$ 解:(1)設(shè)一次函數(shù)為y=kx+b$
$把(0,331)和(5,334)分別代入函數(shù)有$
$\begin{cases}{ b=331 }\ \\ {\ 5k+b=334} \end{cases}解得\begin{cases}{ k=\frac {3}{5} }\ \\ { b=331 } \end{cases}$
$∴y=\frac {3}{5}x+331$
$(2)把x=22代入函數(shù)有,y=\frac {3}{5}×22+331=344\frac {1}{5}$
$∴5×344\frac {1}{5}=1721(m)$
$∴相距1721m$
