Question

calculo II ......................

Answer 1

Integrando a função velocidade:

v = 9t²/5 + 12t - 3

$\int\limits^4_0 {(\frac{9t^2}{5} + 12t - 3}) \, dt$


Reescrevendo:

$\int\limits^4_0 {\frac{9t^2}{5}} \, dt \ + \  \int\limits^4_0 {12t} \, dt \ - \ \int\limits^4_0 {3} \, dt$

$\frac{9}{5} \int\limits^4_0 {t^2} \, dt \ + \ 12\int\limits^4_0 {t} \, dt \ - \ 3\int\limits^4_0 \, dt$

$\frac{9}{5} \ . \ \frac{t^{2+1}}{2+1}|^4_0 \ + \ 12 \ . \ \frac{t^{1+1}}{1+1}|^4_0 \ - \ 3 \ . \ t|^4_0$

$\frac{9}{5} \ . \ \frac{t^{3}}{3}|^4_0 \ + \ 12 \ . \ \frac{t^2}{2}|^4_0 \ - \ 3 \ . \ t|^4_0$

$\frac{3}{5} \ . \ t^{3}|^4_0 \ + \ 6 \ . \ t^{2}|^4_0 \ - \ 3 \ . \ t|^4_0$

$\frac{3}{5} \ . \ (4)^{3} \ + \ 6 \ . \ (4)^{2} \ - \ 3 \ . \ (4)$

$\frac{3}{5} \ . \ 64 \ + \ 6 \ . \ 16 \ - \ 12$

$\frac{192}{5} \ + \ 96 \ - \ 12$

$\frac{192}{5} \ + \ 84$

$\frac{192+420}{5}$

$\frac{612}{5}$  = 122,4 m



Como a posição inicial era de 15 m, temos, no instante t = 4 s:

15 + 122,4 = 137,4 m