Question

calcule a derivada da função acima, usando a definição, para mostrar que a velocidade v(t)=ds/dt é dada por v(t)=v0+at.​

Answer 1

$\displaystyle \text {s(t)} = \text s_o + \text v_o.\text t +\frac{\text a.\text t^2 }{2}$

A derivada pela definição  :

$\displaystyle \text {s '(t)} = \lim_{\text h \to 0} \ \frac{\text{s(t+h)}-\text {s(t)}}{\text h } \\\\\\ \text{s '(t)} = \lim_{\text h\to 0} \ \frac{\displaystyle \text s_o+\text v_o.(\text{t+h})+\frac{\text a.(\text{t+h})^2}{2}-\text s_o-\text v_o.\text t-\frac{\text {a(t)}^2}{2} }{\text h}$

$\displaystyle \text{s '(t)} = \lim_{\text h\to 0} \ \frac{\text v_o.\text t+\text v_o.\text h+\frac{\displaystyle \text {a.t}^2+2\text{a.t.h+\text a.\text h }^2}{2}-\text v_o.\text t- \displaystyle \frac{\text {a.t}^2}{2}}{\text h} \\\\\\ \text{s '(t)}= \lim_{\text h\to 0} \ \frac{\text h(\text v_o+\frac{2\text {a.t}+ \text{a.h}}{2})}{\text h} \\\\\\ \text{s '(t)} = \lim_{\text h\to0} \ \text v_o + \frac{2\text {a.t}+\text a.\text h}{2}$

fazendo h = 0

$\displaystyle \text{s '(t)} = \text v_o + \frac{2.\text{a.t}+\text a.0 }{2}\\\\\\ \boxed{\text {s '(t)} = \text v_o + \text {a.t} }$

Isso é a função horária da velocidade, ou seja :

$\boxed{\text{v(t)} = \text v_o + \text {a.t}}$

Portanto, vemos que :

$\huge\boxed{\displaystyle \frac{\text {ds}}{\text {dt}}\text{ s(t)} = \text {v(t)}}\checkmark$