Question

A lift in mine moves down 24m in 87 mins. If it moves in uniforme rate ,find at what distance below the surface ,it will be after 6 mins . If the lift starts from a hileight 10 m above the ground ,find how deep the lift will go from the surface after 70 mins​

Answer 1

Answer:

Ok, first we have the relation:

Speed = Distance/Time.

Here we have:

Distance = 24m

Time = 87 mins

Speed = 24m/87mins = 0.279 m/min.

Now, in 6 minutes, the distance moved will be (remember that the movement is downwards)

D(6min) = (0.279m/min)*6min = 1.66 m.

So in 6 minutes, it moves 1.66 meters down, if the initial position is 10m above the ground, then the position after 6 minutes is:

10m - 1.66m = 8.34m above the ground.

And after 70 minutes, the distance moved is:

D(70min) =  (0.279m/min)*70min = 19.53m

Again, the initial position is 10m above the ground, so after 70 minutes the position is:

10m - 19.53m = -9.53m

So after 70 minutes, the position is 9.53 meters under the ground.

Answer 2

The lift will go 2.76m deep from the surface after 70 mins.

The lift moves down 24m in 87 minutes.

What is the speed of an object that covers 'd' distance in 't' time?

The speed of an object that covers 'd' distance in 't' time is d/t.

So, speed of the lift = 24/87 = 3/29m/min

Distance covered in 6 minutes = 6 * 3/29 = 18/29 m

So, if initially lift is at the surface of the earth, it will go 18/29m below the surface.

Distance covered in 70 minutes = 70* 3/29 =210/29m

So, if the initially lift is at 10 m height, it will go (10-210/29) = -80/29 or 2.76m below the surface of the earth.

Therefore, the lift will go 2.76m deep from the surface after 70 mins.

To get more about time, speed and distance visit:

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