Решение.\[ \begin{align}
& \vec{\upsilon }(t)=A\cdot {{(\frac{t}{\tau })}^{3}}\cdot \vec{i}+B\cdot {{(\frac{t}{\tau })}^{5}}\cdot \vec{j},\vec{\upsilon }(t)=3\cdot {{(\frac{t}{1})}^{3}}\cdot \vec{i}+4\cdot {{(\frac{t}{1})}^{5}}\cdot \vec{j}, \\
& \vec{\upsilon }(t)=3\cdot {{t}^{3}}\cdot \vec{i}+4\cdot {{t}^{5}}\cdot \vec{j}(1). \\
& x={{x}_{0}}+\int\limits_{0}^{t}{3\cdot {{t}^{3}}dt}={{x}_{0}}+\left. 3\cdot \frac{1}{4}\cdot {{t}^{4}} \right|_{0}^{t}={{x}_{0}}+\frac{3\cdot {{t}^{4}}}{4},{{x}_{0}}=0,x=\frac{3\cdot {{t}^{4}}}{4}(2). \\
& y={{y}_{0}}+\int\limits_{0}^{t}{4\cdot {{t}^{5}}dt}={{y}_{0}}+\left. 4\cdot \frac{1}{6}\cdot {{t}^{6}} \right|_{0}^{t}={{y}_{0}}+\frac{4\cdot {{t}^{6}}}{6},{{y}_{0}}=0,y=\frac{4\cdot {{t}^{6}}}{6}=\frac{2\cdot {{t}^{6}}}{3}(3). \\
& s=\sqrt{{{(\frac{3\cdot {{t}^{4}}}{4})}^{2}}+{{(\frac{2\cdot {{t}^{6}}}{3})}^{2}}}(4). \\
& s=\sqrt{{{(\frac{3\cdot {{1}^{4}}}{4})}^{2}}+{{(\frac{2\cdot {{1}^{6}}}{3})}^{2}}}=1. \\
\end{align}
\]
Ответ: д) 1,00 м.