JAMB 1992 · UME · Q38✱✱

Ice forms on a refrigerator ice-box at the rate of (4−0.6t)(4 - 0.6t) g per minute after tt minutes. If initially there are 2 g of ice in the box, find the mass of ice in the box after 5 minutes.

Worked solution (try it first)
  1. The rate is the derivative of the mass, so integrate it to find the ice formed from t=0t = 0 to t=5t = 5.
  2. ∫(4−0.6t) dt=4t−0.3t2\int (4 - 0.6t)\,dt = 4t - 0.3t^2.
  3. At t=5t = 5 this is 20−7.5=12.520 - 7.5 = 12.5, and at t=0t = 0 it is 0.
  4. So 12.5 g of ice forms.
  5. Add the 2 g already there: 2+12.5=14.52 + 12.5 = 14.5 g, option C.

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