The position of a particle moving in the xy plane at any time t is given by `x=(3t^2-6t)` metres, `y=(t^2-2t)` metres. Select the correct statement about the moving particle from the following

Options

A. The acceleration of the particle is zero at t = 0 second
B. The velocity of the particle is zero at t = 0 second
C. The velocity of the particle is zero at t = 1 second
D. The velocity and acceleration of the particle are never zero

Answer

C

Explanation

`v_x=(dx)/(dt)=d/(dt)=(3t^2-6t)=6t-6`. At t=1, `v_x`=0
`v_y=(dy)/(dt)=d/(dt) (t^2-2t) = 2t-2`. At t=1, `v_y`=0
Hence, `v=sqrt(v_x^2+v_y^2)`=0