Chapter 3 - Derivatives - 3.6 Derivatives as Rates of Change - 3.6 Exercises - Page 181: 11

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

$$\eqalign{ & f\left( t \right) = {t^2} - 4t;{\text{ 0}} \leqslant t \leqslant {\text{5}} \cr & \left( a \right){\text{ Graph below}} \cr & \left( b \right){\text{ }} \cr & {\text{Position }}s = {t^2} - 4t \cr & v = \frac{{ds}}{{dt}} \cr & v = \frac{d}{{dt}}\left[ {{t^2} - 4t} \right] \cr & v = 2t - 4 \cr & {\text{The object is stationary when }}v = 0 \cr & v = 2t - 4 \cr & 2t - 4 = 0 \cr & 2t = 4 \cr & t = 2 \cr & {\text{Stationary at }}t = 2 \cr & 2t - 4 > 0 \cr & 2t > 4 \cr & t > 2 \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant {\text{5}},{\text{ then}} \cr & {\text{Is moving to the right on: }}\left( {2,5} \right] \cr & 2t - 4 < 0 \cr & 2t < 4 \cr & t < 2 \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant {\text{5}},{\text{ then}} \cr & {\text{Is moving to the left on: }}\left[ {0,2} \right) \cr & \cr & \left( c \right) \cr & v = 2t - 4 \cr & a = \frac{{dv}}{{dt}} = 2 \cr & {\text{at }}t = 1 \cr & v\left( 1 \right) = 2\left( 1 \right) - 4 = - 2 \cr & a\left( 1 \right) = 2 \cr & \cr & \left( d \right){\text{ The velocity is 0 at }}t = 2 \cr & a\left( 2 \right) = 2 \cr & \left( e \right){\text{ The speed is increasing for }} \cr & {\text{speed}} = \left| {2t - 4} \right| \cr & {\text{Increasing on: }}\left( {2,5} \right] \cr} $$