Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 408: 33

$$s\left( t \right) = \frac{{29}}{4} + 5t - \frac{1}{4}\cos 2t{\text{ and }}v\left( t \right) = 5 + \frac{1}{2}\sin 2t$$

$$\eqalign{ & a\left( t \right) = \cos 2t,\,\,\,\,v\left( 0 \right) = 5,\,\,\,\,s\left( 0 \right) = 7 \cr & {\text{Use theorem 6}}{\text{.2 }} \cr & {\text{Given the acceleration }}a\left( t \right){\text{ of an object moving along a line and its initial velocity }}v\left( 0 \right){\text{, }} \cr & {\text{the velocity of the object for future time }}t \geqslant 0{\text{ is }}v\left( t \right) = v\left( 0 \right) + \int_0^t {a\left( x \right)} dx \cr & {\text{let }}a\left( x \right) = \cos 2x{\text{ and }}v\left( 0 \right) = 5.{\text{ then }}v\left( t \right){\text{ is}} \cr & v\left( t \right) = 5 + \int_0^t {\cos 2x} dx \cr & v\left( t \right) = 5 + \int_0^t {\cos 2x} dx \cr & {\text{integrate using }}\int {\cos ax} dx = \frac{1}{a}\sin ax + C \cr & v\left( t \right) = 5 + \frac{1}{2}\left( {\sin 2x} \right)_0^t \cr & {\text{evaluate the limits}} \cr & v\left( t \right) = 5 + \frac{1}{2}\left( {\sin 2t - \sin 2\left( 0 \right)} \right) \cr & {\text{simplify}} \cr & v\left( t \right) = 5 + \frac{1}{2}\sin 2t \cr & \cr & {\text{Use theorem 6}}{\text{.1 }} \cr & {\text{Given the velocity }}v\left( t \right){\text{ of an object moving along a line and its initial position }}s\left( 0 \right){\text{, }} \cr & {\text{the position function of the object for future time }}t \geqslant 0{\text{ is }}s\left( t \right) = s\left( 0 \right) + \int_0^t {v\left( x \right)} dx \cr & {\text{let }}v\left( x \right) = 5 + \frac{1}{2}\sin 2x{\text{ and }}s\left( 0 \right) = 7.{\text{ then }}s\left( t \right){\text{ is}} \cr & s\left( t \right) = 7 + \int_0^t {\left( {5 + \frac{1}{2}\sin 2x} \right)} dx \cr & {\text{integrate using }}\int {\sin ax} dx = - \frac{1}{a}cosax + C \cr & s\left( t \right) = 7 + \left( {5x + \frac{1}{2}\left( { - \frac{1}{2}\cos 2x} \right)} \right)_0^t \cr & s\left( t \right) = 7 + \left( {5x - \frac{1}{4}\cos 2x} \right)_0^t \cr & {\text{evaluate the limits}} \cr & s\left( t \right) = 7 + \left( {5t - \frac{1}{4}\cos 2t} \right) - \left( {5\left( 0 \right) - \frac{1}{4}\cos 2\left( 0 \right)} \right) \cr & {\text{simplify}} \cr & s\left( t \right) = 7 + 5t - \frac{1}{4}\cos 2t - 0 + \frac{1}{4}\left( 1 \right) \cr & s\left( t \right) = \frac{{29}}{4} + 5t - \frac{1}{4}\cos 2t \cr} $$