An elevator descends into a mine shaft
at the rate of 6 min. If the descent starts
brom 20m above the ground level. how
will it take to reach - 370m?
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Answer:
Answer: 65 minutes or 1 hour and 5 minutes. Descends starts from 20 m above the ground level
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Let the position of elevator above and below the ground level be denoted by +x an -x respectively.
Initial position of elevator = +20m
Final position of elevator =-370m
Distance between the two points
= {20-(-370)}m = {370+20}m = 390m
Speed of the elevator = 6m/min
Thus, time taken by the lift to reach from 20m to -370m = Distance/Speed = (390/6)min
=65 min = 1h 5 min
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