Position of a particle is related to time as given by equation $x=\left(t^4+6t^2+2t\right) \mathrm{m}$. Find its acceleration at $t=5 \mathrm{s}$ in $\mathrm{m} {\mathrm{s}}^{-2}$.

If position vector of a particle is given by $\overrightarrow{r}\left(t\right)=8t\hat{\mathrm{i}}+5t^2\hat{\mathrm{j}}+6\hat{\mathrm{k}}$, then the correct statement about the acceleration of the particle is

A particle of mass $10 \mathrm{g}$moves in a straight line with retardation $2x,$ where $x$ is the displacement in $\mathrm{SI}$ units. Its loss of kinetic energy for above displacement is ${\left(\frac{10}{x}\right)}^{-n}\mathrm{J}.$ The value of $n$ will be $\_\_\_\_\_\_\_\_.$

A particle undergoes a one-dimensional motion such that its velocity varies according to $v\left(x\right)\mathit{=}\beta x^{\mathit{-}2n}$, where $\beta$ and $\mathit{\text{n}}$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$ is given by

In the cube of side $a$ shown in the figure, the vector from the central point of the face $ABOD$ to the central point of the face $BEFO$ will be:

Assertion: The zero velocity of a particle at any instant always implies zero acceleration at that instant

Reason: A body is momentarily at rest when reverses its direction of motion.

The correct option among the following is