt}$ = 2

or (1 + 4t²) $\frac{d\theta}{dt}$= 2

or $\frac{d\theta}{dt}$ = $\frac{2}{1 + 4t^{2}}$

$\frac{d\theta}{dt}$at t = 2 s is $\frac{d\theta}{dt}$=

$\frac{2}{1 + 4(2)^{2}}$= $\frac{2}{17}$ rad/s