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An elevator descends into a mine shaft at the rate of 6 m/min. The descent starts from 10 m above the ground level. How long will it take to reach $-$350 m?
Given:
An elevator descends into a mine shaft at the rate of 6 m/min.
The descent starts from 10 m above the ground level.
To find: Here we have to find the time taken by the elevator to reach $-$350 m.
Solution:
In the above figure, A represents the current position of the elevator which is 10 m above ground level. B is the ground level. And point C is 350 m below the ground level.
Now,
Total distance $=$ AB $+$ BC
Total distance $=$ 10 $+$ 350 m
Total distance $=$ 360 m
Given that speed of elevator $=$ 6 m/min
So,
$Time\ =\ \frac{Distance}{Speed}$
$Time\ =\ \frac{360}{6} \ \ minutes$
$\mathbf{Time\ =\ 60\ \ minutes}$
or,
$\mathbf{Time\ =\ 1\ hour}$
Therefore, the elevator will take 1 hour to reach its destination.