Question

An elevator descends in to a mine shaft at the rate of 6m/min.if the descent starts from 10m above the ground level how long will it take to reach -350m deep from the ground

Answer 1

$\sf{\underline{Given:}}$

Initial height = $\sf{10\:m}$

Final depth = $\sf{-350\:m}$

$\sf{\underline{Note:}}$ Distance descended is denoted by a '-ve' integer.

$\sf{\underline{Now:}}$

Distance to be descended:

$\sf{x = ( - 350) - ( + 10)}$

$\sf{x = - 350 - 10}$

$\boxed{\sf{x = - 360 \: m}}$

$\sf{\underline{So:}}$

Total distance to be descended by the elevator is -360 m.

$\sf{\underline{Thus:}}$

To descend -360 m, time taken by elevator is:

$\sf{x = \frac{ - 360}{ - 6}}$

$\sf{x = 60}$

$\sf{\underline{We\:know\:that:}}$

$\boxed{\sf{60\:minutes\:= 1\:hour}}$

$\sf{\underline{Therefore:}}$

Answer: 1 hour