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