$A$ bullet moving with a velocity of $200 \, cm/s$ penetrates a wooden block and comes to rest after traversing $4 \, cm$ inside it. What velocity is needed for travelling a distance of $9 \, cm$ in the same block? (in $cm/s$)

$A$ body starts from rest with uniform acceleration and its velocity at a time of $n$ seconds is $v$. The total displacement of the body in the $n^{\text{th}}$ and $(n-1)^{\text{th}}$ seconds of its motion is:

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 $m$)

$A$ 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 .............. $m/s$.

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:

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