The acceleration vector along x-axis of a particle having initial velocity $v_0$ changes with distance x as $a=\sqrt{x}$.The distance covered by the particles, When its speed becomes twice that of initial speed is:-

Q. A particle moves along x-axis with an initial speed $v_0$ = $5m{s}^{-1}$. If its acceleration varies with time as shown in a-t graph in the figure.
a. Find the velocity of the particle at t = 4s.
b. Find the time when the particle starts moving along -x direction.

Q. A particle is thrown upwards with speed $v_0$. Find its speed when it reaches $2/3^{rd}$ of the maximum height.

Q. For a particle moving along $x$ - axis, the acceleration $a$ of the particle in terms of its $x$ - coordinate is given by $a=-9x$, where $x$ is in meters and $a$ is in ${\text{m/s}}^2$. Take acceleration, velocity and displacement in positive $x$ - direction as positive. The initial velocity of particle at $x=0$ is $u=+6\text{m/s}$. The maximum distance of the particle along the positive $x$-direction from origin will be

Q. The distance $x$ covered in time $t$ by a body having initial velocity $v_0$ and having constant acceleration $a$ is given by $x=v_{0}t+1/2a{t}^2$. This result follows from :

Q. A point moves rectilinearly with deceleration which depends on the velocity $v$ of the particle as $a=k\sqrt{v}$, where $k$ is a positive constant. At the initial moment the velocity of the point is equal to $v_0$. What distance will it cover before it stops, and what time it will take to cover that distance? Mark this time from the given options.