Kinematics with Differentiation and Integration

A complete AA SL and HL lesson on motion in one dimension: displacement, velocity, acceleration, signed area, total distance and initial conditions.

IB Mathematics AA SL/HLCalculus applicationsDifferentiationIntegrationWorked examples

On this page

  1. The big picture
  2. Definitions and notation
  3. Signs and physical meaning
  4. Worked example 1: starting from displacement
  5. Worked example 2: starting from velocity
  6. Worked example 3: starting from acceleration
  7. Average velocity and average speed
  8. AA HL extension: vector motion
  9. Common mistakes
  10. Related videos
  11. Practice

A bridge between differentiation and integration

Kinematics belongs in both the differentiation and integration pathways. Differentiation moves from position to velocity to acceleration. Integration moves in the reverse direction, but requires constants of integration or initial conditions.

DifferentiationRates of change and gradients. Applications of differentiationTurning points and rates of change. IntegrationAccumulation and antiderivatives. Area under a curveSigned area as accumulation. Trapezoidal ruleApproximate area from data.

1. The big picture

Kinematics is the mathematics of motion. In this lesson, we study motion along a straight line, so displacement, velocity and acceleration are signed quantities.

displacement s(t) velocity v(t) acceleration a(t) differentiate differentiate integrate integrate Motion in one dimension The direction of travel is encoded by the sign of velocity.

Kinematics connects differentiation and integration. You must be able to move in both directions.

\[\begin{aligned}v(t)&=\frac{ds}{dt}\\a(t)&=\frac{dv}{dt}=\frac{d^2s}{dt^2}\end{aligned}\]

Horizontal displacement from an origin

O P s(t) origin particle position x-axis

Displacement is measured from an origin \(O\). If the particle is to the right of the origin, \(s(t)\) is positive.

Vertical displacement from an origin

O P s(t) ground / reference line vertical axis

The same idea works vertically. If the object is above the origin, the vertical displacement is positive.

2. Definitions and notation

Displacement

Signed position relative to an origin. It can be positive, negative or zero.

Distance travelled

Total length of the path travelled. It is always non-negative.

Velocity

Rate of change of displacement. It is signed.

\[v(t)=\frac{ds}{dt}\]

Speed

Magnitude of velocity. It is never negative.

\[\mathrm{speed}=|v(t)|\]

Acceleration

Rate of change of velocity.

\[a(t)=\frac{dv}{dt}\]

3. Signs and physical meaning

For motion along a line, the sign of velocity tells us the direction of motion. The sign of acceleration tells us whether velocity is increasing or decreasing, not necessarily whether the particle is speeding up.

\(v(t)>0\)

The particle is moving in the positive direction.

\(v(t)<0\)

The particle is moving in the negative direction.

\(v(t)=0\)

The particle is momentarily at rest.

\(a(t)>0\)

The velocity is increasing.

\(a(t)<0\)

The velocity is decreasing.

Speeding up?

The particle speeds up when velocity and acceleration have the same sign.

Important: acceleration and speed are not the same thing. A positive acceleration does not always mean the particle is speeding up.

Immediate example: suppose \(v=-5\) and \(a=+2\). The particle is moving in the negative direction because the velocity is negative, but the positive acceleration acts in the opposite direction. After 1 second, the velocity would be \(-3\), so the particle is still moving left but more slowly. That means the speed is decreasing, even though the acceleration is positive.

Positive acceleration, but slowing down

P v = -5 a = +2 Velocity and acceleration have opposite signs moving left accelerating right

Because velocity and acceleration have opposite signs, the particle slows down.

What happens after 1 second?

0 v = -5 v = -3 Same direction, smaller speed The velocity is still negative, but its magnitude is smaller.

The speed falls from \(5\) to \(3\), so the particle is slowing down even though \(a>0\).

4. Worked example 1: starting from displacement

A particle has displacement

\[s(t)=t^2-4t+3,\qquad t\geq 0.\]

Find the velocity, acceleration, and the time when the particle is at rest. Interpret the motion.

Differentiate displacement to find velocity:

\[v(t)=s'(t)=2t-4.\]

Differentiate velocity to find acceleration:

\[a(t)=v'(t)=2.\]

The particle is at rest when \(v(t)=0\):

\[2t-4=0\quad\Rightarrow\quad t=2.\]

For \(0\leq t<2\), \(v(t)<0\), so the particle moves in the negative direction. For \(t>2\), \(v(t)>0\), so the particle moves in the positive direction.

s t 1 2 3 4 s(t)=t²−4t+3 decreasing v(t)<0 increasing v(t)>0 turning point at t = 2 horizontal tangent: ds/dt = 0 so v(t) = 0

The turning point occurs exactly at \(t=2\). At this point the tangent is horizontal, so \(\dfrac{ds}{dt}=0\), which means \(v(t)=0\). The graph decreases before \(t=2\) and increases after \(t=2\).

5. Worked example 2: starting from velocity

A particle has velocity

\[v(t)=6t-12,\qquad 0\leq t\leq 4.\]

Find the displacement and the total distance travelled from \(t=0\) to \(t=4\).

The displacement is the signed area under the velocity-time graph:

\[\begin{aligned}\mathrm{displacement}&=\int_0^4(6t-12)\,dt\\&=\left[3t^2-12t\right]_0^4\\&=0.\end{aligned}\]

For total distance, split where \(v(t)=0\):

\[6t-12=0\quad\Rightarrow\quad t=2.\]

On \(0\leq t\leq2\), the velocity is negative. On \(2\leq t\leq4\), the velocity is positive.

\[\begin{aligned}\mathrm{distance}&=\int_0^2(12-6t)\,dt+\int_2^4(6t-12)\,dt\\&=12+12\\&=24.\end{aligned}\]
Displacement: signed area under v(t) 0 2 4 −12 12 −12 +12 signed areas cancel: −12+12=0 Distance: area under |v(t)| 0 2 4 12 12 areas add: 12+12=24

Displacement uses signed area. Total distance uses positive area, so the interval must be split when velocity changes sign.

6. Worked example 3: starting from acceleration

A particle moves in a straight line with acceleration

\[a(t)=6t-4,\qquad t\geq0.\]

Given that \(v(0)=3\) and \(s(0)=2\), find \(v(t)\), \(s(t)\), and the displacement from \(t=0\) to \(t=2\).

Integrate acceleration to find velocity:

\[\begin{aligned}v(t)&=\int(6t-4)\,dt\\&=3t^2-4t+C.\end{aligned}\]

Use \(v(0)=3\):

\[3=C,\qquad\mathrm{so}\qquad v(t)=3t^2-4t+3.\]

Integrate velocity to find displacement:

\[\begin{aligned}s(t)&=\int(3t^2-4t+3)\,dt\\&=t^3-2t^2+3t+D.\end{aligned}\]

Use \(s(0)=2\):

\[D=2,\qquad\mathrm{so}\qquad s(t)=t^3-2t^2+3t+2.\]

The displacement from \(t=0\) to \(t=2\) is:

\[\begin{aligned}s(2)-s(0)&=(8-8+6+2)-2\\&=6.\end{aligned}\]
a(t) 6t−4 v(t) 3t²−4t+3 s(t) t³−2t²+3t+2 integrate integrate use v(0)=3 use s(0)=2 Initial conditions fix the constants of integration

Without initial conditions, integration gives a family of possible motions. Initial conditions choose the correct one.

7. Average velocity and average speed

Average velocity and average speed are not the same calculation. Average velocity uses displacement, while average speed uses total distance travelled.

\[\mathrm{average\ velocity}=\frac{\mathrm{displacement}}{\mathrm{time\ taken}}=\frac{s(t_2)-s(t_1)}{t_2-t_1}\]
\[\mathrm{average\ speed}=\frac{\mathrm{total\ distance\ travelled}}{\mathrm{time\ taken}}\]

Worked calculation using Example 2: in the velocity example above, \(v(t)=6t-12\) on \(0\leq t\leq 4\). We found that the displacement is \(0\) and the total distance travelled is \(24\).

Average velocity

Average velocity uses the net change in position, so we use the displacement.

\[\begin{aligned} \mathrm{average\ velocity} &=\frac{\mathrm{displacement}}{\mathrm{time\ taken}}\\ &=\frac{0}{4-0}\\ &=0. \end{aligned}\]

The average velocity is \(0\) units per second. This is possible because the particle moved in one direction first, then moved back in the opposite direction, so the signed displacement cancelled.

Average speed

Average speed uses how much ground was actually covered, so we use the total distance travelled.

\[\begin{aligned} \mathrm{average\ speed} &=\frac{\mathrm{total\ distance}}{\mathrm{time\ taken}}\\ &=\frac{24}{4-0}\\ &=6. \end{aligned}\]

The average speed is \(6\) units per second.

Average velocity uses displacement: signed change in position start finishes back at start displacement = 0 Average speed uses total distance: all movement counts positive start turns moves back distance = 24

Average velocity can be zero even when the particle has moved. Average speed measures how much motion actually took place.

Exam tip: if the question asks for average velocity, use displacement. If it asks for average speed, find total distance first, which may require splitting the interval where \(v(t)=0\).

8. AA HL extension: vector motion

For AA HL students, it is useful to connect one-dimensional kinematics to vector motion. A position vector can depend on time, and the velocity vector is the derivative of the position vector.

Suppose

\[\vec{r}(t)=\begin{pmatrix}t^2-1\\2t+1\end{pmatrix}.\]

Find the velocity vector, acceleration vector, and speed when \(t=2\).

\[\vec{v}(t)=\vec{r}\,'(t)=\begin{pmatrix}2t\\2\end{pmatrix}\]
\[\vec{a}(t)=\vec{v}\,'(t)=\begin{pmatrix}2\\0\end{pmatrix}\]

At \(t=2\),

\[\vec{v}(2)=\begin{pmatrix}4\\2\end{pmatrix}.\]

The speed is the magnitude of the velocity vector:

\[\left|\vec{v}(2)\right|=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt5.\]
t=2 velocity vector x y Vector motion extension

For vector motion, velocity points tangent to the path. Speed is the magnitude of the velocity vector.

9. Common mistakes

Displacement vs distance

Displacement can cancel because it is signed. Distance travelled cannot cancel.

Velocity vs speed

Velocity is signed. Speed is \(|v(t)|\).

Forgetting to split

When \(v(t)\) changes sign, split the interval before calculating total distance.

Missing constants

When integrating, use initial conditions to find constants.

Misreading acceleration

\(a(t)>0\) means velocity is increasing, not automatically that speed is increasing.

Units

If velocity is in metres per second and time is in seconds, displacement is in metres.

10. Related Radford Mathematics videos

These videos are from the Radford Mathematics IB AA Kinematics sequence and connect directly to velocity, displacement, distance travelled and integration of velocity.

IB Math AA Kinematics: Velocity Function

Finding Distance from Velocity

11. Practice with solutions

  1. A particle has \(s(t)=t^3-6t^2+9t\). Find \(v(t)\), \(a(t)\), and the times when the particle is at rest.
    Show solution
    \[v(t)=3t^2-12t+9=3(t-1)(t-3)\]
    \[a(t)=6t-12\]

    The particle is at rest when \(t=1\) or \(t=3\).

  2. For \(v(t)=t^2-5t+6\), find the displacement from \(t=0\) to \(t=4\).
    Show solution
    \[\int_0^4(t^2-5t+6)\,dt=\left[\frac{t^3}{3}-\frac{5t^2}{2}+6t\right]_0^4=\frac{16}{3}\]
  3. For \(v(t)=t^2-5t+6\), find the total distance travelled from \(t=0\) to \(t=4\).
    Show solution

    Factor \(v(t)=(t-2)(t-3)\), so split at \(t=2\) and \(t=3\).

    \[\mathrm{distance}=\int_0^2 v(t)\,dt-\int_2^3 v(t)\,dt+\int_3^4 v(t)\,dt\]
    \[=\frac{14}{3}+\frac{1}{6}+\frac{5}{6}=\frac{17}{3}\]
  4. A particle has \(a(t)=4t-6\), \(v(0)=5\). Find \(v(t)\).
    Show solution
    \[v(t)=\int(4t-6)\,dt=2t^2-6t+C\]

    Since \(v(0)=5\), \(C=5\), so \(v(t)=2t^2-6t+5\).

  5. A velocity-time graph is below the axis from \(t=0\) to \(t=3\) with area \(10\), then above the axis from \(t=3\) to \(t=5\) with area \(7\). Find the displacement and total distance.
    Show solution

    Displacement is signed area: \(-10+7=-3\). Total distance is positive area: \(10+7=17\).

  6. HL extension: \(\vec{r}(t)=\begin{pmatrix}t^2\\t^3\end{pmatrix}\). Find \(\vec{v}(t)\), \(\vec{a}(t)\), and the speed when \(t=1\).
    Show solution
    \[\vec{v}(t)=\begin{pmatrix}2t\\3t^2\end{pmatrix},\qquad \vec{a}(t)=\begin{pmatrix}2\\6t\end{pmatrix}\]

    At \(t=1\), \(\vec{v}(1)=\begin{pmatrix}2\\3\end{pmatrix}\), so the speed is \(\sqrt{2^2+3^2}=\sqrt{13}\).

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