Question

An elevator descends into a mine shaft at the rate of 5 m/mm. What will be its position after one hour? If it begins to descend from 35m above the ground, what will be its position after 30 minutes?

Give answer with formula

Please give answer ​

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.

Brainliest please