An elevator descends into a mine shaft at the rate of 5 m/min. If the descend starts from 15m above the ground, by how long will it take to reach 350 m deep from the ground?
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Answers
Given data :
- An elevator descends into a mine shaft at the rate of 5 m/min.
- The descend starts from 15m above the ground.
- Depth of mine shaft is 350 m deep.
Solution : We have to find the distance traveled by the elevator. Here we know according to question, descend starts from 15m above the ground.
Hence,
- Total distance covered by elevator = 350 + 15
- Total distance covered by elevator = 365 m
Now we use formula to find time taken by elevator to covered 365 m distance.
Here, we know
- Displacement = 365 m
- Velocity of elevator = 5 m/min
- Time taken by elevator = ?
Now,
⟹ Velocity = displacement/time
⟹ 5 = 365/time
⟹ time = 365/5
⟹ time = 73 minutes
Answer : Time taken by elevator to cover distance of deep mine shaft when it starts 15 m above the ground is 73 minutes.
{More info :
- 73 minutes = 73 * 60 = 4380 seconds}
Given :
An elevator descends into a mine shaft at the rate of 5 m/min.
The descend starts from 15m above the ground.
Depth of mine shaft is 350 m deep.
Solution : We have to find the distance traveled by the elevator. Here we know according to question, descend starts from 15m above the ground.
Hence,
Total distance covered by elevator = 350 + 15
Total distance covered by elevator = 365 m
Now we use formula to find time taken by elevator to covered 365 m distance.
Here, we know
Displacement = 365 m
Velocity of elevator = 5 m/min
Time taken by elevator = ?
Now,
⟹ Velocity = displacement/time
⟹ 5 = 365/time
⟹ time = 365/5
⟹ time = 73 minutes
Answer : Time taken by elevator to cover distance of deep mine shaft when it starts 15 m above the ground is 73 minutes.
Hope it will helps you...