CBSE 2024 · Region 2 · Set 3 · Q37 · 4 marks
Overspeeding increases fuel consumption and decreases fuel economy as a result of tyre rolling friction and air resistance. While vehicles reach optimal fuel economy at different speeds, fuel mileage usually decreases rapidly at speeds above $\displaystyle 80 \mathrm{~km} / \mathrm{h}$.
The relation between fuel consumption $\displaystyle \mathrm{F}(l / 100 \mathrm{~km})$ and speed $\displaystyle \mathrm{V}(\mathrm{km} / \mathrm{h})$ under some constraints is given as $\displaystyle \mathrm{F}=\frac{\mathrm{V}^{2}}{500}-\frac{\mathrm{V}}{4}+14$. On the basis of the above information, answer the following questions :(i)$$ Find F , when $\displaystyle \mathrm{V}=40 \mathrm{~km} / \mathrm{h}$.(ii)Find $\displaystyle \frac{d \mathrm{F}}{d \mathrm{V}}$.(iii)Find the speed V for which fuel consumption F is minimum.Find the quantity of fuel required to travel $\displaystyle 600$ km at the speed V at which $\displaystyle \frac{\mathrm{dF}}{\mathrm{dV}}=-0 \cdot 01$. Case Study - $\displaystyle 3$
Overspeeding increases fuel consumption and decreases fuel economy as a result of tyre rolling friction and air resistance. While vehicles reach optimal fuel economy at different speeds, fuel mileage usually decreases rapidly at speeds above $\displaystyle 80 \mathrm{~km} / \mathrm{h}$.
The relation between fuel consumption $\displaystyle \mathrm{F}(l / 100 \mathrm{~km})$ and speed $\displaystyle \mathrm{V}(\mathrm{km} / \mathrm{h})$ under some constraints is given as $\displaystyle \mathrm{F}=\frac{\mathrm{V}^{2}}{500}-\frac{\mathrm{V}}{4}+14$. On the basis of the above information, answer the following questions :
(i)
$$ Find F , when $\displaystyle \mathrm{V}=40 \mathrm{~km} / \mathrm{h}$.
(ii)
Find $\displaystyle \frac{d \mathrm{F}}{d \mathrm{V}}$.
(iii)
Find the speed V for which fuel consumption F is minimum.
Find the quantity of fuel required to travel $\displaystyle 600$ km at the speed V at which $\displaystyle \frac{\mathrm{dF}}{\mathrm{dV}}=-0 \cdot 01$. Case Study - $\displaystyle 3$
Marking-scheme solution
(i)
When $\displaystyle V = 40$ km/h, $\displaystyle F = \dfrac{36}{5}\ \ell/100\,\text{km}$
(ii)
\[\frac{dF}{dV} = \frac{V}{250} - \frac{1}{4}\]
(iii)
\[\frac{dF}{dV} = 0\]
$\displaystyle \Rightarrow V = 62.5$ km/h
\[\frac{d^2F}{dV^2} = \frac{1}{250} > 0 \text{ at } V = 62.5 \text{ km/h}\]
Hence, $\displaystyle F$ is minimum when $\displaystyle V = 62.5$ km/h
\[\frac{dF}{dV} = -0.01\]
\[\Rightarrow \frac{V}{250} - \frac{1}{4} = \frac{-1}{100}\]
\[\Rightarrow V = 60 \text{ km/h}\]
\[F = \frac{60^2}{500} - \frac{60}{4} + 14 = 6.2\ \ell/100\,km\]
Quantity of fuel required for $\displaystyle 600$ km
$\displaystyle = 6.2 \times 6 = 37.2\ \ell$
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