Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 9

$$\eqalign{ & a.{\text{ Positive direction for }}\left( {0,3} \right] \cr & {\text{ Negative direction for }}\left( {3,6} \right] \cr & b.{\text{ 0m}} \cr & c.{\text{ }}18{\text{m}} \cr} $$

$$\eqalign{ & {\text{Let }}v\left( t \right) = 6 - 2t{\text{ on the interval }}0 \leqslant t \leqslant 6 \cr & \cr & a.{\text{ The graph of the velocity of the given interval is shown}} \cr & {\text{below }}\left( {{\text{Graph using geogebra}}} \right). \cr & *{\text{The motion of the object is in the positive direction when }} \cr & v\left( t \right) > 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) > 0{\text{ on the interval }}\left( {0,3} \right]. \cr & *{\text{The motion of the object is in the negative direction when }} \cr & v\left( t \right) < 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) < 0{\text{ on the interval }}\left( {3,6} \right]. \cr & \cr & b.{\text{ The displacement of the object between }}t = a{\text{ and }}t = b \cr & {\text{where }}a > b{\text{ is :}} \cr & s\left( b \right) - s\left( a \right) = \int_a^b {v\left( t \right)} dt,{\text{ }}\left( {{\text{see page 399 on the book}}} \right) \cr & {\text{Therefore}}{\text{,}} \cr & {\text{For the given interval }}0 \leqslant t \leqslant 6{\text{ }} \cr & s\left( 6 \right) - s\left( 0 \right) = \int_0^6 {\left( {6 - 2t} \right)} dt \cr & {\text{Integrating}} \cr & s\left( 6 \right) - s\left( 0 \right) = \left[ {6t - {t^2}} \right]_0^6 \cr & s\left( 6 \right) - s\left( 0 \right) = \left[ {6\left( 6 \right) - {{\left( 6 \right)}^2}} \right] - \left[ {6\left( 0 \right) - {{\left( 0 \right)}^2}} \right] \cr & s\left( 6 \right) - s\left( 0 \right) = 0 \cr & {\text{The displacement is 0}} \cr & \cr & c.{\text{ The distance traveled of the given interval is given by}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \int_0^6 {\left| {6 - 2t} \right|} dt \cr & {\text{By the definition of absolute value}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \int_0^3 {\left( {6 - 2t} \right)} dt + \int_3^6 {\left( {2t - 6} \right)} dt \cr & {\text{Integrating}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \left[ {6t - {t^2}} \right]_0^3 + \left[ {{t^2} - 6t} \right]_3^6 \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \left[ {6\left( 3 \right) - {{\left( 3 \right)}^2}} \right] + \left[ {{{\left( 6 \right)}^2} - 6\left( 6 \right)} \right] - \left[ {{{\left( 3 \right)}^2} - 6\left( 3 \right)} \right] \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 9 + 0 + 9 \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 18 \cr} $$