Answer
$$v = {\text{14m/s}}$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 2t + 6,{\text{ for 0}} \leqslant t \leqslant 8 \cr
& {\text{The distance traveled by the object between }}t = a{\text{ and }}t = b, \cr
& b > a{\text{ is }}\int_a^b {\left| {v\left( t \right)} \right|} dt.{\text{ }}\left( {{\text{See page 399}}} \right) \cr
& {\text{Therefore}} \cr
& {\text{Let }}{d_t}{\text{ be the distance traveled}}{\text{, }}a = 0{\text{ and }}b = 8 \cr
& {d_t} = \int_0^8 {\left| {2t + 6} \right|} dt \cr
& {\text{By the definition of absolute value}} \cr
& \left| {2t + 6} \right| = 2t + 6,{\text{ }}t \geqslant -3 \cr
& \left| {2t + 6} \right| = - \left( {2t + 6} \right),{\text{ }}t < -3,{\text{ then}} \cr
& {d_t} = \int_0^8 {\left( {2t + 6} \right)} dt \cr
& {d_t} = \left[ {{t^2} + 6t} \right]_0^8 \cr
& {d_t} = \left[ {{{\left( 8 \right)}^2} + 6\left( 8 \right)} \right] - \left[ {{{\left( 0 \right)}^2} + 6\left( 0 \right)} \right] \cr
& {d_t} = 112{\text{m}} \cr
& \cr
& {\text{Recall from physics formulas that }}d = vt,{\text{ }} \cr
& d = vt \cr
& {\text{Then for a constant velocity}} \cr
& v = \frac{d}{t} \cr
& {\text{Let }}d = {d_t},{\text{ and }}t = \Delta t = 8 - 0 = 8{\text{s}} \cr
& v = \frac{{112m}}{{8{\text{s}}}} \cr
& v = {\text{14m/s}} \cr} $$
