Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 21

$$\eqalign{ & \left( a \right)s\left( t \right) = 2\sin \pi t,{\text{ }}t \geqslant 0 \cr & \left( b \right){\text{Graph}} \cr & \left( c \right)t = \frac{3}{2},{\text{ }}t = \frac{7}{2},{\text{ }}t = \frac{{11}}{2} \cr & \left( d \right)t = \frac{1}{2},{\text{ }}t = \frac{5}{2},{\text{ }}t = \frac{9}{2} \cr} $$

$$\eqalign{ & {\text{Let }}v\left( t \right) = 2\pi \cos \pi t,{\text{ for }}t \geqslant 0,{\text{ }}s\left( 0 \right) = 0 \cr & \cr & \left( a \right){\text{The position is an antiderivative of the velocity}}{\text{, we obtain}} \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {2\pi \cos \pi t} dt \cr & s\left( t \right) = 2\int {\pi \cos \pi t} dt \cr & s\left( t \right) = 2\sin \pi t + C,{\text{ }}\left( {\bf{1}} \right) \cr & {\text{To find the constant }}C{\text{ we use the initial condition }}s\left( 0 \right) = 0 \cr & 0 = 2\sin \pi \left( 0 \right) + C \cr & C = 0 \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & s\left( t \right) = 2\sin \pi t,{\text{ }}t \geqslant 0 \cr & \cr & \left( b \right){\text{Graph }}s\left( t \right) = 2\sin \pi t,{\text{ for 0}} \leqslant t \leqslant {\text{4 is shown below}} \cr & \cr & \left( c \right){\text{From the graph shown below}}{\text{, we can see that the mass}} \cr & {\text{reach its low point at }}t = 1.5{\text{s}}{\text{, }}t = 3.5{\text{s}}{\text{, and then }}t = 5.5{\text{s}} \cr & t = \frac{3}{2},{\text{ }}t = \frac{7}{2},{\text{ }}t = \frac{{11}}{2} \cr & \cr & \left( d \right){\text{From the graph shown below}}{\text{, we can see that the mass}} \cr & {\text{reach its high point at }}t = 0.5{\text{s}}{\text{, }}t = 2.5{\text{s}}{\text{, and then }}t = 4.5{\text{s}} \cr & t = \frac{1}{2},{\text{ }}t = \frac{5}{2},{\text{ }}t = \frac{9}{2} \cr & \cr & {\text{Graph}} \cr} $$