An elevator descends into a mine shaft at the rate of $8$ m/min. If the descend starts from $12$ m above the ground level, how long will it take to reach $- 468$ m?

60

mins

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30

mins

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50

mins

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10

mins

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Solution

The correct option is A $60$ mins
Total distance covered from $A$ to $C$ $= 12 + 468 = 480$ m

Speed $= 8$ m/min

Therefore, time $= \frac{Total distance covered}{Speed}$

$t i m e = \frac{480}{8} = 60$ minutes.

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An elevator descends into a mine shaft at the rate of 8 m/min. If the descend starts from 12 m above the ground level, how long will it take to reach −468 m?

The correct option is A 60 minsTotal distance covered from A to C =12+468=480 mSpeed =8 m/minTherefore, time =Total distance coveredSpeedtime=4808=60 minutes.

An elevator descends into a mine shaft at the rate of $8$ m/min. If the descend starts from $12$ m above the ground level, how long will it take to reach $- 468$ m?

The correct option is A $60$ mins
Total distance covered from $A$ to $C$ $= 12 + 468 = 480$ m

Speed $= 8$ m/min

Therefore, time $= \frac{Total distance covered}{Speed}$

$t i m e = \frac{480}{8} = 60$ minutes.