A particle moves in x-y plane. The position vector of particle at any time t is `vecr={(2t)hati+(2t^2)hatj}` m. The rate of change of `theta` at time t = 2 s (where `theta` is the angle which its velocity vector makes with positive x-axis) is

Options

A. `2/17 "rad s"^(-1)`
B. `1/14 "rad s"^(-1)`
C. `4/7 "rad s"^(-1)`
D. `6/5 "rad s"^(-1)`

Answer

A

Explanation

Given , `vecr={(2t)hati+(2t^2)hatj}`
Comparing it with standard equation of position vector ,
`vecr=xhati+yhatj` , we get x=2t and `y=2t^2`
`rArr v_x=(dx)/(dt)=2` and `v_y=(dy)/(dt)=4t`
`therefore tan theta = v_y/v_x = (4t)/2=2t`
Differentiating with respect to time we get , `(sec^2 theta ) (d theta)/(dt)=2`
or `(1+tan^2 theta) (d theta)/(dt) = 2` or `(d theta)/(dt) =2/(1+4t^2)`
at t=2 s , `((d theta)/(dt))=2/(1+4(2)^2)=2/17 "rad s"^(-1)`