The velocity ( v ) versus time ( t ) graph of a particle moving in...

The acceleration $a$ of a particle is given by the slope of its velocity-time graph, $a = \dfrac{dv}{dt}$.

From $t = 0$ to $t = t_1$: The $v-t$ graph is a parabola opening upwards. This implies $v$ is proportional to $t^2$, so $v = kt^2$ for some positive constant $k$. The acceleration is $a = \dfrac{dv}{dt} = 2kt$, which means the acceleration increases linearly with time.

From $t = t_1$ to $t = t_2$: The $v-t$ graph is a straight line with a positive slope. This implies $v = mt + c$ where $m > 0$. The acceleration is $a = \dfrac{dv}{dt} = m$, which means the acceleration is a positive constant.

From $t = t_2$ to $t = t_3$: The $v-t$ graph is a horizontal line. This implies the velocity is constant. The acceleration is $a = \dfrac{dv}{dt} = 0$.

Comparing these findings with the given options, the correct statement is that from $0$ to $t_1$, $a$ increases linearly with time; from $t_1$ to $t_2$, $a$ is a positive constant; and from $t_2$ to $t_3$, $a$ is zero.

Answer: From $0$ to $t_1$, $a$ increases linearly with time; from $t_1$ to $t_2$, $a$ is a positive constant; from $t_2$ to $t_3$, $a$ is zero.