An elevator descends into a mine shaft at the rate of 7 meters per minute if the descent starts 10 meters above the ground level how long will it take to reach 410 meters below the ground level?

Given:

The speed of the elevator, $v=-7m/min$(since the motion is vertically downwards, the value is taken negative)

Position of the elevator = $-410m$

Finding the distance covered by the elevator:

As the elevator is 10 m above the ground, it had to descend extra 10 metres to reach the ground level.

So, the distance covered by the elevator is given by,

$-410m+(-10m)=-420m$

Finding the time taken by the elevator to reach its position:

We know that $Time=\frac{Dis\tan ce}{Speed}$

$$\begin{gathered} Bysubstitutingthegivenvaluesintheaboveexpression,weget \\ Time=\frac{-420}{-7} \\ =60 \end{gathered}$$

Hence, the elevator takes 60 min to reach 410 m below the ground level.