Question

an elevator descends into a mine shaft at the rate of 6m/ min. if the descent starts from 10m above the ground level, how long it will take to reach 350m deep under the ground​

Answer 1

Answer:

total distance the elevator has to cover = 360 metre

velocity = 6m/min   =  6/60 m/s   =   0.1 m/s

Use the formula t= S/v  

         t = 360/0.1    

          t = 3600 seconds  = 60 minutes   = 1 hour

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Answer 2

Solution:-

From the question,

The initial height of the elevator = 10 m

Final depth of elevator = – 350 m … [∵distance descended is denoted by a negative

integer]

The total distance to descended by the elevator = (-350) – (10)

= – 360 m

Then,

Time taken by the elevator to descend -6 m = 1 min

So, time taken by the elevator to descend – 360 m = (-360) ÷ (-60)

= 60 minutes

= 1 hour