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# An elevator descends into mine shaft at the rate of $6{\text{ m}}/\min$. If the descent starts from $10{\text{ m}}$ above the ground level, how long will it take to reach $350{\text{m}}$?

Last updated date: 13th Jun 2024
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Hint: Here we are given that an elevator descends into mine shaft at the rate of$6{\text{ m}}/\min$ and we need to find the time when it reaches $350{\text{m}}$ below the ground. So we know that
${\text{time taken }} = {\text{ }}\dfrac{{{\text{total distance travelled }}}}{{{\text{speed of travel}}}}$

Here we are given that the elevator descends into a mine shaft. This means that there is a mine shaft and the elevator descends or goes down with the constant rate or with the constant speed of $6{\text{ m/min}}$and it is also given that the descend starts from above the ground and we need to find the time when it reaches $350{\text{m}}$ above the ground.
Here we are assuming that the elevator is moving with the constant rate of $6{\text{ m/min}}$that means in one minute it travels $6{\text{m}}$. Now as we know that the distance travelled is the product of the speed and the time taken to travel. So here we need to find the total time.
${\text{total time}} = {\text{ }}\dfrac{{{\text{total distance travelled }}}}{{{\text{constant speed}}}}$
Here we are given the total time when the elevator reaches $350{\text{m}}$below ground. Here $350$ is not the total distance as it starts to descend from above $10{\text{m}}$ from the ground level and travels to $350{\text{m}}$below ground. So the total distance travelled by the elevator will be$= 350 + 10 = 360{\text{ m}}$
And as we know constant speed is given as $6{\text{ m}}/\min$
So ${\text{time taken }} = {\text{ }}\dfrac{{{\text{total distance travelled }}}}{{{\text{speed of travel}}}}$$= \dfrac{{360}}{6} = 60{\text{min}}$
And as we know that $1{\text{ hour }} = {\text{ 60 minute}}$
So an elevator takes $1{\text{ hour}}$ to reach $350{\text{ m}}$
${\text{total time}} = {\text{ }}\dfrac{{{\text{total distance travelled }}}}{{{\text{constant speed}}}}$ and but if the speed is not constant or acceleration is applied then this formula is not valid.