Velocity at a specific time

Apply the quotient rule: if $s(t) = \dfrac{t^2}{t-2}$, then $v(t) = s'(t) = \dfrac{(2t)(t-2) - t^2(1)}{(t-2)^2}$. Expanding the numerator gives $2t^2 - 4t - t^2 = t^2 - 4t$, so $v(t) = \dfrac{t^2 - 4t}{(t-2)^2}$.

Substitute $t = 5$: $v(5) = \dfrac{25 - 20}{(3)^2} = \dfrac{5}{9}$. So the velocity at $t=5$ is $\dfrac{5}{9}$ meters per second (positive, meaning the particle is moving in the positive direction).

As $t \to 2^+$, the numerator $t^2 \to 4$ (a positive constant), while the denominator $t - 2 \to 0^+$ (a small positive number). Thus $s(t) = \dfrac{t^2}{t-2} \to +\infty$.

This means the particle's position grows without bound as $t$ approaches 2 from the right, so the particle races off toward positive infinity in a finite amount of time as $t\to 2^+$; there is no well-defined finite position or velocity at $t=2$ itself, consistent with the domain restriction $t>2$.