Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 10

$$\eqalign{ & a.{\text{ Positive direction for }}\left( {0,\frac{\pi }{2}} \right]{\text{ and }}\left( {\pi ,\frac{{3\pi }}{2}} \right] \cr & {\text{ Negative direction for }}\left( {\frac{\pi }{2},\pi } \right]{\text{ and }}\left( {\frac{{3\pi }}{2},2\pi } \right] \cr & b.{\text{ 0m}} \cr & c.{\text{ }}80{\text{m}} \cr} $$

$$\eqalign{ & {\text{Let }}v\left( t \right) = 10\sin 2t{\text{ on the interval }}0 \leqslant t \leqslant 2\pi \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 intervals }}\left( {0,\frac{\pi }{2}} \right]{\text{ and }}\left( {\pi ,\frac{{3\pi }}{2}} \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 intervals }}\left( {\frac{\pi }{2},\pi } \right]{\text{ and }}\left( {\frac{{3\pi }}{2},2\pi } \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 2\pi {\text{ }} \cr & s\left( {2\pi } \right) - s\left( 0 \right) = \int_0^{2\pi } {\left( {10\sin 2t} \right)} dt \cr & {\text{Integrating}} \cr & s\left( {2\pi } \right) - s\left( 0 \right) = 5\left[ { - \cos 2t} \right]_0^{2\pi } \cr & s\left( {2\pi } \right) - s\left( 0 \right) = - 5\left[ {\cos 2\left( {2\pi } \right) - \cos 2\left( 0 \right)} \right] \cr & s\left( {2\pi } \right) - s\left( 0 \right) = - 5\left[ {1 - 1} \right] \cr & s\left( {2\pi } \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^{2\pi } {\left| {10\sin 2t} \right|} dt \cr & {\text{By the definition of absolute value and properties of }} \cr & {\text{integration}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 4\int_0^{\pi /2} {\left( {10\sin 2t} \right)} dt \cr & {\text{Integrating}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 40\int_0^{\pi /2} {\left( {\sin 2t} \right)} dt \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 40\left[ { - \cos 2t} \right]_0^{\pi /2} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = - 40\left[ {\cos 2\left( {\frac{\pi }{2}} \right) - \cos 2\left( 0 \right)} \right] \cr & {\text{Simplifying}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = - 40\left[ {\cos \left( \pi \right) - \cos \left( 0 \right)} \right] \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = - 40\left[ { - 1 - 1} \right] \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = 80 \cr} $$