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 $t_2$?
$(a)$ $x(t_2) = x(t_1) + v(t_1)(t_2 - t_1) + (1/2)a(t_2 - t_1)^2$
$(b)$ $v(t_2) = v(t_1) + a(t_2 - t_1)$
$(c)$ $v_{\text{average}} = (x(t_2) - x(t_1)) / (t_2 - t_1)$
$(d)$ $a_{\text{average}} = (v(t_2) - v(t_1)) / (t_2 - t_1)$
$(e)$ $x(t_2) = x(t_1) + v_{\text{average}}(t_2 - t_1) + (1/2)a_{\text{average}}(t_2 - t_1)^2$
$(f)$ $x(t_2) - x(t_1) = \text{area under the } v-t \text{ curve bounded by the } t\text{-axis and the dotted lines shown.}$

(C, D, F) The correct formulae describing the motion of the particle are $(c)$,$(d)$,and $(f)$.
$1$. The given velocity-time graph is non-linear,which means the acceleration $a$ is not constant. Therefore,the kinematic equations of motion for constant acceleration,such as $x(t_2) = x(t_1) + v(t_1)(t_2 - t_1) + (1/2)a(t_2 - t_1)^2$ and $v(t_2) = v(t_1) + a(t_2 - t_1)$,are not applicable. This invalidates $(a)$ and $(b)$.
$2$. The definition of average velocity is the total displacement divided by the total time interval,which is $v_{\text{average}} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$. Thus,$(c)$ is correct.
$3$. The definition of average acceleration is the change in velocity divided by the time interval,which is $a_{\text{average}} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$. Thus,$(d)$ is correct.
$4$. The equation in $(e)$ is incorrect because it attempts to apply a constant-acceleration kinematic form using average values,which does not hold for non-uniform motion.
$5$. The displacement of a particle in one-dimensional motion is always equal to the area under the velocity-time graph between the given time limits. Thus,$(f)$ is correct.