Question

If an elevator descends a mine shaft from a height of 25 m above the ground at the rate of 5 metre per minute what would be its position after 35 minutes


need steps

Answer 1

Answer:

Starting position of mine shaft is = 10m above ground

but,

it moves in opposite direction so it travels the distance (-350 m ) below the ground.

Total distance covered by mine shaft = 10m - (-350m)

= 10 + 350

= 360m

Now, taken to cover a distance of 6m by it = 1 minute

Time taken to cover a distance of 1m by it = \frac{1}{6\ }\min

6

1

min

Therefore, tie taken to a cover a distance of 360m = \frac{1}{6}\times\ 360\ =\ 60\ \min\ \

6

1

× 360 = 60 min

60 minutes = 1 hour

thus, in one hour the mine shraft reaches = 350 below the ground.

Answer 2

$\huge \bigstar \underline \bold{ \: Solution}$

In \: 1 \: minute \: the \: elevator \: goes = 5min1minutetheelevatorgoes=5m

In \: 35 \: minutes \: the \: elevator \: goes = (5 \times 35) = 175min35minutestheelevatorgoes=(5×35)=175m

Final \: position = (175 - 25) = 150m \: under \: the \: ground final position=(175−25)=150m underground

Hope my answer is correct and helps to you