Answer:
a) 2.56 days b) Zero Liters
Step-by-step explanation
let initial volume of tank be V₀ and volume at time t be V.
Rate is proportional to square root of volume of remaining water, therefore;
[tex]\frac{dV}{dt} = K\sqrt{Vo - V}[/tex] where K is the proportionality constant.
[tex]\frac{dV}{\sqrt{Vo - V} } =K dt[/tex] integrating both sides gives
[tex]Sin^{-1} \frac{\sqrt{V} }{\sqrt{Vo} } = Kt[/tex]
at day one, t = 1 ; V = 25 Liters. Therefore;
[tex]Sin^{-1} \frac{\sqrt{25} }{\sqrt{275} } = K X 1[/tex]
K = 17.548
a) when tank is half V = 275/2 = 137.5
[tex]Sin^{-1} \frac{\sqrt{137.5} }{\sqrt{275} } = 17.548 X t[/tex]
t = 2.56 days
b) How much water will remain after 5 days
[tex]Sin^{-1} \frac{\sqrt{V} }{\sqrt{275} } = 17.548 X 5[/tex]
[tex]\frac{\sqrt{V} }{\sqrt{275} } = Sin 87.74[/tex]
V = 275 liters
Remaining water = 275 - 275 = O
