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Question Number 4633 by sanusihammed last updated on 16/Feb/16

The volume V (in liters)  of petrol in a car tank is decreasing   at the rate of                                        ((1/5) − (m/(2000))) liters/km    After travelling m kilometers. Assuming that V = 40  when m = 0. Find an expression for V  in terms of m.

$${The}\:{volume}\:{V}\:\left({in}\:{liters}\right)\:\:{of}\:{petrol}\:{in}\:{a}\:{car}\:{tank}\:{is}\:{decreasing}\: \\ $$$${at}\:{the}\:{rate}\:{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{5}}\:−\:\frac{{m}}{\mathrm{2000}}\right)\:{liters}/{km} \\ $$$$ \\ $$$${After}\:{travelling}\:{m}\:{kilometers}.\:{Assuming}\:{that}\:{V}\:=\:\mathrm{40} \\ $$$${when}\:{m}\:=\:\mathrm{0}.\:{Find}\:{an}\:{expression}\:{for}\:{V}\:\:{in}\:{terms}\:{of}\:{m}. \\ $$$$ \\ $$

Answered by Yozzii last updated on 21/Feb/16

(dV/dm)=−(0.2−0.0005m)  ⇒V=c−0.2m+0.00025m^2   At m=0⇒V=40.  ∴ 40=c⇒ V=40−0.2m+0.00025m^2   (m≥0)

$$\frac{{dV}}{{dm}}=−\left(\mathrm{0}.\mathrm{2}−\mathrm{0}.\mathrm{0005}{m}\right) \\ $$$$\Rightarrow{V}={c}−\mathrm{0}.\mathrm{2}{m}+\mathrm{0}.\mathrm{00025}{m}^{\mathrm{2}} \\ $$$${At}\:{m}=\mathrm{0}\Rightarrow{V}=\mathrm{40}. \\ $$$$\therefore\:\mathrm{40}={c}\Rightarrow\:{V}=\mathrm{40}−\mathrm{0}.\mathrm{2}{m}+\mathrm{0}.\mathrm{00025}{m}^{\mathrm{2}} \:\:\left({m}\geqslant\mathrm{0}\right) \\ $$$$ \\ $$$$ \\ $$

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