Differentiate to pass from displacement to velocity to acceleration, and integrate to reverse the process, fixing constants from initial conditions

What this dot point is asking

SEAB wants you to move between displacement, velocity and acceleration using calculus: differentiate to go from displacement to velocity to acceleration, and integrate to go back the other way. Crucially, when you integrate you must use the initial conditions to find the constant of integration, since that is what pins down the actual motion.

The answer

Differentiating down the chain

Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity:

So differentiating displacement once gives velocity, and twice gives acceleration.

Integrating up the chain

Reversing the arrows, integration recovers velocity from acceleration and displacement from velocity:

Each integration introduces a constant of integration that must be found.

Using initial conditions

The constant is fixed by a known value, usually at $t = 0$: an initial velocity gives the constant in the velocity expression, and an initial displacement gives the constant in the displacement expression. Find each constant before using the expression further.

Distance versus displacement over an interval

Over a time interval, $\int v\,dt$ between the limits gives the change in displacement. If the velocity changes sign in the interval, the particle reverses, so to find the total distance travelled you split at the time when $v = 0$ and add the magnitudes of the pieces.

Examples in context

Example 1. A car braking to a stop. Given a car's deceleration, integrating once with the initial speed gives the velocity, and the time when that velocity reaches zero is when it stops; integrating again gives the braking distance.

Example 2. A lift's smooth ride. A lift programmed with a known acceleration profile has its velocity and position found by successive integration, with the starting floor and starting rest condition fixing the constants, ensuring it stops level with the next floor.

Try this

Q1. Given $s = t^3 - t$, find the acceleration. [2 marks]

• Cue. $v = 3t^2 - 1$, so $a = \dfrac{dv}{dt} = 6t$.

Q2. A particle has $a = 6t$ and starts from rest. Find its velocity. [3 marks]

• Cue. $v = \int 6t\,dt = 3t^2 + C$; $v = 0$ at $t = 0$ gives $C = 0$, so $v = 3t^2$.

Q3. A particle has $v = 2t + 1$ and is at $s = 4$ when $t = 0$. Find $s$. [3 marks]

• Cue. $s = \int (2t + 1)\,dt = t^2 + t + C$; $s = 4$ at $t = 0$ gives $C = 4$, so $s = t^2 + t + 4$.