Answer
$$\left( {\text{a}} \right)d = \frac{{125}}{{12}}{\text{ft}},{\text{ }}\left( {\text{b}} \right)T = \frac{{253}}{{12}}{\text{ft}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the velocity }}v\left( t \right) = {t^3} - 8{t^2} + 15t,{\text{ in feet per second, }} \cr
& {\text{for 0}} \leqslant t \leqslant 5 \cr
& \left( {\text{a}} \right){\text{The displacement in feet is given by:}} \cr
& d = \int_a^b {f\left( t \right)} dt \cr
& d = \int_0^5 {\left( {{t^3} - 8{t^2} + 15t} \right)} dt \cr
& d = \left[ {\frac{{{t^4}}}{4} - \frac{{8{t^3}}}{3} + \frac{{15{t^2}}}{2}} \right]_0^5 \cr
& d = \left[ {\frac{{{{\left( 5 \right)}^4}}}{4} - \frac{{8{{\left( 5 \right)}^3}}}{3} + \frac{{15{{\left( 5 \right)}^2}}}{2}} \right] - \left[ {\frac{{{{\left( 0 \right)}^4}}}{4} - \frac{{8{{\left( 0 \right)}^3}}}{3} + \frac{{15{{\left( 0 \right)}^2}}}{2}} \right] \cr
& {\text{simplifying}} \cr
& d = \frac{{125}}{{12}} - 0 \cr
& d = \frac{{125}}{{12}}{\text{ft}} \cr
& \cr
& \left( {\text{b}} \right){\text{ The total distance traveled is given by:}} \cr
& T = \int_a^b {\left| {f\left( t \right)} \right|} dt \cr
& T = \int_0^5 {\left| {{t^3} - 8{t^2} + 15t} \right|} dt \cr
& {t^3} - 8{t^2} + 15t = 0 \cr
& t\left( {t - 3} \right)\left( {t - 5} \right) = 0 \cr
& {\text{Where:}} \cr
& {t^3} - 8{t^2} + 15t < 0{\text{ for }}\left( { - \infty ,0} \right) \cup \left( {3,5} \right) \cr
& {t^2} - t - 12 \geqslant 0{\text{ for }}\left[ {0,3} \right] \cup \left[ {5,\infty } \right) \cr
& {\text{then}} \cr
& T = \int_0^3 {\left( {{t^3} - 8{t^2} + 15t} \right)} dt - \int_3^5 {\left( {{t^3} - 8{t^2} + 15t} \right)} dt \cr
& {\text{Integrating each one by a calculator :}} \cr
& T = \frac{{63}}{4} + \frac{{16}}{3} \cr
& T = \frac{{253}}{{12}}{\text{ft}} \cr} $$
