Chapter 3 - Derivatives - 3.6 Derivatives as Rates of Change - 3.6 Exercises - Page 181: 13

$$\eqalign{ & \left( a \right){\text{Graphs}} \cr & \left( b \right){\text{Stationary at }}t = \frac{9}{4},{\text{ }} \cr & {\text{Is moving to the righ on: }}\left( {\frac{9}{4},3} \right] \cr & {\text{Is moving to the left on: }}\left[ {0,\frac{9}{4}} \right) \cr & \left( c \right){\text{ }}v\left( 1 \right) = - 5,{\text{ }}a\left( 1 \right) = - 4 \cr & \left( d \right){\text{ }}a\left( {\frac{9}{4}} \right) = - 4 \cr & \left( e \right){\text{ Increasing on: }}\left( {\frac{9}{4},3} \right] \cr} $$

$$\eqalign{ & f\left( t \right) = 2{t^2} - 9t + 12;{\text{ 0}} \leqslant t \leqslant {\text{3}} \cr & \left( a \right){\text{ Graph below}} \cr & \left( b \right){\text{ }} \cr & {\text{Position }}s = 2{t^2} - 9t + 12 \cr & v = \frac{{ds}}{{dt}} \cr & v = \frac{d}{{dt}}\left[ {2{t^2} - 9t + 12} \right] \cr & v = 4t - 9 \cr & {\text{The object is stationary when }}v = 0 \cr & v = 4t - 9 \cr & 4t - 9 = 0 \cr & t = \frac{9}{4} \cr & {\text{Stationary at }}t = \frac{9}{4} \cr & 4t - 9 > 0 \cr & t > \frac{9}{4} \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant 3,{\text{ then}} \cr & {\text{Is moving to the right on: }}\left( {\frac{9}{4},3} \right] \cr & 4t - 9 < 0 \cr & t < \frac{9}{4} \cr & {\text{The object is moving on the interval 0}} \leqslant t \leqslant 3,{\text{ then}} \cr & {\text{Is moving to the left on: }}\left[ {0,\frac{9}{4}} \right) \cr & \cr & \left( c \right) \cr & v = 4t - 9 \cr & a = \frac{{dv}}{{dt}} = 2 \cr & {\text{at }}t = 1 \cr & v\left( 1 \right) = 4\left( 1 \right) - 9 \cr & v = - 5 \cr & a\left( 1 \right) = 4 \cr & \cr & \left( d \right){\text{ The velocity is 0 at }}t = \frac{9}{4} \cr & a\left( {\frac{9}{4}} \right) = 4 \cr & \left( e \right){\text{ The speed is}} \cr & {\text{speed}} = \left| v \right| = \left| {4t - 9} \right| \cr & {\text{Increasing on: }}\left( {\frac{9}{4},3} \right] \cr} $$