Answer
$$\left( {\text{a}} \right)d = \frac{3}{2}{\text{ft}},{\text{ }}\left( {\text{b}} \right)T = \frac{{113}}{{10}}{\text{ft}}$$
Work Step by Step
$$\eqalign{
& {\text{Let the velocity }}v\left( t \right) = 5t - 7,{\text{ in feet per second, for 0}} \leqslant t \leqslant {\text{3}} \cr
& \left( {\text{a}} \right){\text{The displacement in feet is given by:}} \cr
& d = \int_a^b {f\left( t \right)} dt \cr
& d = \int_0^3 {\left( {5t - 7} \right)} dt \cr
& d = \left[ {\frac{{5{t^2}}}{2} - 7t} \right]_0^3 \cr
& d = \left[ {\frac{{5{{\left( 3 \right)}^2}}}{2} - 7\left( 3 \right)} \right] - \left[ {\frac{{5{{\left( 0 \right)}^2}}}{2} - 7\left( 0 \right)} \right] \cr
& {\text{simplifying}} \cr
& d = \left( {\frac{{45}}{2} - 21} \right) - \left( 0 \right) \cr
& d = \frac{3}{2}{\text{ft}} \cr
& \cr
& \left( {\text{b}} \right){\text{ The total distance traveled is given by:}} \cr
& T = \int_a^b {\left| {f\left( t \right)} \right|} dt \cr
& T = \int_0^3 {\left| {5t - 7} \right|} dt \cr
& {\text{Where }}5t - 7 < 0{\text{ for }}t > \frac{7}{5}{\text{ and }}5t - 7 \geqslant {\text{ for }}t \geqslant \frac{7}{5},{\text{ then}} \cr
& T = \int_0^3 {\left| {5t - 7} \right|} dt = \int_0^{7/5} {\left[ { - \left( {5t - 7} \right)} \right]} dt + \int_{7/5}^3 {\left( {5t - 7} \right)} dt \cr
& T = \int_0^{7/5} {\left( {7 - 5t} \right)} dt + \int_{7/5}^3 {\left( {5t - 7} \right)} dt \cr
& T = \left[ {7t - \frac{{5{t^2}}}{2}} \right]_0^{7/5} + \left[ {\frac{{5{t^2}}}{2} - 7t} \right]_{7/5}^3 \cr
& {\text{Evaluating}} \cr
& T = \left( {\frac{{49}}{{10}} - 0} \right) + \left( {\frac{3}{2} + \frac{{49}}{{10}}} \right) \cr
& T = \frac{{113}}{{10}}{\text{ft}} \cr} $$
