Question

An elevator descends into a mine shaft at the rate of 8m/min. What will be
its position after 120 min?

If it begins to descend from 30 m above the ground, what will be its position
after 50 mins?

Answer 1

Answer:

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\ }\min

6

1

min

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

6

1

× 360 = 60 min

60 minutes = 1 hour

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

Answer 2

Answer:

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 = 1 / 6 min

Therefore, tie taken to a cover a distance of 360m = 1 / 6 × 360 = 60 min  

60 minutes = 1 hour

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