Answer to Question #143137 in Math for Bernita

Question

A gold mine had two elevators, one for equipment and one for miners. One day, the equipment elevator begins to descend. After 28 seconds, the elevator for the miners begins to descend. What is the position of each elevator relative to the surface after another 14seconds? At that time how much deeper is the elevator for the miners

Answer 1

Let $d_{eq}(t)=$ the depth of descend of the equipment elevator relative to the surface, $d_{min}(t)=$ 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

$$d_{eq}(t)=4\cdot28+4t$$

$$d_{min}(t)=15t$$

What is the position of each elevator relative to the surface after another 14seconds? 

$$d_{eq}(14)=112+4\cdot14=168(ft)$$

$$d_{min}(14)=15\cdot14=210(ft)$$

$$d_{min}(14)-d_{eq}(14)=210ft-168ft=42ft$$

At that time the elevator for the miners is 42 feet deeper than the equipment elevator.