Answer
$v\left( t\right) =v_{0}+\dfrac {1}{m}\int ^{t}_{0}F\left( t\right) \partial t=v_{0}+\dfrac {1}{m}\left( A_{1}-A_{2}\right) =3\dfrac {m}{s}+\dfrac {1}{3kg}\times15N\times s=8\dfrac {m}{s}$
Work Step by Step
Lets write Newton's second law for the ice $\sum \overrightarrow {F}=m\overrightarrow {a}\Rightarrow \overrightarrow {a}=\dfrac {\sum \overrightarrow {F}}{m}(1)$ İn order to find change in speed of ice : $\Delta v=\int ^{t}_{0}a\left( t\right) \partial t=v\left( t\right) -v_{0}\Rightarrow v\left( t\right) =v_{0}+\int ^{t}_{0}a\left( t\right) \partial t(2)$ İf we combine (1) and (2) we get: $v\left( t\right) =v_{0}+\int ^{t}_{0}a\left( t\right) \partial t=v_{0}+\dfrac {1}{m}\int ^{t}_{0}F\left( t\right) \partial t(3)$ $\int ^{t}_{0}F\left( t\right) \partial t$ will equal to area under the graph. The area above th $x$ line will be marked as positive and area below the $x$ line will be marked as negative so lets simplify (3) and we get $v\left( t\right) =v_{0}+\dfrac {1}{m}\int ^{t}_{0}F\left( t\right) \partial t=v_{0}+\dfrac {1}{m}\left( A_{1}-A_{2}\right) $ $A_{1}$ and $A_{2}$ are the areas shown in the picture ($A_{1}=x+y;A_{2}=\alpha +\beta $) $v_{0}=3\dfrac {m}{s};t=11s;m=3kg$ So if we calculate ve get $v\left( t\right) =v_{0}+\dfrac {1}{m}\int ^{t}_{0}F\left( t\right) \partial t=v_{0}+\dfrac {1}{m}\left( A_{1}-A_{2}\right) =3\dfrac {m}{s}+\dfrac {1}{3kg}\times15N\times s=8\dfrac {m}{s}$
