At an instant t ,the coordinates of a particle are x=at2,y=bt2,z=0,then its velocity at that instant when its acceleration is zero will be

Q. A particle is moving so that its displacement s s given as $s=t^3-6t^2+3t+4$ meter. Its velocity at the instant when its acceleration is zero will be -

Q. A particle moves in the $x-y$ plane and at time $t$, it is at the point whose coordinates are $(t^2,t^3-2t)$. Then at what instant of time will its velocity and acceleration vectors be perpendicular to each other?

Q. Velocity and acceleration of a particle at some instant of time are $v=(3\hat{i}+4\hat{j}){ms}^{-1}$ and $a=-(6\hat{i}+8\hat{j}){ms}^{-2}$ respectively. At the same instant particle is at origin. Maximum x-coordinate of particle will be

Q. The position of a particle as a function of time $t,$ is given by $x\left(t\right)=at+b{t}^2-c{t}^3$

Where, $a,b$ and $c$ are constants. When the particle attains zero acceleration, then its velocity will be-

Q. Assertion :A particle may be momentarily at rest and yet have non-zero acceleration Reason: The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant.