SECHION-D16. An elevator descends into a mine shaft at the rate of 6 m

1.	SECHION-D16. An elevator descends into a mine shaft at the rate of 6 m/min. Ifthe descend starts from 10 m above the ground level, how longwill it take to reach the shaft 350 m below the ground level?
Answer» Considering the elevator to be at constant speed and the destination to be 350 m below ground level, The elevator has to travel 10m till the ground, then 350 m more. So, the total distance to travel is 360 m If the elevator's speed is 6 metres/minute,Then Time= Distance/Speed = 360/6 =60 minutes =1 hour Thus the elevator would take 1 hour to reach its destination. The descend starts from 10 m above ground level and the elevator has to travel 350 m, so total distance that is travelled by the elevator= (350+10)m =360m.Now, speed =6m/minSo, time required= distance/speed=(360/6) mins=60 mins= 1hr

SECHION-D16. An elevator descends into a mine shaft at the rate of 6 m/min. Ifthe descend starts from 10 m above the ground level, how longwill it take to reach the shaft 350 m below the ground level?

Answer»

Considering the elevator to be at constant speed and the destination to be 350 m below ground level,

The elevator has to travel 10m till the ground, then 350 m more.

So, the total distance to travel is 360 m

If the elevator's speed is 6 metres/minute,Then

Time= Distance/Speed = 360/6 =60 minutes =1 hour

Thus the elevator would take 1 hour to reach its destination.

The descend starts from 10 m above ground level and the elevator has to travel 350 m, so total distance that is travelled by the elevator= (350+10)m =360m.Now, speed =6m/minSo, time required= distance/speed=(360/6) mins=60 mins= 1hr

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