A lift travels at the rate of 4 m/sec the descent starts from 20 m above ground level, how long will it take reach-220 m?
Pls ans it correct only if u know. Plsss someone,Urgent for exam tomorrow
Answers
Answer:
Starting position of mine shaft is = 10m above ground
but,
it moves in opposite direction so it travels the distance (-350 m ) below the ground.
Total distance covered by mine shaft = 10m - (-350m)
= 10 + 350
= 360m
Now, taken to cover a distance of 6m by it = 1 minute
Time taken to cover a distance of 1m by it = \frac{1}{6\ }\min
Therefore, tie taken to a cover a distance of 360m = \frac{1}{6}\times\ 360\ =\ 60\ \min\ \
60 minutes = 1 hour
thus, in one hour the mine shraft reaches = 350 below the ground.
OR
Distance descended is denoted by a negative integer,
Initial height =+10 m
Final Depth =−350 m
Total distance to be descended by the elevator =(−350)−(+10)=−360 m
Given, the time is taken by the elevator to descend −6 m=1 minute
Thus, time taken by the elevator to descend −360 m=
(−6)
(−360)
------- [from the formula : speed=
time
distance
]
=60 minutes