Welding fumes contain a dangerous amount of particulates. To protect workers in a workshop, an extra

Question

Welding fumes contain a dangerous amount of particulates. To protect workers in a workshop, an extraction device remove particulates in the air continuously. Let $V$ mg represent the mass of particulates in the air in the workshop at time $t$ min after the extraction device is activated. Particulates are produced at a constant rate of $0.32$ mg/min. The rate at which particulates are removed is proportional to the mass of particulates in the air. When the mass of particulates in the air in the workshop is $18.75$ mg, the mass of particulates in the air increases at a rate of $0.12$ mg/min. Show that $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{4}{375}(30-V)$.[2] Solve the differential equation, given that the workshop is initially free of particulates.[5] Sketch the graph of $V$ against $t$.[2] It is recommended that the mass of particulates in this workshop should be kept below $32$ mg. Comment on whether the extraction device is effective enough to meet the recommendation.[1] The extraction device becomes less efficient and now removes particulates at half its original rate. Write down a new differential equation to model this.[1] The mass of particulates in the air in the workshop is said to reach a steady-state when it remains constant. Find the value of $V$ at steady-state.[1]

Accepted Answer

(ii)$V=30-30e^{-\frac{4}{3775}t}$ (v)$\frac{2}{375}(60-V)$ (vi)V=60

2025 SAJC P1 Q12

2025 SAJC P1 Q12