Answer
$$v = \frac{2}{3}{\text{m/s}}$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = 1 - \frac{{{t^2}}}{{16}},{\text{ for 0}} \leqslant t \leqslant 4 \cr
& {\text{The distance traveled by the object between }}t = a{\text{ and }}t = b, \cr
& b > a{\text{ is }}\int_a^b {\left| {v\left( t \right)} \right|} dt.{\text{ }}\left( {{\text{See page 399}}} \right) \cr
& {\text{Therefore}} \cr
& {\text{Let }}{d_t}{\text{ the distance traveled}}{\text{, }}a = 0{\text{ and }}b = 4 \cr
& {d_t} = \int_0^4 {\left| {1 - \frac{{{t^2}}}{{16}}} \right|} dt \cr
& {\text{By the definition of absolute value}} \cr
& {d_t} = \int_0^4 {\left( {1 - \frac{{{t^2}}}{{16}}} \right)} dt \cr
& {\text{Integrating}} \cr
& {d_t} = \left[ {t - \frac{{{t^3}}}{{48}}} \right]_0^4 \cr
& {d_t} = \left[ {\left( 4 \right) - \frac{{{{\left( 4 \right)}^3}}}{{48}}} \right] - \left[ {\left( 0 \right) - \frac{{{{\left( 0 \right)}^3}}}{{48}}} \right] \cr
& {d_t} = \frac{8}{3}{\text{m}} \cr
& \cr
& {\text{Recall from physics formulas that }}d = vt,{\text{ }} \cr
& d = vt \cr
& {\text{Then for a constant velocity}} \cr
& v = \frac{d}{t} \cr
& {\text{Let }}d = {d_t},{\text{ and }}t = \Delta t = 4 - 0 = 4{\text{s}} \cr
& v = \frac{{\frac{8}{3}{\text{m}}}}{{{\text{4s}}}} \cr
& v = \frac{2}{3}{\text{m/s}} \cr} $$
