Answer
$$\frac{{13}}{2}$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = \left| {t - 3} \right|;{\text{ 0}} \leqslant t \leqslant {\text{5}} \cr
& {\text{The displacement is given by:}} \cr
& s = \int_0^5 {\left| {t - 3} \right|dt} \cr
& {\text{By the definition of the absolute value}} \cr
& s = - \int_0^3 {\left( {t - 3} \right)} dt + \int_3^5 {\left( {t - 3} \right)} dt \cr
& {\text{Integrating}} \cr
& s = - \left[ {\frac{1}{2}{t^2} - 3t} \right]_0^3 + \left[ {\frac{1}{2}{t^2} - 3t} \right]_3^5 \cr
& s = - \left[ {\frac{1}{2}{{\left( 3 \right)}^2} - 3\left( 3 \right)} \right] + \left[ {\frac{1}{2}{{\left( 5 \right)}^2} - 3\left( 5 \right)} \right] - \left[ {\frac{1}{2}{{\left( 3 \right)}^2} - 3\left( 3 \right)} \right] \cr
& {\text{Simplifying}} \cr
& s = \frac{9}{2} - \frac{5}{2} + \frac{9}{2} \cr
& s = \frac{{13}}{2} \cr} $$
