An elevator descends into a mine shaft at the rate of 5 metres per minute. If it begins to descend from 15 metres above the ground, what will be its position after 45 minutes?

Given:

The speed of the elevator, $v=-5m/min$(since the motion is vertically downwards, the value is taken negative)

The time taken by the elevator, $t=45min$

Finding the distance covered by the elevator in the downward direction:

We know that,

$Dis\tan ce=speed\times time$

$$\begin{gathered} Bysubstitutingthegivenvaluesintheaboveexpression,weget \\ Dis\tan ce=-5\times 45 \\ =-225m \end{gathered}$$

Finding the final position of the elevator:

As the elevator is 15 m above the ground, we have to add this to the distance covered by it.

So, the final position of the elevator is given by,

$-225m+15m=-210m$

Hence, the elevator is 210 metres below the ground level after 45 minutes.