Question

Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the seven seconds that I’m watching it (so from t = 0 to t = 7). (a) What is the velocity function, v(t), for the motion of the squirrel? (b) What is the acceleration function, a(t), for the motion of the squirrel? (c) At the four second mark, is the squirrel moving up the tree or down the tree? Justify your answer.

Answer 1

Answer:

Explanation:

Given

Position of squirrel is given by

$s(t)=t^3-12t^2+36t$

Velocity is given by

$v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}$

$v(t)=3t^2-12\times 2t+36$

$v(t)=3t^2-24t+36$

(b) acceleration is given by

$a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}$

$a(t)=6t-24$

(c)at $s(3)=3^3-12(3)^2+36(3)$

$s(3)=27\ m$

at $s(4)=4^3-12(4)^2+36(4)$

$s(4)=16\ m$

at $t=3\ s$ Position is $27\ m$ and at $t=4\ s$ position is $16\ m$

therefore squirrel is moving down