A cage carries miners down a vertical shaft.

Step 1. Going down. $v_{\text{av}} = \dfrac{-250\ \text{m} - (-40\ \text{m})}{25\ \text{s} - 5\ \text{s}} = -10.5\ \text{m/s}$: $10.5\ \text{m/s}$ downwards.

Step 2. The whole trip, velocity. $\dfrac{-100\ \text{m} - (-40\ \text{m})}{60\ \text{s} - 5\ \text{s}} = -1.09\ \text{m/s}$.

Step 3. The whole trip, speed. Path length $210\ \text{m} + 150\ \text{m} = 360\ \text{m}$, so $\dfrac{360\ \text{m}}{55\ \text{s}} = 6.55\ \text{m/s}$.