Given the position function $x = (t + 5)^{-1}$.

1. Find Velocity ($v$): Velocity is the rate of change of position . $v = \frac{dx}{dt} = \frac{d}{dt}(t + 5)^{-1}$ $v = -1(t + 5)^{-2}$ From this, we can write $(t + 5)^{-1} = v^{1/2}$ (considering magnitude) or $(t+5) = v^{-1/2}$.
2. Find Acceleration ($a$): Acceleration is the rate of change of velocity . $a = \frac{dv}{dt} = \frac{d}{dt}[-(t + 5)^{-2}]$ $a = -(-2)(t + 5)^{-3} = 2(t + 5)^{-3}$
3. Relate Acceleration to Velocity: Substitute $(t + 5)$ from the velocity equation into the acceleration equation: $a = 2 [v^{-1/2}]^{-3}$ $a = 2 v^{3/2}$ Therefore, acceleration is proportional to $(velocity)^{3/2}$. (Note: In terms of distance $x$, since $x = (t+5)^{-1}$, then $a = 2x^3$, which is not among the options).