Mathematics/Lessons/Variable Acceleration/Using Integration
Mechanics · Variable Acceleration · Year 1

Using Integration

Integration reverses the calculus chain — from acceleration back to velocity, and velocity back to displacement — provided you know one value along the way to pin down each constant of integration.

12 min read 3 worked examples

Integrating: acceleration back to velocity and displacement

Differentiation goes from $s$ to $v$ to $a$; integration reverses each step. Integrating acceleration gives velocity, and integrating velocity gives displacement — but each integration introduces a constant that must be pinned down using extra information, such as an initial velocity or starting position.

Integrating acceleration

$$v = \int a\, dt \qquad s = \int v\, dt$$

Each integral carries a constant of integration. Without an initial condition — a known value of $v$ or $s$ at a specific $t$, almost always $t=0$ — the constant cannot be found, and $v(t)$ or $s(t)$ is only known up to an unknown shift.

Worked example 1

Integrating from acceleration to displacement
A particle starts at the origin with initial velocity $2\ \text{m s}^{-1}$, and its acceleration at time $t$ seconds is $a = 4 - 6t\ \text{m s}^{-2}$. Find expressions for its velocity and displacement at time $t$.
MethodIntegrate twice, fixing a constant each time from an initial condition
1
$v = \displaystyle\int(4-6t)\,dt = 4t - 3t^2 + c$.
Integrate a(t) term by term.
2
At $t=0$, $v=2$: $2 = 0-0+c$, so $c=2$. Hence $v = 4t-3t^2+2$.
Use the initial velocity to find c.
3
$s = \displaystyle\int(4t-3t^2+2)\,dt = 2t^2-t^3+2t+k$.
Integrate v(t) term by term.
4
At $t=0$, $s=0$ (starts at the origin): $0=0-0+0+k$, so $k=0$. Hence $s = 2t^2-t^3+2t$.
Use the starting position to find k.
▪ solved

Displacement as a definite integral

If the acceleration or velocity function is already known, and the constants of integration aren't needed for the question being asked, a definite integral gives the displacement over an interval directly — this is the same "area under a velocity–time graph" idea from constant acceleration, now applied to a curved graph.

Worked example 2

Displacement over an interval
A particle's velocity at time $t$ seconds is $v = 3t^2+2\ \text{m s}^{-1}$. Find its displacement between $t=1$ and $t=3$.
MethodIntegrate v between the two limits
1
$\displaystyle\int_1^3 (3t^2+2)\,dt = \Big[t^3+2t\Big]_1^3$.
Find the integral first, then apply limits.
2
$= (27+6)-(1+2) = 33-3 = 30\ \text{m}$.
Upper limit minus lower limit.
▪ solved

Worked example 3

Total distance when velocity changes signChallenge
A particle's velocity at time $t$ seconds ($0 \le t \le 5$) is $v = t^2 - 6t + 8\ \text{m s}^{-1}$. Find the total distance travelled by the particle in this interval.
MethodFind where v changes sign, then integrate each piece separately
1
$t^2-6t+8=0 \Rightarrow (t-2)(t-4)=0$, so $v$ changes sign at $t=2$ and $t=4$, splitting $[0,5]$ into three pieces.
Zeros of v mark changes of direction.
2
$\displaystyle\int_0^2(t^2-6t+8)\,dt = \Big[\tfrac{t^3}{3}-3t^2+8t\Big]_0^2 = \tfrac83-12+16 = \tfrac{20}{3}$.
v > 0 here — genuine forward distance.
3
$\displaystyle\int_2^4(t^2-6t+8)\,dt = \Big[\tfrac{t^3}{3}-3t^2+8t\Big]_2^4 = \tfrac{16}{3}-\tfrac{20}{3} = -\tfrac43$.
Negative: the particle moves backward here.
4
$\displaystyle\int_4^5(t^2-6t+8)\,dt = \Big[\tfrac{t^3}{3}-3t^2+8t\Big]_4^5 = \tfrac{20}{3}-\tfrac{16}{3} = \tfrac43$.
Positive again: forward motion resumes.
5
Total distance $= \tfrac{20}{3}+\left|-\tfrac43\right|+\tfrac43 = \tfrac{20}{3}+\tfrac83 = \tfrac{28}{3} = 9.33\ \text{m}$ (3 s.f.).
Sum the magnitudes, never the signed values.
▪ solved
Compare with the net displacement

A single integral $\displaystyle\int_0^5(t^2-6t+8)\,dt = \tfrac{20}{3}-\tfrac43+\tfrac43 = \tfrac{20}{3} \approx 6.67\ \text{m}$ gives the net displacement — smaller than the total distance of $9.33\ \text{m}$, because the backward and forward pieces partly cancel when added with their signs.

Common mistake

Forgetting the constant of integration entirely, or assuming it is always zero — it represents whatever initial condition the question gives, and is only zero when the particle genuinely starts from rest or from the origin. The second frequent error is reporting $\displaystyle\int_a^b v\,dt$ as the "distance travelled" without first checking whether $v$ changes sign anywhere in $[a,b]$.

Check yourself

A particle starts at rest at the origin, and its acceleration at time $t$ seconds is $a = 6t-8\ \text{m s}^{-2}$. Find expressions for $v(t)$ and $s(t)$, and find the displacement when $t=3$.
(Hint: "starts at rest at the origin" fixes both constants of integration to zero.)

$v = \displaystyle\int(6t-8)\,dt = 3t^2-8t+c$. Since $v=0$ at $t=0$, $c=0$, so $v=3t^2-8t$.

$s = \displaystyle\int(3t^2-8t)\,dt = t^3-4t^2+k$. Since $s=0$ at $t=0$, $k=0$, so $s=t^3-4t^2$. At $t=3$: $s = 27-36 = -9\ \text{m}$.

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