$A$ body of mass $10 \, kg$ is moving with a constant velocity of $10 \, m/s$. When a constant force acts for $4 \, s$ on it,it moves with a velocity of $2 \, m/s$ in the opposite direction. The acceleration produced in it is .......... $m/s^2$.

$A$ body travels $102.5 \ m$ in the $n^{\text{th}}$ second and $115.0 \ m$ in the $(n+2)^{\text{th}}$ second. The acceleration is: (in $m/s^2$)

The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,then what is the retardation?

$A$ body of mass $10 \,kg$ has a constant velocity of $10 \,m \,s^{-1}$. $A$ constant force is applied for a duration of $4 \,s$ on the object, which gives it a speed of $2 \,m \,s^{-1}$ in the opposite direction. Find the acceleration produced in it.

Particle $A$ is moving along the $x$-axis. At time $t=0$,it has a velocity of $10\,m/s$ and an acceleration of $-4\,m/s^2$. Particle $B$ has a velocity of $20\,m/s$ and an acceleration of $-2\,m/s^2$. Initially,both particles are at the origin. At time $t=2\,s$,the distance between the two particles is $.............\,m$.

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)