Question

A gold mine has two​ elevators, one for equipment and another for the miners. The equipment elevator descends 4 feet per second. The elevator for the miners descends 15 feet per second. One​ day, the equipment elevator begins to descend. After 30 ​seconds, the elevator for the miners begins to descend. What is the position of each elevator relative to the surface after another 14 seconds? At that​ time, which elevator is​ deeper?

Answer 1

Let $d_m,\ d_e$ represent the depth of the miner and equipments elevator, respectively. We know that equipment elevator descends with a velocity of $v_e = 4$ feet per second, and the miners one descends with a velocity of $v_m = 15$ feet per second.

If we start counting time ($t=0$) when the equipment elevator begins to descend, after $t$ seconds its depth will be

$d_e(t) = v_et = 4t$

On the other hand, the miner elevator follows the same rule, but we have to use its velocity, and remember that it starts with a delay of thirty seconds:

$d_m(t) = v_m(t-30) = 15(t - 30)$

Now, we have to wait the 30 seconds of delay, and then another 14 seconds. This means that we want to know the positions of both elevators when $t = 44$. Let's plug this value into the two equations:

$d_e(44) = 4\cdot 44 = 176$

$d_m(44) = 15(44-30) = 15\cdot 14 = 210$

So, the equipment elevator is 176 feet deep, and the miner elevator is 210 feet deep, and thus this is the deeper one.