Chapter 3 - Derivatives - 3.6 Derivatives as Rates of Change - 3.6 Exercises - Page 181: 14

$$\eqalign{ & \left( a \right){\text{Graphs}} \cr & \left( b \right){\text{Stationary at }}t = 3,{\text{ }} \cr & {\text{Is moving to the righ on: }}\left[ {0,8} \right) \cr & {\text{Is moving to the left on: }}\left( {3,8} \right] \cr & \left( c \right){\text{ }}v\left( 1 \right) = 12,{\text{ }}a\left( 1 \right) = - 6 \cr & \left( d \right){\text{ }}a\left( 3 \right) = - 6 \cr & \left( e \right){\text{ Increasing on: }}\left( {3,8} \right] \cr} $$

$$\eqalign{ & f\left( t \right) = 18t - 3{t^2};{\text{ 0}} \leqslant t \leqslant 8 \cr & \left( a \right){\text{ Graph below}} \cr & \left( b \right){\text{ }} \cr & {\text{Position }}s = 18t - 3{t^2} \cr & v = \frac{{ds}}{{dt}} \cr & v = \frac{d}{{dt}}\left[ {18t - 3{t^2}} \right] \cr & v = 18 - 6t \cr & {\text{The object is stationary when }}v = 0 \cr & v = 18 - 6t \cr & 18 - 6t = 0 \cr & t = 3 \cr & {\text{Stationary at }}t = 3 \cr & 18 - 6t > 0 \cr & t < 3 \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant {\text{8}},{\text{ then}} \cr & {\text{Is moving to the righ on: }}\left[ {0,8} \right) \cr & 18 - 6t < 0 \cr & t > 3 \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant {\text{8}},{\text{ then}} \cr & {\text{Is moving to the left on: }}\left( {3,8} \right] \cr & \cr & \left( c \right) \cr & v = 18 - 6t \cr & a = \frac{{dv}}{{dt}} = - 6 \cr & {\text{at }}t = 1 \cr & v\left( 1 \right) = 18 - 6\left( 1 \right) \cr & v = 12 \cr & a\left( 1 \right) = - 6 \cr & \cr & \left( d \right){\text{ The velocity is 0 at }}t = 3 \cr & a\left( 3 \right) = - 6 \cr & \left( e \right){\text{ The speed is}} \cr & {\text{speed}} = \left| v \right| = \left| {18 - 6t} \right| \cr & {\text{Increasing on: }}\left( {3,8} \right] \cr} $$