Question

Preciso de ajuda.
Se possível com resolução.

E


D


B


A


C

Answer 1

$a=\dfrac{dV}{dt}\;\;\Rightarrow\;\;dV=a\left(t \right )\,dt\\ \\ \int_{V_{0}}^{V}{dV}=\int_{t_{0}}^{t}a\left(\tau \right )\,d\tau\\ \\ V-V_{0}=\int_{t_{0}}^{t}a\left(\tau \right )\,d\tau$


Para $V_{0}=7\text{ m/s}$, $t_{0}=0$ e $t=7\text{ s}$, temos

$V-7=\int_{0}^{7}{\left[\dfrac{18\tau^{2}}{7}+\dfrac{6\tau}{5}+\dfrac{3}{7} \right ]d \tau}\\ \\ \\ V-7=\left[\dfrac{18}{7}\cdot \dfrac{\tau^{3}}{3}+\dfrac{6}{5}\cdot\dfrac{\tau^{2}}{2}+\dfrac{3\tau}{7} \right ]_{0}^{7}\\ \\ \\ V-7=\left[\dfrac{18}{7}\cdot \dfrac{\left(7 \right )^{3}}{3}+\dfrac{6}{5}\cdot\dfrac{\left(7 \right )^{2}}{2}+\dfrac{3\cdot\left(7 \right )}{7} \right ]-\left[\dfrac{18}{7}\cdot \dfrac{\left(0 \right )^{3}}{3}+\dfrac{6}{5}\cdot\dfrac{\left(0 \right )^{2}}{2}+\dfrac{3\cdot\left(0 \right )}{7} \right ]\\ \\ \\ V-7=\left[294+\dfrac{147}{5}+3 \right ]-0\\ \\ \\ V-7=\dfrac{1\,470+147+15}{5}\\ \\ \\ V-7=\dfrac{1\,632}{5}\\ \\ V=\dfrac{1\,632}{5}+7\\ \\ V=\dfrac{1\,632+35}{5}\\ \\ V=\dfrac{1\,667}{5}\\ \\ V=333,4\text{ m/s}$


Resposta: alternativa $\text{A) }333,4\text{ m/s.}$