## Calculus: Early Transcendentals 8th Edition

a. $t = 0$ and $t = 5$ b. $t \approx 3.08$
$s = t^{4} - 4t^{3} - 20t^{2} + 20t, t\geq0$ $s' = 20 \frac{m}{s}$ 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 $t\geq0$. $\begin{array}{*{20}c} {t = \frac{{ 6 + \sqrt {156} }}{{6}}} \\ \end{array}$ $\begin{array}{*{20}c} {t = \frac{{ 6 + \sqrt {39(4)} }}{{6}}} \\ \end{array}$ $\begin{array}{*{20}c} {t = \frac{{ 6 + 2\sqrt {39} }}{{6}}} \\ \end{array}$ $\begin{array}{*{20}c} {t = \frac{{ 3 + \sqrt {39} }}{{2}}} \\ \end{array}\approx3.08$