Calculus for motion: differentiating and integrating

8 minAS Level4.2.3

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can differentiate a displacement function to get velocity and acceleration, integrate an acceleration function to recover velocity and displacement, and use the starting conditions to fix the constants.

Key to the symbols

Symbol	Meaning
$s(t)$	displacement at time $t$ , a formula in $t$
$v(t)$	velocity at time $t$ , in $\text{m s}^{-1}$
$a(t)$	acceleration at time $t$ , in $\text{m s}^{-2}$
$\dfrac{ds}{dt}$	the derivative of $s$ with respect to $t$
$\dfrac{d^2s}{dt^2}$	the second derivative: differentiate $s$ twice
$C$	a constant of integration

The same motion, three functions

When a particle moves, its displacement changes with time. Write it as $s(t)$ .

The rate of change of displacement is velocity, and the rate of change of velocity is acceleration. In calculus language:

$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

So one function contains all the information. Differentiate once for velocity, twice for acceleration.

Going the other way, integrate: acceleration gives velocity, and velocity gives displacement. Each integration brings an arbitrary constant, fixed by what the particle was doing at $t = 0$ .

The rules you need

Differentiate and integrate powers of $t$ term by term:

$\frac{d}{dt}(t^n) = nt^{n-1}, \qquad \int t^n\,dt = \frac{t^{n+1}}{n+1} + C \quad (n \ne -1)$

A constant differentiates to $0$ , and $kt$ differentiates to $k$ .

What the derivatives tell you

Once you have $v(t)$ and $a(t)$ , the physics is quick to read off:

At rest: $v = 0$ . Solve the velocity equation.
Moving in the positive direction: $v > 0$ ; negative direction: $v < 0$ .
Reversing: $v$ changes sign, which happens where $v = 0$ and the graph of $v$ crosses the axis.
Constant velocity: $a = 0$ throughout an interval of time. (If $a = 0$ at just one instant, the velocity is at a maximum or minimum there, not constant.)
Maximum or minimum displacement: $v = 0$ (a stationary point of $s$ ).

Notice that "at rest" is a statement about $v$ , not about $a$ . A ball at the top of its flight has $v = 0$ but $a = -10\ \text{m s}^{-2}$ (taking upwards as positive).

Worked example: From displacement to velocity and acceleration

A particle moves so that its displacement is $s = t^3 - 6t^2 + 9t \quad (\text{metres, } t \text{ in seconds}).$

Find $v$ and $a$ , the velocity and acceleration at $t = 0$ , and the times when the particle is at rest.

Have a go first?

Worked example: Integrating back from acceleration

A particle has acceleration $a = 6t - 4\ \text{m s}^{-2}$ . At $t = 0$ its velocity is $3\ \text{m s}^{-1}$ .

Find its velocity at $t = 2\ \text{s}$ .

Have a go first?

Never lose the constant

Integrating acceleration gives a whole family of velocity functions, differing by a constant. The starting velocity picks out the one that describes this particle.

The same happens again when you integrate velocity to get displacement: you need the initial displacement.

In an exam, write the " $+C$ " line explicitly, then substitute the initial condition. Skipping that step is the most common way to lose marks in this topic.

Also check the units of your answer. Differentiating a displacement in metres with respect to time in seconds gives metres per second; differentiating again gives metres per second squared.

Common mistakes

To get displacement from velocity, you differentiate.

It is the other way round with derivatives: differentiating $s$ gives $v$ , and differentiating $v$ gives $a$ . Integration goes back.

If $v = 0$ then $a = 0$ as well.

In the exam

Show the differentiation or integration clearly, then answer the question that was asked. If the question says "find the time at rest", do not stop at the velocity expression: solve $v = 0$ .

When a displacement function has powers of $t$ , differentiate term by term and keep track of coefficients.

For an integration question, write " $+C$ ", substitute the given condition, and state the final function before evaluating at the required time.

Check yourself

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