Ice forms on a refrigerator ice-box at the rate of (4 - 06t)g per minute after t minutes. If initially there are 2g of ice in the box, find the mass of ice formed in 5 minutes
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Correct Answer: Option C
Explanation:
\(\frac{dm}{dt}\) = 4 - 0.6t
\(\int\)dm = \(\int\)(4 - 0.6t)dt
m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g
− c = 2
m = \(4t - 0.3t^2 + 2\), when t = 5 minutes
m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\)
= 14.5
\(\frac{dm}{dt}\) = 4 - 0.6t
\(\int\)dm = \(\int\)(4 - 0.6t)dt
m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g
− c = 2
m = \(4t - 0.3t^2 + 2\), when t = 5 minutes
m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\)
= 14.5