Velocity from position function

Write $s(t) = (t+4)^{1/2}$. By the chain rule, $s'(t) = \frac{1}{2}(t+4)^{-1/2}\cdot 1 = \frac{1}{2\sqrt{t+4}}$. This is the velocity function, $v(t) = \dfrac{1}{2\sqrt{t+4}}$.

Substituting $t = 5$ gives $v(5) = \dfrac{1}{2\sqrt{9}} = \dfrac{1}{2\cdot 3} = \dfrac{1}{6}$ meters per second.

Since $v(t) = \frac{1}{2}(t+4)^{-1/2} > 0$ for all $t > -4$, the particle always moves in the positive direction, so speed equals $v(t)$ itself.

Differentiating again, $a(t) = v'(t) = -\frac{1}{4}(t+4)^{-3/2}$, which is negative for all $t > -4$. Since $v(t) > 0$ and $a(t) < 0$ have opposite signs, the particle is slowing down — its speed continuously decreases as $t$ increases (approaching 0 as $t \to \infty$).