Question

an elevator descends into a mine shaft at the rate of 5m/min . what will be it's position after 1 hour q?if it began to descends to from 35 m above the ground, what will be it's position after 30 mins​

Answer 1

i) 300 m

ii) 115 m below the ground

Step-by-step explanation:

i) 1 hours = 60 min

speed = distance/time

5 m/min = distance/60 min

distance = 60×5 m

distance = 300 m

ii) Let's assume that "+" means above and "-" means below

then the distance after 30 min will decrease by

distance = 30× -5 m

= -150 m

Then, final position = 35m - 150m = -115

Thus, the final position is 115 m below the ground

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