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?
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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?
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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:
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.