The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depend on time?

$A$ particle is projected with velocity $v_0$ along the $x$-axis. The deceleration of the particle is proportional to the square of the distance from the origin,i.e.,$a = -\alpha x^2$. The distance at which the particle stops is:

The relation between position $(x)$ and time $(t)$ is given below for a particle moving along a straight line. Which of the following equations represents uniformly accelerated motion? [where $\alpha$ and $\beta$ are positive constants]

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

$A$ particle starts from rest at $x=0 \, m$ with an acceleration of $1 \, m/s^2$. At $t = 5 \, s$,it receives an additional acceleration in the same direction as its motion. At $t = 10 \, s$,its speed and position are $v$ and $x$,respectively. Had the additional acceleration not been provided,its speed and position would have been $v_0$ and $x_0$,respectively. It is found that $x - x_0 = 12.5 \, m$. Then one can conclude that $v - v_0$ is .............. $m/s$.

An object moving along the $x$-axis with a uniform acceleration has a velocity $\vec{v} = (12 \ cm \ s^{-1}) \hat{i}$ at $x = 3 \ cm$. After $2 \ s$,if it is at $x = -5 \ cm$,then its acceleration is: