The relation between time $t$ and distance $x$ for a moving body is given as $t = m x^{2} + n x$,where $m$ and $n$ are constants. The retardation of the motion is:

$A$ rifle bullet loses $(1/20)^{th}$ of its velocity in passing through a plank. The least number of such planks required just to stop the bullet is

In some appropriate units,the time $(t)$ and position $(x)$ relation of a moving particle is given by $t = x^2 + x$. The acceleration of the particle is

$A$ $100\, m$ long train is moving with a uniform velocity of $45\, km/hr$. The time taken by the train to cross a bridge of length $1\, km$ is.........$s$

$A$ particle moves according to the equation $y = a + bt + ct^2 - dt^4$. What are the initial velocity and initial acceleration of the particle?

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