A gold mine has two elevators, one for equipment and another for the miners. The equipment elevator descends 66 feet per second. The elevator for the miners descends 1616 feet per second. One day, the equipment elevator begins to descend. After 2323 seconds, the elevator for the miners begins to descend. What is the position of each elevator relative to the surface after another 1616 seconds? At that time, which elevator is deeper? The position of the equipment elevator relative to the surface is negative 234-234 feet. The position of the elevator for the miners relative to the surface is nothing feet.

Question

Leland Hunter

Mathematics

13 July, 13:46

A gold mine has two elevators, one for equipment and another for the miners. The equipment elevator descends 66 feet per second. The elevator for the miners descends 1616 feet per second. One day, the equipment elevator begins to descend. After 2323 seconds, the elevator for the miners begins to descend. What is the position of each elevator relative to the surface after another 1616 seconds? At that time, which elevator is deeper? The position of the equipment elevator relative to the surface is negative 234-234 feet. The position of the elevator for the miners relative to the surface is nothing feet.

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Alice Lyons · Answer 1

Given that the equipment elevator descends at 6 feet per second.

The total number of seconds travelled by the equipment elevator is 23 + 16 = 39 seconds.

Thus the position of the equipment elevator relative to the surface is given by - (6 x 39) = - 234 feet.

Given that the miners elevator descends at 16 feet per second.

The total number of seconds travelled by the miners elevator is 16 seconds.

Thus, the position of the miners elevator relative to the surface is given by - (16 x 16) = - 256 feet.

At that time the miners' elevator is deeper.