A particle starts from rest at the origin and moves along the x...

Given acceleration $a = 12t + 6$

Velocity $v = \int a dt = \int (12t + 6) dt = 6t^2 + 6t + C_1$

Since the particle starts from rest, $v = 0$ at $t = 0$, which gives $C_1 = 0$.

$v = 6t^2 + 6t$

Displacement $x = \int v dt = \int (6t^2 + 6t) dt = 2t^3 + 3t^2 + C_2$

Since the particle starts from the origin, $x = 0$ at $t = 0$, which gives $C_2 = 0$.

$x = 2t^3 + 3t^2$

At $t = 2\text{ s}$:

$v = 6(2)^2 + 6(2) = 24 + 12 = 36\text{ m/s}$

$x = 2(2)^3 + 3(2)^2 = 16 + 12 = 28\text{ m}$

Ratio of displacement to velocity is $\dfrac{x}{v} = \dfrac{28}{36} = \dfrac{7}{9}$

Answer: $7 : 9$