The distance covered by a body of mass $5\,g$ having linear momentum $0.3\,kg\,m/s$ in $5\,s$ is $..........\,m$.

$A$ bullet is fired into a fixed target and loses one-third of its velocity after traveling $4 \ cm$. It penetrates a further $D \times 10^{-3} \ m$ before coming to rest. The value of $D$ is:

$A$ particle moving along the $x-$axis has acceleration $f$ at time $t$,given by $f = f_0(1 - t/T)$,where $f_0$ and $T$ are constants. The particle at $t = 0$ has zero velocity. In the time interval between $t = 0$ and the instant when $f = 0$,the particle's velocity $(v_x)$ is:

The distance $x$ covered in time $t$ by a body having initial velocity $u$ and having constant acceleration $a$ is given by $x=ut+\frac{1}{2}at^2$. This result follows from

An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement $(s)-$ velocity $(v)$ graph of this object is

The acceleration $a$ in $m/s^2$ of a particle is given by $a = 3t^2 + 2t + 2$,where $t$ is the time. If the particle starts with an initial velocity $u = 2\,m/s$ at $t = 0$,then the velocity at the end of $2\,s$ is: