Answer
$$v = \frac{4}{\pi }{\text{m/s}}$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 2\sin t,{\text{ for 0}} \leqslant t \leqslant \pi \cr
& {\text{The distance traveled by the object between }}t = a{\text{ and }}t = b, \cr
& b > a{\text{ is }}\int_a^b {\left| {v\left( t \right)} \right|} dt.{\text{ }}\left( {{\text{See page 399}}} \right) \cr
& {\text{Therefore}} \cr
& {\text{Let }}{d_t}{\text{ be the distance traveled}}{\text{, }}a = 0{\text{ and }}b = \pi \cr
& {d_t} = \int_0^\pi {\left| {2\sin t} \right|} dt \cr
& {\text{By the definition of absolute value}} \cr
& {d_t} = \int_0^\pi {2\sin t} dt \cr
& {\text{Integrating}} \cr
& {d_t} = \left[ { - 2\cos t} \right]_0^\pi \cr
& {d_t} = - 2\cos \left( \pi \right) + 2\cos \left( 0 \right) \cr
& {d_t} = 2 + 2 \cr
& {d_t} = 4{\text{m}} \cr
& \cr
& {\text{Recall from physics formulas that }}d = vt,{\text{ }} \cr
& d = vt \cr
& {\text{Then for a constant velocity}} \cr
& v = \frac{d}{t} \cr
& {\text{Let }}d = {d_t},{\text{ and }}t = \Delta t = \pi - 0 = \pi {\text{s}} \cr
& v = \frac{{4{\text{m}}}}{{\pi {\text{s}}}} \cr
& v = \frac{4}{\pi }{\text{m/s}} \cr} $$