An elevator descends into a mine shaft at the rate of 6 m/min. If the

1.	An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 above the ground level, how long will it take to reach -350 m?
Answer» Starting position of mine shaft is 10 m above the ground but it moves in opposite direction so it travels the distance (–350) m below the ground. So total distance covered by mine shaft = 10 m – (–350) m = 10 + 350 = 360 m Now, time taken to cover a distance of 6 m by it = 1 minute So, time taken to cover a distance of 1 m by it = 1/6minute Therefore, time taken to cover a distance of 360 m = 1/6 x 360 = 60 minutes = 1 hour (Since 60 minutes = 1 hour) Thus, in one hour the mine shaft reaches –350 below the ground.

An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 above the ground level, how long will it take to reach -350 m?

Answer»

Starting position of mine shaft is 10 m above the ground but it moves in opposite direction so it travels the distance (–350) m below the ground.

So total distance covered by mine shaft = 10 m – (–350) m = 10 + 350 = 360 m

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

Therefore, time taken to cover a distance of 360 m = 1/6 x 360 = 60 minutes = 1 hour

(Since 60 minutes = 1 hour)

Thus, in one hour the mine shaft reaches –350 below the ground.

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An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 above the ground level, how long will it take to reach -350 m?