An elevator descends into a mine shaft at the rate of $6$ meters per minute. If the decent starts from $10$ m above the ground level, how long will it take to reach $- 350$ m (in hours)

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Solution

Consider the elevator to be at constant speed and the destination to be $350$ m below ground level, the elevator has to travel $10$ m till the ground, then $350$ m more
So, total distance to travel is $360$ m
If the elevators speed is $6$ m/min,
Time $= \frac{D i s t a n c e}{S p e e d}$
$= \frac{360}{6}$
$= 60$ minutes
$= 1$ hour.

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An elevator descends into a mine shaft at the rate of 6 meters per minute. If the decent starts from 10 m above the ground level, how long will it take to reach −350 m (in hours)

Consider the elevator to be at constant speed and the destination to be 350m below ground level, the elevator has to travel 10m till the ground, then 350m moreSo, total distance to travel is 360mIf the elevators speed is 6m/min,Time=DistanceSpeed=3606=60 minutes=1 hour.

An elevator descends into a mine shaft at the rate of $6$ meters per minute. If the decent starts from $10$ m above the ground level, how long will it take to reach $- 350$ m (in hours)