Answer
$$28{\text{ units}}$$
Work Step by Step
$$\eqalign{
& x\left( t \right) = {t^3} - 6{t^2} + 9t - 2,{\text{ 0}} \leqslant t \leqslant {\text{5}} \cr
& {\text{Differentiating}} \cr
& x'\left( t \right) = \frac{d}{{dt}}\left[ {{t^3} - 6{t^2} + 9t - 2} \right] \cr
& x'\left( t \right) = 3{t^2} - 12t + 9 \cr
& {\text{Let }}x\left( t \right) = 0 \cr
& 3{t^2} - 12t + 9 = 0 \cr
& {t^2} - 4t - 3 = 0 \cr
& \left( {t - 3} \right)\left( {t - 1} \right) = 0 \cr
& 3{t^2} - 12t + 9 < 0{\text{ for }}\left( {1,3} \right) \cr
& 3{t^2} - 12t + 9 > 0{\text{ for }}\left( { - \infty ,1} \right] \cup \left[ {3,\infty } \right) \cr
& {\text{The total distance traveled is given by:}} \cr
& {T_d} = \int_0^5 {\left| {x'\left( t \right)} \right|} dt \cr
& {\text{Therefore,}} \cr
& {T_d} = \int_0^5 {\left| {3{t^2} - 12t + 9} \right|} dt \cr
& {T_d} = \int_0^1 {\left( {3{t^2} - 12t + 9} \right)} dt - \int_1^3 {\left( {3{t^2} - 12t + 9} \right)} dt \cr
& + \int_3^5 {\left( {3{t^2} - 12t + 9} \right)dt} \cr
& {\text{Integrating}} \cr
& {T_d} = \left[ {{t^3} - 6{t^2} + 9t} \right]_0^1 - \left[ {{t^3} - 6{t^2} + 9t} \right]_1^3 + \left[ {{t^3} - 6{t^2} + 9t} \right]_3^5 \cr
& {T_d} = 4 - \left( { - 4} \right) + 20 \cr
& {T_d} = 28 \cr} $$
