Let deq(t)= the depth of descend of the equipment elevator relative to the surface, dmin(t)= the depth of descend of the miner's elevator relative to the surface, t= the time of the descend of the miner's elevator. Then
deq(t)=4⋅28+4t
dmin(t)=15tWhat is the position of each elevator relative to the surface after another 14seconds?
deq(14)=112+4⋅14=168(ft)
dmin(14)=15⋅14=210(ft)
dmin(14)−deq(14)=210ft−168ft=42ft At that time the elevator for the miners is 42 feet deeper than the equipment elevator.
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