Question

an elevator descends into a mine shaft at the rate of 5m/min lf the descent starts from 8m above the ground level how long will it take to reach 42m ( means - 42m) below the ground level​

Answer 1

Answer:

I can't becuz I also have problem in this question

Answer 2

Given:

$\text { Total distance elevator has to travel = Initial position - final position }$

$\begin{array}{l}=8-(-42) \\=8+42 \\=50 \mathrm{~m}\end{array}$

$\text { Now, }$

$\text { Elevator descends at rate } 5 \mathrm{~m} / \mathrm{min}$

$\therefore \text { Elevator travels } 5 \mathrm{~m} \text { in }=1 \text { minute }$

$\text { Elevator travels } 1 \mathrm{~m} \text { in }=\frac{1}{5} \text { minute }$

$\text { Elevator travels } 360 \mathrm{~m} \text { in }=\frac{1}{5} \times 360 \text { minutes }$

$=72 \text { minutes }$

$=1 \text { hour } 12 minutes$

$\therefore \text { It will take the elevator } 1 \text { hour} 12 minutes$