Resposta:
54.75 metros.
Explicação:
[tex]v(t)=2t+1\ \therefore\ \dfrac{dx}{dt}=2t+1\ \therefore\ dx=(2t+1)dt[/tex]
Definindo os limites de integração de [tex]x_0[/tex] a [tex]x[/tex] e de [tex]t_0[/tex] a [tex]t[/tex], teremos:
[tex]\int\limits_{x_0}^{x}{dx}=\int\limits_{t_0}^t{(2t+1)}dt\ \therefore\ \bigg(x\bigg)\bigg|^{x}_{x_0}=\bigg(2\int{tdt}+\int{dt}\bigg)\bigg|^{t}_{t_0}\ \therefore[/tex]
[tex]x-x_0=\bigg(2\bigg(\dfrac{t^2}{2}\bigg)+t\bigg)\bigg|^{t}_{t_0}\ \therefore\ \boxed{x-x_0=\bigg(t^2+t\bigg)\bigg|^{t}_{t_0}}[/tex]
Para [tex]t_0=0\ s[/tex]:
[tex]x-x_0=\bigg(t^2+t\bigg)\bigg|^t_0\ \therefore\ x-x_0=t^2+t-(0^2+0)\ \therefore[/tex]
[tex]\boxed{x=x_0+t^2+t}[/tex]
Para [tex]x_0=30\ m[/tex]:
[tex]\boxed{x(t)=30+t^2+t}[/tex]
Para [tex]t=4.5\ s[/tex]:
[tex]x(4.5)=30+(4.5)^2+4.5\ \therefore\ \boxed{x(4.5)=54.75\ m}[/tex]
