Question

A particle moves in x-y plane. The position vector of particle at any time t is =$\overset{\rightarrow}{r}${(2t)$\widehat{i}$+ (2t²)$\widehat{j}$} m. The rate of change of q at time t = 2 s. (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,

(sec² q)$\frac{d\theta}{dt}$= 2

or (1 + tan² q)$\frac{d\theta}{dt}$= 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