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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 $-350\ m$.**

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