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

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.

1. minimum

2. 6400

3. eight

hope this !