Question

An elevator in a mine shaft is stopped at a point 25 feet below ground level. It then travels 28 feet downward. What equation can be written to find the final position, in feet, of the elevator in relation to ground level?

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.