Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 14

$$\eqalign{ & a.{\text{ Positive direction for }}\left[ {0,4} \right] \cr & b.{\text{ }}25\left( {1 - {e^{ - 8}}} \right){\text{m}} \cr & c.{\text{ }}25\left( {1 - {e^{ - 8}}} \right){\text{m}} \cr} $$

$$\eqalign{ & {\text{Let }}v\left( t \right) = 50{e^{ - 2t}}{\text{ on the interval }}0 \leqslant t \leqslant 4 \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 the }}v\left( t \right){\text{ is always}} \cr & {\text{positive}} \cr & v\left( t \right) > 0{\text{ on the intervals }}\left[ {0,4} \right] \cr & *{\text{The motion of the object is in the negative direction when }} \cr & v\left( t \right) < 0.{\text{ }}v\left( t \right){\text{ never is negative}}{\text{.}} \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 4 \cr & s\left( 4 \right) - s\left( 0 \right) = \int_0^4 {50{e^{ - 2t}}} dt \cr & {\text{Integrating}} \cr & s\left( 4 \right) - s\left( 0 \right) = - \frac{{50}}{2}\left[ {{e^{ - 2t}}} \right]_0^4 \cr & s\left( 4 \right) - s\left( 0 \right) = - 25\left[ {{e^{ - 2\left( 4 \right)}} - {e^{ - 2\left( 0 \right)}}} \right] \cr & s\left( 4 \right) - s\left( 0 \right) = - 25\left( {{e^{ - 8}} - 1} \right) \cr & s\left( 4 \right) - s\left( 0 \right) = 25\left( {1 - {e^{ - 8}}} \right) \cr & {\text{The displacement is }}25\left( {1 - {e^{ - 8}}} \right){\text{m}} \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_0^4 {\left| {50{e^{ - 2t}}} \right|} dt \cr & {\text{By the definition of absolute value and using the graph}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \int_0^4 {50{e^{ - 2t}}} dt \cr & {\text{Integrating }} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 25\left( {1 - {e^{ - 8}}} \right){\text{m}} \cr} $$