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

$A$ particle is projected with velocity $v_0$ along the $x-$axis,and its deceleration is given by $a = -\alpha x^2$. The distance at which the particle stops is

$A$ body travels in a straight line from point $A$ to point $B$ with an initial velocity zero and uniform acceleration, covering $1 \,m$ during the first second and $39 \,m$ during the last second. The distance between $A$ and $B$ in metre is

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

$A$ particle moving along the $X$-axis has acceleration $f$ at time $t$ given by $f=f_0\left(1-\frac{t}{T}\right)$, 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 is

The path of a particle moving under the influence of a force that is fixed in magnitude and direction is: