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Question Number 171908 by ajfour last updated on 21/Jun/22

Commented by ajfour last updated on 21/Jun/22

$M a n h a s t o w a t e r t h e t r e e$ $o n c e w i t h v o l u m e V_{0} . H i s$ $b u c k e t l e a k s a t a r a t e \frac{d V}{d t} = - k V .$ $h i s s p e e d = u (2 - \frac{m}{m + ρ V})$ $ρ V < m . F i n d x s u c h t h a t t h e$ $h e c a n r e a c h w i t h V_{0} v o l u m e$ $o f w a t e r i n b u c k e t i n m i n .$ $t i m e . (m i s m a s s o f b u c k e t)$
Commented by mr W last updated on 22/Jun/22

$s_{1} = \sqrt{a^{2} + x^{2}}$ $v_{1} = u$ $t_{1} = \frac{s_{1}}{u} = \frac{\sqrt{a^{2} + x^{2}}}{u}$ $s a y t h e v o l u m e o f w a t e r i n t h e b u c k e t$ $a t t h e r i v e r i s V_{1} .$ $s_{2} = \sqrt{b^{2} + {(c - x)}^{2}}$ $\frac{d V}{d t} = - k V$ $\int_{V_{1}}^{V} \frac{d V}{V} = - k \int_{0}^{t} d t$ $\ln \frac{V}{V_{1}} = - k t$ $V = V_{1} e^{- k t}$ $V_{0} = V_{1} e^{- {k t}_{2}}$ $\frac{d s}{d t} = v = u (2 - \frac{m}{m + ρ V})$ $\frac{d s}{d t} = u (2 - \frac{m}{m + ρ V_{1} e^{- k t}})$ $\int_{0}^{s_{2}} d s = u \int_{0}^{t_{2}} (2 - \frac{1}{1 + \frac{ρ V_{1}}{m} e^{- k t}}) d t$ $s_{2} = u {[t - \frac{\ln (1 + \frac{ρ V_{1}}{m} e^{- k t})}{k}]}_{0}^{t_{2}}$ $\sqrt{b^{2} + {(c - x)}^{2}} = u (t_{2} - \frac{1}{k} \ln \frac{m + ρ V_{1} e^{- {k t}_{2}}}{m + ρ V_{1}})$ $\sqrt{b^{2} + {(c - x)}^{2}} = u (t_{2} - \frac{1}{k} \ln \frac{m + ρ V_{0}}{m + ρ V_{0} e^{{k t}_{2}}})$ $t o t a l t i m e T = t_{1} + t_{2}$ $t_{2} = T - t_{1} = T - \frac{\sqrt{a^{2} + x^{2}}}{u}$ $\sqrt{b^{2} + {(c - x)}^{2}} = u [T - \frac{\sqrt{a^{2} + x^{2}}}{u} - \frac{1}{k} \ln \frac{m + ρ V_{0}}{m + ρ V_{0} e^{k (T - \frac{\sqrt{a^{2} + x^{2}}}{u})}}]$ $\frac{\sqrt{a^{2} + x^{2}} + \sqrt{b^{2} + {(c - x)}^{2}}}{u} = T - \frac{1}{k} \ln \frac{\frac{m}{ρ V_{0}}}{\frac{m}{ρ V_{0}} + e^{k (T - \frac{\sqrt{a^{2} + x^{2}}}{u})}}$ $\frac{\sqrt{a^{2} + x^{2}} + \sqrt{b^{2} + {(c - x)}^{2}}}{\frac{u}{k}} = k T - \ln \frac{\frac{m}{ρ V_{0}}}{\frac{m}{ρ V_{0}} + e^{(k T - \frac{\sqrt{a^{2} + x^{2}}}{\frac{u}{k}})}}$ $l e t λ = \frac{ρ V_{0}}{m}, Φ = k T, μ = \frac{u}{k}$ $\frac{\sqrt{a^{2} + x^{2}} + \sqrt{b^{2} + {(c - x)}^{2}}}{μ} = Φ + \ln (1 + λ e^{Φ - \frac{\sqrt{a^{2} + x^{2}}}{μ}})$ $\frac{d Φ}{d x} = 0 f o r m i n i m u m T .$ $s e e e x a m p l e .$
Commented by ajfour last updated on 22/Jun/22

$T h a n k y o u v e r y m u c h s i r . . .$
Commented by mr W last updated on 22/Jun/22

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