TITLE: calculus, velocity, time acceleration
QUESTION [0 upvotes]: The motion of a body is modelled by the following relationship:
$$s = t^3 - 3t^2 + 3t + 8$$
Where: s = distance in meters, and t = time in seconds
Use calculus to determine the following:
a) the velocity of the body at the end of 3 seconds
b) the time when the body has zero velocity
c) By finding the second derivative of the above relationship the acceleration of the body after 2 seconds
d) when  the bodies acceleration is zero
This is my answer to a). I am trying to complete all the questions.. I am trying to learn this. Is my answer correct? If not how have i gone wrong?
$$ v(t) = s(t) = t^3 - 3t^2 + 3t + 8$$
$$ v(t) = s(t) = 3t^2 - 6t + 3$$
$$ v(3) = s(3) = 3(3)^2 - 6(3) + 3$$
$$ = 15m/sec$$

REPLY [0 votes]: $a)$ $v(t) = \dot{s}(t) = 3t^2 - 6t + 3 $, so $ v(t = 3) = 3 \cdot 9 - 6 \cdot 3 + 3 =  12$ (with units $ms^{-1}$)
$b)$ Solve: $v(t) = 0$, which is $3t^2 - 6t + 3 =0$ or equivalently $t^2 - 2t + 1 = 0$. Factorisation leads to $(t - 1)^2 = 0$, so the solution is $t = 1$ (with units $s$) 
$c)$ $a(t) = \ddot{s}(t) = 6t - 6$, so $a(t = 2) = 6 \cdot 2 - 6 = 6$ (with units $ms^{-2}$) 
$d)$ Solve: $a(t) = 0$, which is $6t - 6 = 0$. So the solution is $t = 1$ (with units $s$)
Note: I use the dot to denote once differentiating $s$ with respect to time $t$.
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