Applications of anti-differentiation to motion in astraight line
In Year 12 Mathematical Methods (Queensland, QCAA), motion is recovered by anti-differentiation — the reverse of the velocity/acceleration chain. From the acceleration \(a(t)\), anti-differentiate for the velocity \(v=\int a\,dt\); from the velocity, anti-differentiate for the displacement \(x=\int v\,dt\). Each step adds a constant of integration, fixed by an initial value — \(v(0)\) for the velocity and \(x(0)\) for the displacement.
Anti-differentiation is the reverse of differentiation. Since differentiating displacement gives velocity and differentiating velocity gives acceleration, anti-differentiating runs the chain the other way: \(a \to v \to x\).
Velocity from acceleration. \(v=\displaystyle\int a\,dt\). The anti-derivative is only known up to a constant, so use the initial velocity \(v(0)\) to find it.
Displacement from velocity. \(x=\displaystyle\int v\,dt\). Again a constant appears; use the initial displacement \(x(0)\) to find it.
Given only the acceleration, anti-differentiate twice, using both initial values (velocity then displacement). The displacement over an interval is the change in \(x\); the distance travelled adds the legs between the times where \(v=0\), because the particle reverses direction there.
Recovering velocity and displacement by anti-differentiation:
Standard-form anti-derivatives (each adds a constant \(c\)):
Displacement and distance over \(t\in[t_{1},t_{2}]\):
How to recover motion by anti-differentiation
- Anti-differentiate once. From \(v\), find \(x=\int v\,dt\); or from \(a\), find \(v=\int a\,dt\). Do not drop the \(+c\).
- Apply the initial value. Substitute \(v(0)\) (for a velocity constant) or \(x(0)\) (for a displacement constant) and solve for the constant.
- Anti-differentiate again if needed. Given the acceleration, integrate a second time and apply the second initial value.
- Evaluate. Substitute the required time into the finished function for the position or velocity.
- Distance travelled. Solve \(v=0\) for the direction changes, evaluate \(x\) at the start, at each change and at the end, then add the absolute change over each leg.
Anti-differentiate, then use \(x(0)=0\).
| \(x=\int(4t+3)\,dt\) | \(=\) | \(2t^{2}+3t+c\) |
| \(x(0)=0\) | \(\Rightarrow\) | \(c=0\) |
| \(x(2)=8+6\) | \(=\) | \(14\;\text{m}\) |
Anti-differentiate, then use \(v(0)=5\).
| \(v=\int(2t-6)\,dt\) | \(=\) | \(t^{2}-6t+c\) |
| \(v(0)=5\) | \(\Rightarrow\) | \(c=5\) |
| \(v\) | \(=\) | \(t^{2}-6t+5\) |
Anti-differentiate twice, using \(v(0)=12\) then \(x(0)=2\).
| \(v=\int(-6)\,dt\) | \(=\) | \(-6t+12\) |
| \(x=\int(-6t+12)\,dt\) | \(=\) | \(-3t^{2}+12t+2\) |
| \(x(2)=-12+24+2\) | \(=\) | \(14\;\text{m}\) |
Anti-differentiate (with \(x(0)=0\)); the particle reverses where \(v=0\).
| \(x=\int(3t^{2}-12t+9)\,dt\) | \(=\) | \(t^{3}-6t^{2}+9t\) |
| \(v=0\) | \(\Rightarrow\) | \(t=1,\ 3\) |
| \(x(0)=0,\ x(1)=4,\ x(3)\) | \(=\) | \(0\) |
| displacement | \(=\) | \(0\;\text{m}\) |
| distance \(=4+4\) | \(=\) | \(8\;\text{m}\) |
Common pitfalls
Frequently asked questions
How do you find displacement from velocity?
Anti-differentiate: \(x=\int v\,dt\), then use \(x(0)\) to fix the constant and substitute the required time.
How do you find displacement from acceleration?
Integrate twice: \(v=\int a\,dt\) using \(v(0)\), then \(x=\int v\,dt\) using \(x(0)\).
Why do you need the constant of integration?
Anti-differentiating only determines the function up to \(+c\); the initial value pins down \(c\) and the one correct function.
What is the difference between displacement and distance travelled?
Displacement is the signed change in position; distance is the total path length. When the particle reverses, add the legs between successive rest positions.
How do you find the total distance travelled?
Find \(x\) by anti-differentiation, solve \(v=0\) for the direction changes, then add the absolute change in position over each leg.