$A$ particle moving along the $x-$axis has acceleration $f$ at time $t$,given by $f = f_0(1 - t/T)$,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 $(v_x)$ is:

Two particles $A$ and $B$ start from rest and move for equal time on a straight line. Particle $A$ has an acceleration of $2\,m/s^2$ for the first half of the total time and $4\,m/s^2$ for the second half. The particle $B$ has acceleration $4\,m/s^2$ for the first half and $2\,m/s^2$ for the second half. Which particle has covered a larger distance?

An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement $(s)-$ velocity $(v)$ graph of this object is

$A$ body starts from rest with uniform acceleration $a$. Its velocity after $n \ s$ is $v$. The displacement of the body in the last $3 \ s$ is: (Assume total time of journey is $n \ s$)

$A$ vehicle starts moving in a straight line with an acceleration $a = 4 \,m/s^2$, with initial velocity equal to zero. After accelerating for time $t_1$, the vehicle moves uniformly for time $t_2$, and finally decelerates for time $t_1$, eventually coming to a stop. The total time taken during the motion is $10 \,s$ and the average velocity during the motion is $5.1 \,m/s$. The time taken by the vehicle during acceleration is (in $\,s$)

$A$ body moves on a frictionless plane starting from rest. If $S_{n}$ is the distance moved between $t=n-1$ and $t=n$,and $S_{n-1}$ is the distance moved between $t=n-2$ and $t=n-1$,then the ratio $\frac{S_{n-1}}{S_n}$ is $\left(1-\frac{2}{x}\right)$ for $n=10$. The value of $x$ is: