Question

Questão 10/10 - Matemática Aplicada
Suponha que um objeto esteja se movendo com velocidade v(t) = t2-t-6 em metros por segundo. No período de 1sts 4, encontre a distância
percorrida pelo objeto
Distância: $ \u(|dt
-10,2
O
-6.4
64
1012​

Answer 1

A distância percorrida será dada pela integral do módulo da função velocidade:

$\boxed{d=\int\limits_{t_1}^{t_2}{|v(t)|dt}}$

Analisando o sinal de v(t):

$v(t)=t^2-t-6\ \to\ v(t)=0\ \therefore\ t^2-t-6=0\ \therefore\\\\ t=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}=\dfrac{1\pm5}{2}=\boxed{-2}\ \text{ou}\ \boxed{3}$

$v(t)>0\ \iff\ \boxed{t<-2}\ \text{ou}\ \boxed{t>3}\\\\ v(t)<0\ \iff\ \boxed{-2</p><p>Para [tex]t_1=1\ s$ e $t_2=4\ s$, a distância percorrida será determinada por:

$d=\int\limits_{1}^{4}{|v(t)|dt}$

Os intervalos de integração considerando a função modular serão:

$10$

Dessa forma:

$d=\int\limits_{1}^{4}{|v(t)|dt}=\int\limits_{1}^3{-v(t)dt}+\int\limits^4_3{v(t)dt}=$

$=\bigg(\int{v(t)dt}\bigg)\bigg|^{4}_{3}-\bigg(\int{v(t)dt}\bigg)\bigg|_{1}^{3}=$

$=\bigg(\int{(t^2-t-6)dt}\bigg)\bigg|^{4}_{3}-\bigg(\int{(t^2-t-6)dt}\bigg)\bigg|_{1}^{3}=$

$=\bigg(\dfrac{t^3}{3}-\dfrac{t^2}{2}-6t\bigg)\bigg|^{4}_{3}-\bigg(\dfrac{t^3}{3}-\dfrac{t^2}{2}-6t\bigg)\bigg|_{1}^{3}=$

$=\bigg[\bigg(\dfrac{4^3}{3}-\dfrac{4^2}{2}-6(4)\bigg)-\bigg(\dfrac{3^3}{3}-\dfrac{3^2}{2}-6(3)\bigg)\bigg]-\\\\ -\bigg[\bigg(\dfrac{3^3}{3}-\dfrac{3^2}{2}-6(3)\bigg)-\bigg(\dfrac{1^3}{3}-\dfrac{1^2}{2}-6(1)\bigg)\bigg] =$

$=\bigg[-\dfrac{32}{3}-\bigg(-\dfrac{27}{2}\bigg)\bigg]-\bigg[-\dfrac{27}{2}-\bigg(-\dfrac{37}{6}\bigg)\bigg]=$

$=\dfrac{17}{6}-\bigg[-\dfrac{22}{3}\bigg]=\dfrac{17}{6}+\dfrac{22}{3}=\boxed{\dfrac{61}{6}\ m}$

Portanto, a distância percorrida pelo objeto entre 1 e 4 segundos é:

$d=\dfrac{61}{6}\ m\ \therefore\ \boxed{d\approx10.2\ m}$