# Solución

Dado que $v (t) = s ′ (t)$ y $a (t) = v ′ (t) = s '' (t)$, comenzamos por encontrar la derivada de $s (t)$: $$\begin{aligned} s ' (t) &= \mathop {\lim }\limits_{h \to 0} \frac{s (t + h) −s (t)} {h }\\ &= \mathop {\lim }\limits_{h \to 0} \frac{3 ((t + h)^2−4 (t + h) + 1)− (3t^2−4t + 1)} {h}\\ &= 6t −4. \end{aligned}$$

Ahora calculamos la derivada segunda $$\begin{aligned} s '' (t) &= \mathop {\lim }\limits_{h \to 0} \frac{s ' (t + h) −s ' (t)} {h }\\ &= \mathop {\lim }\limits_{h \to 0} \frac{6 (t + h) −4− (6t − 4)} {h}\\ &= 6. \end{aligned} $$ Por tanto, $a = 6 m /s^2$.