Answer
a. $t = 0$ and $t = 5$
b. $t \approx 3.08$
Work Step by Step
$s = t^{4} - 4t^{3} - 20t^{2} + 20t, t\geq0$ $s' = 20 \frac{m}{s}$
a. $s' = \frac{d(t^{4})}{dt} - \frac{d(4t^{3})}{dt} - \frac{d(20t^{2})}{dt} + \frac{d(20t)}{dt}$
$s' = 4t^{3} - 12t^{2} - 40t + 20$
$20 = 4(t)^{3} - 12(20)^{2} - 40t + 20$
$4t^{3} - 12(t)^{2} - 40t + 20-20 = 0$
$4t^{3} - 12t^{2} - 40t = 0$
Now simplify by $4t$:
$4t(t^{2} - 3t - 10)= 0$
$t^{2} - 3t - 10 = 0$
Now factor to zero:
$4t(t - 5)(t+2)$
$t = 0, t = 5, t = -2$
We eliminate $t = -2$ because the problem says that $t\geq0$
So the answers are: $t = 0$ and $t = 5$
b. Find the time of acceleration at 0:
First find the second derivative of $f(x)$
$f'(x) = 4t^{3} - 12t^{2} - 40t$
$f''(x) = \frac{d(4t^{3})}{dt} - \frac{d(12t^{2})}{dt} - \frac{d(40t)}{dt}$
$f''(x) = 12t^{2} - 24t - 40$
Now divide by $4$:
$f''(x) = 4(3t^{2} - 6t - 10)$
Now equal to $0$:
$4(3t^{2} - 6t - 10) = 0$
To factorize use the Quadratic formula:
$\begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}$
$a = 3$, $b = -6$ and $c = -10$
$\begin{array}{*{20}c} {t = \frac{{ - (-6) \pm \sqrt {(-6)^2 - 4(3)(-10)} }}{{2(3)}}} \\ \end{array}$
$\begin{array}{*{20}c} {t = \frac{{ 6 \pm \sqrt {36 - 4(-30)} }}{{6}}} \\ \end{array}$
$\begin{array}{*{20}c} {t = \frac{{ 6 \pm \sqrt {36 +120} }}{{6}}} \\ \end{array}$
$\begin{array}{*{20}c} {t = \frac{{ 6 \pm \sqrt {156} }}{{6}}} \\ \end{array}$
Before simplifying we eliminate the negative solution because the problem says that $t\geq0$.
$\begin{array}{*{20}c} {t = \frac{{ 6 + \sqrt {156} }}{{6}}} \\ \end{array}$
$\begin{array}{*{20}c} {t = \frac{{ 6 + \sqrt {39(4)} }}{{6}}} \\ \end{array}$
$\begin{array}{*{20}c} {t = \frac{{ 6 + 2\sqrt {39} }}{{6}}} \\ \end{array}$
$\begin{array}{*{20}c} {t = \frac{{ 3 + \sqrt {39} }}{{2}}} \\ \end{array}\approx3.08$
