Rates of Change and Proportionality in Mechanics

Problem Statement

In the real world, forces are rarely constant. Mechanics heavily involves considering proportionality (e.g., air resistance proportional to velocity, spring force proportional to extension) and interpreting rates of change (velocity as the rate of change of displacement, acceleration as the rate of change of velocity).

Understanding how to translate a physical statement of proportionality into a mathematical differential equation using rates of change is a core skill in HSC Mathematics Extension 1 and 2.

Consider a boat of mass $M$ moving in a straight line across a calm lake. The boat's engine provides a constant driving force $F$. However, the water provides a resistive force that is directly proportional to the square of the boat's velocity $v$.

(a) Set up the differential equation of motion for the boat, defining any constants of proportionality you introduce. Let $x$ be the displacement of the boat.

(b) Write down the terminal velocity $V_T$ of the boat in terms of the given constants.

(c) Using the rate of change form for acceleration $a = v \frac{dv}{dx}$, show that the distance $D$ travelled by the boat as it accelerates from rest to half its terminal velocity ($\frac{1}{2}V_T$) is given by: $D = \frac{M}{2k} \ln \left( \frac{4}{3} \right)$ where $k$ is the constant of proportionality for the resistive force.

Hints

• Part (a): Use Newton's Second Law. The net force is the driving force minus the resistive force. Remember that a resistive force "proportional to the square of velocity" can be written as $R = kv^2$.
• Part (b): Terminal velocity occurs when the boat stops accelerating (i.e., when acceleration $a = 0$). Set your differential equation to zero and solve for $v$.
• Part (c): Substitute $a = v \frac{dv}{dx}$ into your equation of motion. Rearrange to separate the variables $x$ and $v$. You will need to integrate both sides. The limits of integration for velocity will be from $0$ to $\frac{1}{2}V_T$, and for displacement from $0$ to $D$.

Solutions

Part (a): Setting up the Differential Equation

1. Let the driving force be $F$.
2. The resistive force is proportional to $v^2$, so $R = kv^2$, where $k > 0$ is a constant of proportionality.
3. The net force acting on the boat is $F_{net} = F - kv^2$.
4. Using Newton's Second Law ($F_{net} = Ma$, where $a = \frac{dv}{dt}$): $M \frac{dv}{dt} = F - kv^2$ This is the differential equation of motion.

Part (b): Finding Terminal Velocity

1. Terminal velocity $V_T$ is reached when the net force is zero, meaning acceleration is zero ($\frac{dv}{dt} = 0$).
2. Substitute $\frac{dv}{dt} = 0$ into the equation of motion: $0 = F - k(V_T)^2$
3. Solve for $V_T$: $k(V_T)^2 = F$ $(V_T)^2 = \frac{F}{k}$ $V_T = \sqrt{\frac{F}{k}}$ (taking the positive root as speed is positive).

Part (c): Calculating Distance Travelled

1. We need distance, so we use the alternate form of acceleration: $a = v \frac{dv}{dx}$.
2. Substitute this into the equation of motion: $M \left( v \frac{dv}{dx} \right) = F - kv^2$
3. Rearrange to separate variables $x$ and $v$: $\frac{dx}{dv} = \frac{Mv}{F - kv^2}$ $dx = \frac{Mv}{F - kv^2} dv$
4. Integrate both sides to find the distance $D$. The boat starts from rest ($x=0, v=0$) and reaches half terminal velocity ($x=D, v=\frac{1}{2}V_T$): $\int_0^D 1 \, dx = \int_0^{\frac{1}{2}V_T} \frac{Mv}{F - kv^2} \, dv$ $D = M \int_0^{\frac{1}{2}V_T} \frac{v}{F - kv^2} \, dv$
5. Notice that the derivative of the denominator is $-2kv$. We adjust the numerator to match this: $D = M \left( -\frac{1}{2k} \right) \int_0^{\frac{1}{2}V_T} \frac{-2kv}{F - kv^2} \, dv$ $D = -\frac{M}{2k} \left[ \ln |F - kv^2| \right]_0^{\frac{1}{2}V_T}$
6. Substitute the limits of integration: $D = -\frac{M}{2k} \left( \ln \left| F - k\left(\frac{1}{2}V_T\right)^2 \right| - \ln |F - k(0)^2| \right)$
7. We know from (b) that $(V_T)^2 = \frac{F}{k}$. Substitute this into our expression: $k\left(\frac{1}{2}V_T\right)^2 = k \left( \frac{1}{4} (V_T)^2 \right) = \frac{k}{4} \left( \frac{F}{k} \right) = \frac{F}{4}$
8. Now substitute this back into the distance equation: $D = -\frac{M}{2k} \left( \ln \left| F - \frac{F}{4} \right| - \ln |F| \right)$ $D = -\frac{M}{2k} \left( \ln \left( \frac{3F}{4} \right) - \ln (F) \right)$
9. Use logarithm laws to simplify: $D = -\frac{M}{2k} \ln \left( \frac{\frac{3F}{4}}{F} \right)$ $D = -\frac{M}{2k} \ln \left( \frac{3}{4} \right)$
10. Finally, apply the logarithm power rule ($- \ln(x) = \ln(x^{-1})$): $D = \frac{M}{2k} \ln \left( \left(\frac{3}{4}\right)^{-1} \right)$ $D = \frac{M}{2k} \ln \left( \frac{4}{3} \right)$

Takeaways

• Language of Proportionality: Recognizing phrases like "proportional to velocity squared" and translating them to $R = kv^2$ is the vital first step in modelling physical systems.
• Choosing the Right Rate of Change: Acceleration can be represented as $\frac{dv}{dt}$, $\frac{d^2x}{dt^2}$, or $v \frac{dv}{dx}$. The choice depends entirely on what you are trying to find. If the question involves time, use $\frac{dv}{dt}$. If it links velocity and displacement (like finding distance), use $v \frac{dv}{dx}$.
• Terminal Velocity Dynamics: Terminal velocity is a specific rate of change state where $\frac{dv}{dt} = 0$. Identifying this allows you to define constants in terms of maximum speeds.

Further Readings

• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Differential Equations: https://vumaths.com/booklets/hsc-differential-equations/
• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/

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