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Example 1.10: Computing Quantities from Measurement Results and Known Mathematical Relations

While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.
(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
(b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?

Solution

\(\mathrm{distance}\) \(= 1250.\ \mathrm{km}\)


\(V_{\mathrm{gasoline}}\) \(= 213.\ \mathrm{L}\)


(a) We calculate the mileage (which comes out in weird units) and then convert the units:

\(\mathrm{mileage}\) \(= \dfrac{\mathrm{distance}}{V_{\mathrm{gasoline}}}\)

\(\ \ \ =\dfrac{1250.\ \mathrm{km}}{213.\ \mathrm{L}}\)

\(\ \ \ =5.87\times 10^{12}\frac{1}{\mathrm{km}^{2}}\)


\(\mathrm{mileage}\) \(= 5.87\times 10^{12}\frac{1}{\mathrm{km}^{2}}\)

\(\ \ \ =5.87\times 10^{12}\frac{1}{\mathrm{km}^{2}}\)



\(\mathrm{mile}\) \(= \dfrac{1}{0.62137} \cdot 1\ \mathrm{km}\)

\(\ \ \ =1.609347 \cdot 1\ \mathrm{km}\)

\(\ \ \ =1.60935\ \mathrm{km}\)


\(\mathrm{gallon}\) \(= 3.78541\ \mathrm{L}\)


\(\mathrm{mileage}\) \(= \dfrac{\mathrm{mileage}}{\dfrac{\mathrm{mile}}{\mathrm{gallon}}}\mathrm{\ \mathrm{(mile/gallon)}}\)

\(\ \ \ =\dfrac{5.87\times 10^{12}\frac{1}{\mathrm{km}^{2}}}{\dfrac{1.60935\ \mathrm{km}}{3.78541\ \mathrm{L}}}\mathrm{\ \mathrm{(mile/gallon)}}\)

\(\ \ \ =\dfrac{5.87\times 10^{12}\frac{1}{\mathrm{km}^{2}}}{4.251447\times 10^{11}\frac{1}{\mathrm{km}^{2}}}\mathrm{\ \mathrm{(mile/gallon)}}\)

\(\ \ \ =13.80\mathrm{\ \mathrm{(mile/gallon)}}\)


(b) We find:

\(\mathrm{price}_{\mathrm{fuel}}\) \(= \dfrac{3.80\ \mathrm{$}}{\mathrm{gallon}}\)

\(\ \ \ =\dfrac{3.80\ \mathrm{$}}{3.78541\ \mathrm{L}}\)

\(\ \ \ =1.004\ \frac{\mathrm{$}}{\mathrm{L}}\)


\(\mathrm{cost}_{\mathrm{fuel}}\) \(= \mathrm{price}_{\mathrm{fuel}} \cdot V_{\mathrm{gasoline}}\)

\(\ \ \ =1.004\ \frac{\mathrm{$}}{\mathrm{L}} \cdot 213.\ \mathrm{L}\)

\(\ \ \ =214.\ \mathrm{$}\)