Answer
$$v = \frac{{25}}{3}{\text{m/s}}$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = t{\left( {25 - {t^2}} \right)^{1/2}},{\text{ for 0}} \leqslant t \leqslant 5 \cr
& {\text{The distance traveled by the object between }}t = a{\text{ and }}t = b, \cr
& b > a{\text{ is }}\int_a^b {\left| {v\left( t \right)} \right|} dt.{\text{ }}\left( {{\text{See page 399}}} \right) \cr
& {\text{Therefore}} \cr
& {\text{Let }}{d_t}{\text{ the distance traveled}}{\text{, }}a = 0{\text{ and }}b = 5 \cr
& {d_t} = \int_0^5 {\left| {t{{\left( {25 - {t^2}} \right)}^{1/2}}} \right|} dt \cr
& {\text{By the definition of absolute value}} \cr
& {d_t} = \int_0^5 {t{{\left( {25 - {t^2}} \right)}^{1/2}}} dt \cr
& {d_t} = - \frac{1}{2}\int_0^5 {{{\left( {25 - {t^2}} \right)}^{1/2}}} \left( { - 2t} \right)dt \cr
& {\text{Integrating}} \cr
& {d_t} = - \frac{1}{2}\left[ {\frac{{{{\left( {25 - {t^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^5 \cr
& {d_t} = - \frac{1}{3}\left[ {{{\left( {25 - {t^2}} \right)}^{3/2}}} \right]_0^5 \cr
& {d_t} = - \frac{1}{3}\left[ {{{\left( {25 - {5^2}} \right)}^{3/2}} - {{\left( {25 - {0^2}} \right)}^{3/2}}} \right] \cr
& {d_t} = - \frac{1}{3}\left[ {{{\left( 0 \right)}^{3/2}} - 125} \right] \cr
& {d_t} = \frac{{125}}{3}{\text{m}} \cr
& \cr
& {\text{Recall from physics formulas that }}d = vt,{\text{ }} \cr
& d = vt \cr
& {\text{Then for a constant velocity}} \cr
& v = \frac{d}{t} \cr
& {\text{Let }}d = {d_t},{\text{ and }}t = \Delta t = 5 - 0 = 5{\text{s}} \cr
& v = \frac{{\frac{{125}}{3}{\text{m}}}}{{{\text{5s}}}} \cr
& v = \frac{{25}}{3}{\text{m/s}} \cr} $$
