A submarine can descend from the water surface at s rate of 12 m per minute After decending for 20 minutes it again rises by 30 meters what is its new position?

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Solution

$Descend of submarine from the water surface in 1 \min = - 12 m Descend of submarine from the water surface in 20 \min = 20 \times (- 12) = - 240 m Now, the submarine rises by 30 m . So, final position of the submarine = - 240 + 30 = - 210 m So, submarine is 210 m below the water surface .$

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A submarine can descend from the water surface at s rate of 12 m per minute After decending for 20 minutes it again rises by 30 meters what is its new position?

Descend of submarine from the water surface in 1 min = -12 mDescend of submarine from the water surface in 20 min = 20 × -12 = -240 mNow, the submarine rises by 30 m.So, final position of the submarine = -240 + 30 = -210 mSo, submarine is 210 m below the water surface.

$Descend of submarine from the water surface in 1 \min = - 12 m Descend of submarine from the water surface in 20 \min = 20 \times (- 12) = - 240 m Now, the submarine rises by 30 m . So, final position of the submarine = - 240 + 30 = - 210 m So, submarine is 210 m below the water surface .$