The position-time $(x-t)$ graph for positive acceleration is

$A$ particle starts from rest and moves in a straight line. It travels a distance $2L$ with uniform acceleration and then moves with a constant velocity for a further distance of $L$. Finally,it comes to rest after moving a distance of $3L$ under uniform retardation. Then the ratio of average speed to the maximum speed $\left(\frac{\bar{V}}{V_m}\right)$ of the particle is

The acceleration-time graph of a body is shown. The corresponding velocity-time graph of the same body is

From the given $v-t$ graph,find the ratio of distance to displacement in $25\,s$ of motion.

The velocity-displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by:

The velocity-time graph of a particle in one-dimensional motion is shown in the figure. Which of the following formulae are correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$?
$(a)$ $x(t_2) = x(t_1) + v(t_1)(t_2 - t_1) + (1/2)a(t_2 - t_1)^2$
$(b)$ $v(t_2) = v(t_1) + a(t_2 - t_1)$
$(c)$ $v_{\text{average}} = (x(t_2) - x(t_1)) / (t_2 - t_1)$
$(d)$ $a_{\text{average}} = (v(t_2) - v(t_1)) / (t_2 - t_1)$
$(e)$ $x(t_2) = x(t_1) + v_{\text{average}}(t_2 - t_1) + (1/2)a_{\text{average}}(t_2 - t_1)^2$
$(f)$ $x(t_2) - x(t_1) = \text{area under the } v-t \text{ curve bounded by the } t\text{-axis and the dotted lines shown.}$