Question

An elevator descends into a mine shaft at the rate of 5 m per minute .what will be its position after one hours??

Answer 1

speed=distance÷time
distance=speed ×time
=5m/m×1h
=5m/m×60m
=120meters ans.pls follow as it takes hard work to answer so pls appreciated the efforts

Answer 2

since the elevator is going down , so the distance covered by it will be represented by a negative integer.
change of position of elevator in one min--(-5)m
position of the elevator after (1hour --60mins)=(-5)×60={-300) m ground level ...