解：$(1)$全线长约$s=40\ \mathrm {km}$，若地铁全程的平均速度为$40\ \mathrm {km/h}$，由$v=\frac {s}{t} $可得，全程的运行时间$t=\frac {s}{v}=\frac {40\ \mathrm {km}}{40\ \mathrm {km/h}}=1\ \mathrm {h} $
$(2)$最高速度$ v' =80\ \mathrm {km/h}$，地铁以最高速度行驶时间$ t ' =3 \mathrm {\mathrm {min}} =\frac {1}{20}\ \mathrm {h}$，由$v=\frac {s}{t} $可得，通过的路程$s'=v't'=80\ \mathrm {km/h}× \frac {1}{20}\ \mathrm {h}=4\ \mathrm {km }$
$(3)A$站到$B$站运行时间$t_{AB}=3 \mathrm {\mathrm {min}} 20\ \mathrm {s}=\frac {1}{18}\ \mathrm {h}$，由$v=\frac {s}{t} $可得，$A$站到$B$站的路程$s_{AB}=v_{AB}t_{AB}=36\ \mathrm {km/h}× \frac {1}{18}\ \mathrm {h}=2\ \mathrm {km}$；$B$站到$C$站运行时间$t_{BC}=2 \mathrm {\mathrm {min}} 40\ \mathrm {s}=\frac {2}{45}\ \mathrm {h}$，由$ v=\frac {s}{t} $可得，$B$站到$C$站的路程$s_{BC}=v_{BC}t_{BC}=54\ \mathrm {km/h}× \frac {2}{45}\ \mathrm {h}=2.4\ \mathrm {km}$；总路程$s_{总}=s_{AB}+s_{BC}=2\ \mathrm {km}+2.4\ \mathrm {km}=4.4\ \mathrm {km}$，在$B$站停车$t''=40\ \mathrm {s}=\frac {1}{90}\ \mathrm {h}$，总时间$t_{总}=t_{AB}+t''+t_{BC}=\frac {1}{18}\ \mathrm {h}+ \frac {1}{90}\ \mathrm {h}+ \frac {2}{45}\ \mathrm {h}=\frac {1}{9}\ \mathrm {h}$，地铁从$A$站到$C$站的平均速度$v $平均$=\frac {s_{总}}{t_{总}}=\frac {4.4\ \mathrm {km}}{\frac {1}{9}\ \mathrm {h}}=39.6\ \mathrm {kmh}$
