Solved Problem on Kinematics
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A steamboat, sails at a constant speed v (km/h), and consumes 0.3 + 0.001v3 tonnes of coal per hour. Calculate:
a) The speed that should have a 1000 km route to have a minimum consumption;
b) The amount of coal consumed on this trip.


Data problem:
  • Coal consume rate:    \( c=0.3+0.001v^{3}\;\frac{\text{t}}{\text{h}} \).
Solution

a) The total consumption of coal CT during the trip will be the consumption rate per unit of time c, given in the problem, multiplied by the time of travel Δt
\[ \begin{gather} C_{T}=c\Delta t \tag{I} \end{gather} \]
as the speed of the ship is constant the time of trip can be obtained from the expression of the average speed
\[ \bbox[#99CCFF,10px] {v=\frac{\Delta x}{\Delta t}} \]
\[ \begin{gather} \Delta t=\frac{\Delta x}{v} \tag{II} \end{gather} \]
Substituting the consumption of coal given in the problem and the time obtained from (II) into the expression (I)
\[ C_{T}=\left(0.3+0.001v^{3}\right)\frac{\Delta x}{v} \]
for the distance given in the problem, Δx = 1000 km
\[ \begin{gather} C_{T}=\left(0.3+0.001v^{3}\right)\frac{1000}{v}\\ C_{T}=\frac{300}{v}+\frac{v^{3}}{v}\\ C_{T}=\frac{300}{v}+v^{2} \tag{III} \end{gather} \]
To find the speed at which consumption is minimum, we take the derivative with respect to time of the expression (III) and equal to zero.

Differentiation of    \( C_{T}=\dfrac{300}{v}+v^{2} \)
\[ \begin{gather} \frac{dC_{T}}{dv}=300 v^{-1}+v^{2}\\ \frac{dC_{T}}{dv}=-1\times 300 v^{-1-1}+2 v^{2-1}\\ \frac{dC_{T}}{dv}=-300 v^{-2}+2 v^{1}\\ \frac{dC_{T}}{dv}=-{\frac{300}{v^{2}}}+2v \end{gather} \]
\[ -{\frac{300}{v^{2}}}+2v=0 \]
multiplying this expression by v2
\[ \begin{gather} \qquad \qquad\quad -\frac{300}{v^{2}}+2v=0\qquad (\times\;v^{2})\\ -{\frac{300}{\cancel{v^{2}}}}\cancel{v^{2}}+2v\;v^{2}=0\\ -300+2v^{3}=0\\ 2v^{3}=300\\ v^{3}=\frac{300}{2}\\ v^{3}=150\\ v=\sqrt[{3\;}]{150\;} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {v\simeq 5.3\;\text{km/h}} \]
To verify if this is the minimum point of the function, we calculate the second derivative.

Differentiation of    \( \dfrac{dC_{T}}{dv}=-{\dfrac{300}{v^{2}}}+2v \)
\[ \begin{gather} \frac{d^{2}C_{T}}{dv^{2}}=-300v^{-2}+2v\\ \frac{d^{2}C_{T}}{dv^{2}}=-(-2)\times 300v^{-2-1}+2v^{1-1}\\ \frac{d^{2}C_{T}}{dv^{2}}=600v^{-3}+2v^{0}\\ \frac{d^{2}C_{T}}{dv^{2}}=\frac{600}{v^{3}}+2 \end{gather} \]

substituting the speed
\[ \begin{gather} \frac{d^{2}C_{T}}{dv^{2}}=\frac{600}{(\sqrt[{3\;}]{150})^{3}}+2\\ \frac{d^{2}C_{T}}{dv^{2}}=\frac{600}{150}+2\\ \frac{d^{2}C_{T}}{dv^{2}}=6>0 \end{gather} \]
as the second derivative is greater than zero, the speed found represents a point of minimum of the function.

b) The amount of coal consumed is obtained by substituting the result of the previous item in the expression (III) for total consumption
\[ \begin{gather} C_{T}=\frac{300}{\sqrt[{3\;}]{150\;}}+(\sqrt[{3\;}]{150\;})^{2}\\ C_{T}=\frac{300}{5.3}+(5.3)^{2}\\ C_{T}=56.6+28.1 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {C_{T}=84.7\;\text{t}} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .