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 -

(Where $v$ stands for velocity)

A monkey climbs up a slippery pole for $3$ and subsequently slips for $3$. Its velocity at time $t$ is given by $v (t) = 2t \,(3s -t)$ ;  $0 < t < 3$ and $v(t) =\,-\, (t -3)\,(6 -t)$ ; $3 < t < 6$ $s$ in $m/s$. It repeats this cycle till it reaches the height of $20\, m$.

$(a)$ At what time is its velocity maximum ?

$(b)$ At what time is its average velocity maximum ?

$(c)$ At what time is its acceleration maximum in magnitude ?

$(d)$ How many cycles (counting fractions) are required to reach the top ?

A man goes $10\,m $ towards North, then $ 20\,m$ towards east then displacement is........$m$

The velocity $(v)$ of a particle moving along $x$-axis varies with its position $x$ as shown in figure. The acceleration $(a)$ of particle varies with position $(x)$ as

A particle moves along $x$-axis as $x=4(t-2)+a(t-2)^2$. Which of the following statements is true?

A particle starts from rest at $x=0\; m$ with an acceleration of $1 \,m / s ^2$. At $t = 5\;s _{ s }$ 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 $x - x _0$ is $12.5 \,m$. Then one can conclude that $v - v _0$ is .............. $\,m / s$