Answer
$$ 0.74 \mathrm{m} \cdot \mathrm{s}^{-1}=\quad v_{b}$$
Work Step by Step
As the belt is moving with a constant speed, the time that the belt takes is the distance that the belt move over the time of this displacement:
$$
v_{b} t_{b}=d_{b}
$$
where letter $b$ indicates the belt. A Clifford can cover the same distance that the belt moves, in a time of $t_{b} / 4=t_{c};$ since it mores with an acceleration of $0.37 \mathrm{m} \cdot \mathrm{s}^{-2}=a$ from rest $0=\mathrm{v}_{0},$ the motion of the Clifford is described by this equation of motion:
$$
\begin{array}{c}
v_{0} t_{c}+\frac{1}{2} a t_{c}^{2}=d_{c} \\
\frac{1}{2} a t_{c}^{2}=d_{c} \\
\frac{1}{2} a\left(\frac{t_{b}}{4}\right)^{2}=d_{c}
\end{array}
$$
where letter $b$ indicates the Clifford, but since they cover the same distance $d_{c_{9}}=d_{b}$ so:
$$
\begin{array}{c}
d_{c}=d_{b} \\
\frac{1}{2} a\left(\frac{t_{b}}{4}\right)^{2}=v_{b} t_{b} \\
\frac{1}{32} a t_{b}=v_{b}
\end{array}
$$
Thus:
$$
\begin{aligned}
\frac{1}{32}\left(0.37 \mathrm{m} \cdot \mathrm{s}^{-2}\right)(64 \mathrm{s})=\tau_{b} & \\
&=0.74 \mathrm{m} \cdot \mathrm{s}^{-1} \\
& 0.74 \mathrm{m} \cdot \mathrm{s}^{-1}=\quad v_{b}
\end{aligned}
$$
