Kinematics and rectilinear motion: VCE Specialist Mathematics Unit 4

What this dot point is asking

VCAA wants you to apply calculus to a particle moving in a straight line: differentiate position to get velocity and velocity to get acceleration, integrate to reverse those steps, and choose the right form of acceleration when it is given as a function of time, velocity or position. You must also distinguish displacement from distance travelled and interpret signs of velocity and acceleration physically.

The chain of derivatives

For a particle moving along a straight line, let $x(t)$ be its position at time $t$. Then

Differentiation moves down the chain (position to velocity to acceleration); integration moves back up. The signs carry meaning: positive velocity means motion in the positive direction, and acceleration with the same sign as velocity means speeding up while opposite signs mean slowing down.

The three forms of acceleration

Acceleration can be expressed three equivalent ways, and choosing the right one is the central skill:

The second and third forms come from the chain rule $\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}t} = v\frac{\mathrm{d}v}{\mathrm{d}x}$, and $v\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(\tfrac12 v^2)$ since $\frac{\mathrm{d}}{\mathrm{d}x}(\tfrac12 v^2) = v\frac{\mathrm{d}v}{\mathrm{d}x}$. Choose:

• $a = \frac{\mathrm{d}v}{\mathrm{d}t}$ when acceleration is given as a function of time;
• $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ when acceleration is given as a function of velocity;
• $a = \frac{\mathrm{d}}{\mathrm{d}x}(\tfrac12 v^2)$ when acceleration is given as a function of position $x$, because integrating it directly with respect to $x$ recovers $\tfrac12 v^2$.

Displacement versus distance

Displacement over $[t_1, t_2]$ is the net change in position, $\int_{t_1}^{t_2} v\,\mathrm{d}t = x(t_2) - x(t_1)$, and can be negative or zero. Distance travelled is the total path length, $\int_{t_1}^{t_2} |v|\,\mathrm{d}t$, always non-negative. They agree only when the particle never reverses direction. To compute distance when direction changes, find the times where $v = 0$, integrate $|v|$ piecewise, and add the magnitudes.

Working from acceleration back to position

When acceleration is a function of time, integrate twice. From $a(t)$, velocity is $v(t) = \int a\,\mathrm{d}t + c_1$, with $c_1$ fixed by the initial velocity; then position is $x(t) = \int v\,\mathrm{d}t + c_2$, with $c_2$ fixed by the initial position. Each integration introduces a constant that an initial condition resolves, so two pieces of initial data are needed to recover position fully.

Examples in context

Example 1. Velocity to displacement. If $v = 3t^2 - 12$, the displacement from $t = 0$ to $t = 3$ is $\int_0^3 (3t^2 - 12)\,\mathrm{d}t = [t^3 - 12t]_0^3 = 27 - 36 = -9$, a net move of $9$ in the negative direction.

Example 2. Resistive force. If $a = -2v$ (resistance proportional to speed), use $a = \frac{\mathrm{d}v}{\mathrm{d}t} = -2v$, which separates to $v = v_0 e^{-2t}$, decaying towards rest.

Try this

Q1. A particle has $x = t^3 - 6t^2 + 9t$. Find its velocity and the times it is at rest. [3 marks]

• Cue. $v = 3t^2 - 12t + 9 = 3(t-1)(t-3)$; rest at $t = 1$ and $t = 3$.

Q2. Given $a = 6t$ with $v(0) = 2$ and $x(0) = 0$, find $x(t)$. [3 marks]

• Cue. $v = 3t^2 + 2$; $x = t^3 + 2t$.

Q3. A particle has $v^2 = 16 - x^2$. Find its acceleration in terms of $x$. [2 marks]

• Cue. $\frac{\mathrm{d}}{\mathrm{d}x}(\tfrac12 v^2) = \frac{\mathrm{d}}{\mathrm{d}x}(8 - \tfrac12 x^2) = -x$, so $a = -x$.