Question

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)

Answer 1

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$=\frac{Distance}{Speed}$

$=\frac{360}{6}$

$=60$ minutes

$=1$ hour.