The velocity of a particle moving along the $x$-axis is given by $V = x^2 - 5x + 4$ (in $\text{m/s}$), where $x$ denotes the $x$-coordinate of the particle in metres. The magnitude of the acceleration of the particle when the velocity is zero is:

$A$ particle of mass $m$ moves on the $x$-axis as follows: it starts from rest at $t = 0$ from the point $x = 0$ and comes to rest at $t = 1$ at the point $x = 1$. No other information is available about its motion at intermediate time $(0 < t < 1)$. If $\alpha$ denotes the instantaneous acceleration of the particle,then

The equation of motion of a body is $\frac{dv}{dt} = -4v + 8$,where $v$ is the velocity in $m/s$ and $t$ is the time in seconds. The initial velocity of the particle was zero. Then,

For any arbitrary motion in space,which of the following relations are true?
$(a)$ $v_{\text{average}} = (1/2) (v(t_1) + v(t_2))$
$(b)$ $v_{\text{average}} = [r(t_2) - r(t_1)] / (t_2 - t_1)$
$(c)$ $v(t) = v(0) + at$
$(d)$ $r(t) = r(0) + v(0)t + (1/2)at^2$
$(e)$ $a_{\text{average}} = [v(t_2) - v(t_1)] / (t_2 - t_1)$
(The 'average' stands for the average of the quantity over the time interval $t_1$ to $t_2$.)

$A$ body is at rest at $x=0$. At $t=0$,it starts moving in the positive $x-$direction with a constant acceleration. At the same instant,another body passes through $x=0$ moving in the positive $x$ direction with a constant speed. The position of the first body is given by $x_{1}(t)$ after time $t$ and that of the second body by $x_{2}(t)$ after the same time interval. Which of the following graphs correctly describes $(x_{1}-x_{2})$ as a function of time $t$?

$A$ bead is moving in medium $1$ with a uniform speed of $1\, m/s$ for $2.5\, s$. Then it enters into air and falls freely under gravity for $2\, m$. Finally, it enters medium $2$ and immediately moves with uniform speed for $1.5\, s$. The total distance the bead has traveled is.........$m$ $(g = 10\, m/s^2)$: (in $.1$)