a shaft in a mine moves at a speed of 71/2 m/min. if it is 15m above the ground and then it starts descending in the mine. find the position after 1 hour.
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Solution :-
Speed of the shaft = 71/2 meter per minute
Position of the shaft = 15 meter above the ground level
⇒ In 1 minute the shaft moves = 71/2 meters
⇒ So, in 60 minutes the shaft will move = (71*60)/2
⇒ = 4260/2
⇒ = 2130 meter
As the shaft is 15 meter above the ground level and then it starts descending in the mine, so its position after 60 minutes = 2130 - 15
= 2115 m deep in the mine.
Hence after 60 minutes the shaft will be 2115 meter deep in the mine.
Speed of the shaft = 71/2 meter per minute
Position of the shaft = 15 meter above the ground level
⇒ In 1 minute the shaft moves = 71/2 meters
⇒ So, in 60 minutes the shaft will move = (71*60)/2
⇒ = 4260/2
⇒ = 2130 meter
As the shaft is 15 meter above the ground level and then it starts descending in the mine, so its position after 60 minutes = 2130 - 15
= 2115 m deep in the mine.
Hence after 60 minutes the shaft will be 2115 meter deep in the mine.
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