For uniformly accelerated linear motion starting from rest ($v_0 = 0$), the time-independent kinematic equation relates final velocity $v_f$, acceleration $a$, and displacement $s$:

$$v_f^2 = v_0^2 + 2as$$ $$v_f^2 = 2as$$

This implies that displacement is proportional to the square of the final velocity, $s \propto v_f^2$.

Let's apply this relationship to the two scenarios. In the first case, the velocity is $v$ and the displacement is $s$:

$$v^2 = 2as \quad (1)$$

In the second case, the velocity is $3v$ and the displacement is $s'$:

$$(3v)^2 = 2as'$$ $$9v^2 = 2as' \quad (2)$$

To find $s'$ in terms of $s$, we can substitute the expression for $v^2$ from equation (1) into equation (2):

$$9(2as) = 2as'$$

Dividing both sides by $2a$ (since $a eq 0$ for acceleration to occur):

$$9s = s'$$

Therefore, the displacement is $9s$ when the velocity is $3v$.