Question

An elevator descends at the rate of 5 m/min. If the elevator is present at 20 m above the ground level, how long will it take to reach – 125 m.

Answer 1

Starting position of mine shaft is = 10m above ground

but,

it moves in opposite direction so it travels the distance (-350 m ) below the ground.

Total distance covered by mine shaft = 10m - (-350m)

= 10 + 350

= 360m

Now, taken to cover a distance of 6m by it = 1 minute

Time taken to cover a distance of 1m by it = $\frac{1}{6\ }$

Therefore, tie taken to a cover a distance of 360m = $\frac{1}{6}\times\ 360$= 60 min

60 minutes = 1 hour

thus, in one hour the mine shraft reaches = 350 below the ground.