The figure displays the graph of the x -coordinate of a particle...

The instantaneous velocity of a particle is given by the slope of its position-time ($x-t$) graph, $v = \dfrac{dx}{dt}$.

At $t = 2\text{ s}$, the particle is on the line segment from $t = 0$ to $t = 2.5\text{ s}$. The slope is:
$v_1 = \dfrac{50 - 0}{2.5 - 0} = 20\text{ m/s}$

At $t = 5\text{ s}$, the particle is on the horizontal line segment from $t = 2.5\text{ s}$ to $t = 7.5\text{ s}$. The slope is:
$v_2 = \dfrac{50 - 50}{7.5 - 2.5} = 0$

At $t = 8\text{ s}$, the particle is on the line segment from $t = 7.5\text{ s}$ to $t = 10.0\text{ s}$. The slope is:
$v_3 = \dfrac{100 - 50}{10.0 - 7.5} = \dfrac{50}{2.5} = 20\text{ m/s}$

At $t = 12\text{ s}$, the particle is on the line segment from $t = 10.0\text{ s}$ to $t = 15.0\text{ s}$. The slope is:
$v_4 = \dfrac{0 - 100}{15.0 - 10.0} = \dfrac{-100}{5.0} = -20\text{ m/s}$

Therefore, the instantaneous velocities are $20\text{ m/s}$, zero, $20\text{ m/s}$, and $-20\text{ m/s}$ respectively.