Answer
$$\left( {\text{a}} \right)d = 0{\text{ft}},{\text{ }}\left( {\text{b}} \right)T = 6{\text{ft}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the velocity }}v\left( t \right) = \cos t,{\text{ in feet per second, }} \cr
& {\text{for 0}} \leqslant t \leqslant 3\pi \cr
& \left( {\text{a}} \right){\text{The displacement in feet is given by:}} \cr
& d = \int_a^b {f\left( t \right)} dt \cr
& d = \int_0^{3\pi } {\cos t} dt \cr
& d = \left[ {\sin t} \right]_0^{3\pi } \cr
& d = \sin \left( {3\pi } \right) - \sin \left( 0 \right) \cr
& {\text{simplifying}} \cr
& d = 0 \cr
& \cr
& \left( {\text{b}} \right){\text{ The total distance traveled is given by:}} \cr
& T = \int_a^b {\left| {f\left( t \right)} \right|} dt \cr
& T = \int_0^{3\pi } {\left| {\cos t} \right|} dt \cr
& {\text{Where:}} \cr
& \cos t < 0{\text{ for }}\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right) \cup \left( {\frac{{5\pi }}{2},3\pi } \right) \cr
& \cos t \geqslant 0{\text{ for }}\left[ {0,\frac{\pi }{2}} \right] \cup \left[ {\frac{{3\pi }}{2},\frac{{5\pi }}{2}} \right] \cr
& {\text{then}} \cr
& T = \int_0^{\pi /2} {\cos t} dt - \int_{\pi /2}^{3\pi /2} {\cos t} dt + \int_{3\pi /2}^{5\pi /2} {\cos t} dt - \int_{5\pi /2}^{3\pi } {\cos t} dt \cr
& {\text{Integrating }} \cr
& T = \left[ {\sin t} \right]_0^{\pi /2} - \left[ {\sin t} \right]_{\pi /2}^{3\pi /2} + \left[ {\sin t} \right]_{3\pi /2}^{5\pi /2} - \left[ {\sin t} \right]_{5\pi /2}^{3\pi } \cr
& T = 1 - \left( { - 1 - 1} \right) + \left( {1 + 1} \right) - \left( {0 - 1} \right) \cr
& T = 6{\text{ft}} \cr} $$
