Question

The consumption of coal by a locomotive varies as the square of the velocity; when the speed is 16 miles an hour the consumption of coal per hour is 2 tons. If the price of coal is 10s. per ton, and other expenses of the engine be 11s.3d. And hour, find the least cost of a journey of 100 miles.

Answer 1

Hint: The equation connecting velocity and quantity of fuel is \[q=k{{v}^{2}}\]. Find the total cost of the journey, which includes the cost of fuel per mile and other expenses. Then find the total cost for a journey of 100 miles.

Complete step-by-step answer:
Let \[v\] be the velocity of the train, which is in miles/hour.
Let \[q\] be the quantity of the fuel used per hour. It is given in tons.
The equation connecting velocity and quantity of fuel is,
\[q=k{{v}^{2}}\],
Here \[k=\dfrac{2}{{{\left( 16 \right)}^{2}}}\], fuel consumption constant
\[\therefore k=\dfrac{2}{256}\] , that is \[k=\dfrac{\text{fuel used}}{{{\left( \text{distance} \right)}^{\text{2}}}}\]
That is,
\[q=\dfrac{2}{256}{{v}^{2}}\]
\[\therefore \]Cost of the fuel per hour, it is given than 2 tons are used per hour, so
\[\therefore \]Cost\[=\dfrac{1}{2}\times \dfrac{2}{256}{{v}^{2}}=\dfrac{{{v}^{2}}}{256}\]
\[\therefore \]Cost of fuel per mile will be
\[=\dfrac{1}{v}\times \dfrac{{{v}^{2}}}{256}=\dfrac{v}{256}\]
Cost per mile of other expenses \[=\dfrac{\text{other expenses}}{\text{fuel usage per hour}}=\dfrac{116.3d\times 1v}{2\times 10}=\dfrac{11\dfrac{1}{4}}{20}\times 1v=\dfrac{11.25v}{20}=\dfrac{9}{16}v\]
Taking the root for cost of fuel per mile \[=\sqrt{\dfrac{v}{256}}=\dfrac{\sqrt{v}}{16}\]
Similarly, cost per mile of other expenses \[=\sqrt{\dfrac{9}{16v}}=\dfrac{3}{4\sqrt{v}}\]
\[\therefore \]Total cost of journey \[={{\left( \dfrac{\sqrt{v}}{16}+\dfrac{3}{4\sqrt{v}} \right)}^{2}}\text{ }\because {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
We know, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], so above equation can be written as,
\[={{\left( \dfrac{\sqrt{v}}{16} \right)}^{2}}\text{+}{{\left( \dfrac{3}{4\sqrt{v}} \right)}^{2}}+\dfrac{2\sqrt{v}\times 3}{16\times 4\times \sqrt{4}}\]
Solving this we get,
\[={{\left( \dfrac{\sqrt{v}}{16} \right)}^{2}}\text{+}{{\left( \dfrac{3}{4\sqrt{v}} \right)}^{2}}+\dfrac{3}{32}\]
Since the least value of square is zero.
\[\therefore \]Least cost of the journey for 100 miles \[=\dfrac{3}{32}\times 100=\dfrac{300}{32}\]

Note: Here 10s. denoted money converted to pounds 10s. is 10 shilling. Similarly, 11s.3d, means 11 shilling and 3 pennies. It is important that you know the equation connecting the velocity and quantity of fuel usage or else you won't get the required answer.