Calculus relationships between displacement, velocity and acceleration, derivation of the equations of motion by integration, and motion under constant and varying acceleration.

What this key area is asking

The SQA wants you to treat motion with calculus: to know that velocity is the rate of change of displacement and acceleration the rate of change of velocity, to derive the equations of motion by integration, and to use differentiation or integration to find instantaneous displacement, velocity and acceleration for straight-line motion under constant or varying acceleration. This is the calculus foundation the whole Advanced Higher course rests on.

Velocity and acceleration as derivatives

On a displacement-time graph, the gradient at a point is the instantaneous velocity, found by differentiation. On a velocity-time graph the gradient is the instantaneous acceleration, and the area under the line between two times is the displacement, found by integration. The calculus simply makes these graph rules exact for any curve, not just straight lines.

If you are given displacement as a function of time, differentiate once for velocity and again for acceleration. If you are given acceleration, integrate once for velocity and again for displacement. Each integration introduces a constant of integration that you find from the initial conditions (the velocity or displacement at $t = 0$).

Deriving the equations of motion

Start from constant acceleration $a$. Integrating with respect to time:

At $t = 0$ the velocity is the initial velocity $u$, so $c = u$ and:

Now integrate the velocity to get displacement:

Taking displacement as zero at $t = 0$ gives $c = 0$, so:

The third equation, $v^2 = u^2 + 2as$, follows by eliminating $t$ between these two. The crucial point the SQA tests is that these results hold only for constant acceleration, because the integration treated $a$ as a constant.

Motion under varying acceleration

When acceleration is not constant, the equations of motion fail and you must work from the calculus directly.

For example, if $s = 2t^4$ then $v = 8t^3$ and $a = 24t^2$, so the acceleration grows with time and no single equation of motion applies. Reading the question to decide whether acceleration is constant (use the equations of motion) or varying (use calculus) is the key first step.

Examples in context

A car pulling away has a roughly constant acceleration, so the equations of motion fit, but a rocket burns fuel and loses mass, so its acceleration rises and only calculus describes it. A damped oscillator has an acceleration that depends on position and time, so its motion is found by solving the relationship $a = \frac{dv}{dt}$ directly. Engineers fitting a polynomial to motion-sensor data differentiate it to recover velocity and acceleration, exactly the operation this key area drills.

Try this

Q1. State what physical quantity the second derivative of displacement with respect to time represents. [1 mark]

• Cue. The acceleration.

Q2. The displacement of an object is $s = 5t^2 + 2t$. Find an expression for its velocity. [2 marks]

• Cue. $v = \frac{ds}{dt} = 10t + 2\ \text{m s}^{-1}$.

Q3. State why the equation $s = ut + \tfrac{1}{2}at^2$ cannot be used when the acceleration varies with time. [1 mark]

• Cue. It is derived by integrating a constant acceleration, so it only holds for constant $a$.