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An elevator descends into a mine shaft at the rate of $6\ m/min$. If 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 do:
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 descended by the elevator $=AB +BC$
$= (10+350)\ m$
$=360\ m$
Rate of descent of elevator $= 6\ m/min$
Time taken by the elevator to descend $-6\ m = 1\ min$
This implies,
The time taken by the elevator to descend $-360\ m = (-360) \div (-6)$
$= 60\ min$
$= 1$ hour (Since $1$ hour $=60$ minutes)
The elevator takes $1$ hour to reach $-350\ m$.