Answer:
Explanation:
Given
Position of squirrel is given by
[tex]s(t)=t^3-12t^2+36t[/tex]
Velocity is given by
[tex]v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}[/tex]
[tex]v(t)=3t^2-12\times 2t+36[/tex]
[tex]v(t)=3t^2-24t+36[/tex]
(b) acceleration is given by
[tex]a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}[/tex]
[tex]a(t)=6t-24[/tex]
(c)at [tex]s(3)=3^3-12(3)^2+36(3)[/tex]
[tex]s(3)=27\ m[/tex]
at [tex]s(4)=4^3-12(4)^2+36(4)[/tex]
[tex]s(4)=16\ m[/tex]
at [tex]t=3\ s[/tex] Position is [tex]27\ m[/tex] and at [tex]t=4\ s[/tex] position is [tex]16\ m[/tex]
therefore squirrel is moving down