An elevator descends into a mine shaft at the rate of 8m/min. If it starts descending from 5m above the ground level, how long will it take to reach -195m?

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Solution

Dear student,

speed of descent of the elevator = $8 m / m i n$
initial position of the elevator: 5 m above the ground level
final position of the elevator = 195 m below the ground level
time taken to reach there = $\frac{t o t a l d i s \tan c e}{s p e e d}$ $(s p e e d = \frac{d i s \tan c e}{t i m e})$
$= \frac{5 + 195}{8} = 25 m i n$

Hence, it will take 25 minute to reach -195 m form +5 m.

Regards

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