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.
Key idea. \(a \xrightarrow{\;\int dt\;} v \xrightarrow{\;\int dt\;} x\). Anti-differentiate to go from acceleration to velocity to displacement, fixing each \(+c\) from an initial value.
Displacement-time \(x=t^{3}-6t^{2}+9t\) (from \(v=3t^{2}-12t+9\), \(x(0)=0\)): out to \((1,4)\), back to the origin at \(t=3\) — displacement \(0\), distance \(8\) m
Velocity-time \(v=t^{2}-6t+5\) (from \(a=2t-6\), \(v(0)=5\)): the anti-derivative starts at \(v=5\) and is zero at \(t=1\) and \(t=5\)
Recovering velocity and displacement by anti-differentiation:
\[v=\int a\,dt,\qquad x=\int v\,dt\]
Standard-form anti-derivatives (each adds a constant \(c\)):
Fixing the constant. After anti-differentiating, substitute the initial value: \(v(0)\) determines the constant in \(v\), and \(x(0)\) determines the constant in \(x\).
Displacement and distance over \(t\in[t_{1},t_{2}]\):
\[\text{displacement}=x(t_{2})-x(t_{1});\qquad \text{distance}=\text{sum of the legs between the times where } v=0\]
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.
Displacement vs distance. If the particle changes direction (where \(v=0\)), the distance is the sum of the distances between successive positions, not just \(x(t_{2})-x(t_{1})\).
Example 1 — Displacement from velocity
A particle has velocity \(v=4t+3\) m/s and starts at the origin. Find its displacement when \(t=2\) s.
Solution
Displacement — anti-differentiate the velocity, \(x=\displaystyle\int v\,dt\):
\(x\)
\(=\)
\(\displaystyle\int(4t+3)\,dt\)
\(=\)
\(2t^{2}+3t+c\)
Constant — the particle starts at the origin, so \(x(0)=0\):
\(x(0)\)
\(=\)
\(2(0)^{2}+3(0)+c\)
\(0\)
\(=\)
\(c\)
Evaluate — substitute \(t=2\) into \(x=2t^{2}+3t\):
\(x(2)\)
\(=\)
\(2(2)^{2}+3(2)\)
\(=\)
\(8+6\)
\(=\)
\(14\)
\(\therefore\) Displacement at \(t=2\): \(x=14\;\text{m}\)
Example 2 — Velocity from acceleration
A particle has acceleration \(a=2t-6\) m/s\(^{2}\) with initial velocity \(5\) m/s. Find \(v(t)\).
Solution
Velocity — anti-differentiate the acceleration, \(v=\displaystyle\int a\,dt\):
\(v\)
\(=\)
\(\displaystyle\int(2t-6)\,dt\)
\(=\)
\(t^{2}-6t+c\)
Constant — use the initial velocity \(v(0)=5\):
\(v(0)\)
\(=\)
\((0)^{2}-6(0)+c\)
\(5\)
\(=\)
\(c\)
Write \(v(t)\) — substitute \(c=5\):
\(v\)
\(=\)
\(t^{2}-6t+5\)
\(\therefore\) Velocity: \(v=t^{2}-6t+5\)
Example 3 — Two initial values
A particle has constant acceleration \(a=-6\) m/s\(^{2}\). At \(t=0\) it is at \(x=2\) m with velocity \(12\) m/s. Find its displacement at \(t=2\) s.
Solution
Velocity — anti-differentiate \(a\), then use \(v(0)=12\):
\(v\)
\(=\)
\(\displaystyle\int(-6)\,dt=-6t+c_{1}\)
\(v(0)=12\)
\(\Rightarrow\)
\(c_{1}=12\)
\(v\)
\(=\)
\(-6t+12\)
Displacement — anti-differentiate \(v\), then use \(x(0)=2\):
Do not drop the constant of integration. Every anti-differentiation adds a \(+c\). Forgetting it — or not applying \(v(0)\) or \(x(0)\) — gives the wrong velocity or displacement.
Recovering position needs anti-differentiation, not substitution. Putting a time into \(v\) or \(a\) gives a velocity or acceleration, not the displacement; you must integrate first.
Distance is not just the change in displacement. When the particle reverses direction (where \(v=0\)), add the distances between successive rest positions rather than subtracting the end displacements.
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.