Relation between velocity and displacement is $v = x^2$. Find acceleration at $x = 3m$ :- ............. $\mathrm{m/s}^{2}$

The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as $\mathrm{x}^{2}=\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c.}$ If the acceleration of the particle depends on $\mathrm{x}$ as $\mathrm{x}^{-\mathrm{n}},$ where $\mathrm{n}$ is an integer, the value of $\mathrm{n}$ is

A car is moving with speed of $150\,km / h$ and after applying the brake it will move $27\,m$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling $....m$ distance.

Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_0)$ and the braking capacity, or deceleration, $-a$ that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of $v_0 $ and $a$.

A particle of unit mass undergoes one­ dimensional motion such that its velocity varies according to $ v(x)= \beta {x^{ - 2n}}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by