Solución

Dado que v(t)=s′(t)v (t) = s ′ (t) y a(t)=v′(t)=s′′(t)a (t) = v ′ (t) = s '' (t), comenzamos por encontrar la derivada de s(t)s (t): s′(t)=lim⁡h→0s(t+h)−s(t)h=lim⁡h→03((t+h)2−4(t+h)+1)−(3t2−4t+1)h=6t−4.\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 s′′(t)=lim⁡h→0s′(t+h)−s′(t)h=lim⁡h→06(t+h)−4−(6t−4)h=6.\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=6m/s2a = 6 m /s^2.