Finding velocity and acceleration

Differentiate $s(t) = t^2 e^{-t}$ using the product rule: $v(t) = s'(t) = 2t\,e^{-t} + t^2(-e^{-t}) = e^{-t}(2t - t^2)$. This can be factored as $v(t) = t(2-t)e^{-t}$.

Now differentiate $v(t) = e^{-t}(2t - t^2)$ using the product rule again: $a(t) = -e^{-t}(2t-t^2) + e^{-t}(2-2t) = e^{-t}(-2t + t^2 + 2 - 2t) = e^{-t}(t^2 - 4t + 2)$.

Since $e^{-t} > 0$ always, the sign of $v(t) = t(2-t)e^{-t}$ depends on $t(2-t)$, which is zero at $t=0,2$ and positive for $0<t<2$, negative for $t>2$. The sign of $a(t) = e^{-t}(t^2-4t+2)$ depends on $t^2-4t+2$, whose roots are $t = 2\pm\sqrt{2}$ (approximately $0.586$ and $3.414$); this quadratic is positive for $t<2-\sqrt2$, negative for $2-\sqrt2<t<2+\sqrt2$, and positive for $t>2+\sqrt2$.

Comparing signs on the intervals $[0, 2-\sqrt2)$, $(2-\sqrt2, 2)$, $(2,2+\sqrt2)$, and $(2+\sqrt2,\infty)$: on $[0,2-\sqrt2)$ both $v,a>0$ (speeding up); on $(2-\sqrt2,2)$, $v>0,a<0$ (slowing down); on $(2,2+\sqrt2)$, $v<0,a<0$ (speeding up); on $(2+\sqrt2,\infty)$, $v<0,a>0$ (slowing down). So the particle speeds up for $0 \le t < 2-\sqrt2$ and for $2 < t < 2+\sqrt2$.