Chapter 3 - Derivatives - 3.6 Derivatives as Rates of Change - 3.6 Exercises - Page 181: 15

$$\eqalign{ & \left( a \right){\text{Graphs}} \cr & \left( b \right){\text{Stationary at }}t = 2{\text{ and }}t = 5 \cr & {\text{Is moving to the righ on: }}\left[ {0,2} \right) and \left[ {5,6} \right)\cr & {\text{Is moving to the left on: }}\left( {2,5} \right) \cr & \left( c \right){\text{ }}v\left( 1 \right) = 24,{\text{ }}a\left( 1 \right) = - 30 \cr & \left( d \right){\text{ }}a\left( 5 \right) = 18 \cr & \left( e \right){\text{ Increasing on: }}\left( {2,\frac{7}{2}} \right){\text{ and }}\left( {5,6} \right] \cr} $$

$$\eqalign{ & f\left( t \right) = 2{t^3} - 21{t^2} + 60t;{\text{ 0}} \leqslant t \leqslant 6 \cr & \cr & \left( a \right){\text{ Graph below}} \cr & \cr & \left( b \right){\text{ }} \cr & {\text{Position }}s = 2{t^3} - 21{t^2} + 60t \cr & v = \frac{{ds}}{{dt}} \cr & v = \frac{d}{{dt}}\left[ {2{t^3} - 21{t^2} + 60t} \right] \cr & v = 6{t^2} - 42t + 60 \cr & {\text{The object is stationary when }}v = 0 \cr & 6{t^2} - 42t + 60 = 0 \cr & {t^2} - 7t + 10 = 0 \cr & \left( {t - 5} \right)\left( {t - 2} \right) = 0 \cr & t = 2,{\text{ }}t = 5 \cr & {\text{Stationary at }}t = 2{\text{ and }}t = 5 \cr & \cr & 6{t^2} - 42t + 60 > 0 \cr & \left( {t - 5} \right)\left( {t - 2} \right) > 0 \cr & {\text{Solving}} \cr & t < 2,{\text{ }}t > 5 \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant 6,{\text{ then}} \cr & 0 \leqslant t < 2,{\text{ }}5 < t \leqslant 6 \cr & {\text{Is moving to the righ on: }}\left[ {0,2} \right){\text{ and }}\left[ {5,6} \right),{\text{ and is moving}} \cr & {\text{to the left into the interval }}\left( {2,5} \right) \cr & \cr & \left( c \right) \cr & v = 6{t^2} - 42t + 60 \cr & a = \frac{{dv}}{{dt}} = 12t - 42 \cr & {\text{at }}t = 1 \cr & v\left( 1 \right) = 6{\left( 1 \right)^2} - 42\left( 1 \right) + 60 \cr & v\left( 1 \right) = 24 \cr & a\left( 1 \right) = 12\left( 1 \right) - 42 \cr & a\left( 1 \right) = - 30 \cr & \cr & \left( d \right){\text{ The velocity is 0 at }} \cr & 6{t^2} - 42t + 60 = 0 \cr & \left( {t - 5} \right)\left( {t - 2} \right) = 0 \cr & t = 2{\text{ and }}t = 5 \cr & a\left( 2 \right) = - 18,{\text{ }} \cr & a\left( 5 \right) = 18 \cr & \cr & \left( e \right){\text{ The speed is}} \cr & {\text{speed}} = \left| v \right| = \left| {6{t^2} - 42t + 60} \right| \cr & {\text{From the graph:}} \cr & {\text{Increasing on: }}\left( {2,\frac{7}{2}} \right){\text{ and }}\left( {5,6} \right] \cr} $$