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Question: The consumption of coal by a locomotive varies as the square of the velocity; when the speed is 16 m...

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.

Explanation

Solution

Hint: The equation connecting velocity and quantity of fuel is q=kv2q=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 vv be the velocity of the train, which is in miles/hour.
Let qq be the quantity of the fuel used per hour. It is given in tons.
The equation connecting velocity and quantity of fuel is,
q=kv2q=k{{v}^{2}},
Here k=2(16)2k=\dfrac{2}{{{\left( 16 \right)}^{2}}}, fuel consumption constant
k=2256\therefore k=\dfrac{2}{256} , that is k=fuel used(distance)2k=\dfrac{\text{fuel used}}{{{\left( \text{distance} \right)}^{\text{2}}}}
That is,
q=2256v2q=\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=12×2256v2=v2256=\dfrac{1}{2}\times \dfrac{2}{256}{{v}^{2}}=\dfrac{{{v}^{2}}}{256}
\therefore Cost of fuel per mile will be
=1v×v2256=v256=\dfrac{1}{v}\times \dfrac{{{v}^{2}}}{256}=\dfrac{v}{256}
Cost per mile of other expenses =other expensesfuel usage per hour=116.3d×1v2×10=111420×1v=11.25v20=916v=\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 =v256=v16=\sqrt{\dfrac{v}{256}}=\dfrac{\sqrt{v}}{16}
Similarly, cost per mile of other expenses =916v=34v=\sqrt{\dfrac{9}{16v}}=\dfrac{3}{4\sqrt{v}}
\therefore Total cost of journey =(v16+34v)2 (a+b)2=a2+b2+2ab={{\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, (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab, so above equation can be written as,
=(v16)2+(34v)2+2v×316×4×4={{\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,
=(v16)2+(34v)2+332={{\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 =332×100=30032=\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.