Variable Acceleration

What you'll learn

How displacement, velocity and acceleration are linked using calculus.
How to use initial conditions to find constants of integration.
How to find when a particle is instantaneously at rest or changes direction.
How to calculate total distance travelled, including when velocity becomes negative.

Why this topic needs calculus

In constant acceleration questions, you used SUVAT. In variable acceleration, acceleration is not fixed; it changes with time. That means the SUVAT formulae usually do not apply.

Instead, you use differentiation and integration.

Definition

Variable acceleration

A particle has variable acceleration when its acceleration is given as a function of time, such as $a = 4 t - 8$ , rather than as one constant value.

The big idea is:

A flow diagram showing how displacement, velocity and acceleration are connected by differentiation and integration.

\frac{ds}{dt}, \qquad a = \frac{dv}{dt}

So:

Differentiate displacement to get velocity.
Differentiate velocity to get acceleration.
Integrate acceleration to get velocity.
Integrate velocity to get displacement.

Diagram showing displacement, velocity and acceleration linked by differentiation and integration

Key Idea

The calculus chain

If you move from displacement to velocity to acceleration, you differentiate. If you move backwards from acceleration to velocity to displacement, you integrate and include a constant.

From acceleration to velocity

If you are given acceleration as a function of time, use

\frac{dv}{dt}

So to find velocity, integrate acceleration:

\int a \, dt

Because integration introduces a constant, you then use information like “initially the velocity is 6 m s^-1” to find that constant.

Example

Finding velocity from acceleration

A particle moves in a straight line. Its acceleration at time $t$ seconds is $a = 6 t - 12$ . Initially, its velocity is 9 m s^-1. Find $v$ in terms of $t$ .

Start from the relationship between acceleration and velocity:
$\frac{dv}{dt}$
Integrate $a = 6 t - 12$ with respect to $t$ :
$\int (6t - 12)\,dt$
Carry out the integration, remembering the constant:
$v = 3t^2 - 12t + C$
Use the initial condition. Initially means $t = 0$ , and the velocity is 9:
$9 = 3(0)^2 - 12(0) + C$
Hence $C = 9$ , so the velocity is:
$v = 3t^2 - 12t + 9$

Common Mistake

Forgetting the constant

When you integrate, always include $+ C$ . Many mechanics marks come from using the given initial condition correctly.

From velocity to displacement

Definition

Displacement

Displacement is the signed position of a particle from a fixed point, such as the origin $O$ . It can be positive, negative or zero depending on direction.

A number-line model of displacement showing positions to the left of the origin as negative and positions to the right as positive.

Since velocity is the rate of change of displacement,

\frac{ds}{dt}

So to find displacement, integrate velocity:

\int v \, dt

Again, you must use a given displacement value to find the constant.

Example

Finding displacement from velocity

A particle has velocity $v = 5 t - 3$ m s^-1. When $t = 4$ , its displacement from $O$ is 52 m. Find an expression for $s$ , then find its displacement when $t = 8$ .

Use $\frac{ds}{dt}$ , so integrate velocity:
$\int (5t - 3)\,dt$
Integrate term by term:
$\frac{5}{2}t^2 - 3t + C$
Use the condition $s = 52$ when $t = 4$ :
$\frac{5}{2}(4)^2 - 3(4) + C$
Simplify to find $C$ :
$52 = 40 - 12 + C$
So $C = 24$ , giving:
$\frac{5}{2}t^2 - 3t + 24$
Substitute $t = 8$ :
$\frac{5}{2}(8)^2 - 3(8) + 24 = 160$

Tip

Check your units

Velocity has units m s^-1, acceleration has units m s^-2, and displacement has units m. If your answer is a distance or displacement, it should end in metres.

Instantaneously at rest

Definition

Instantaneously at rest

A particle is instantaneously at rest when its velocity is zero at that instant. This means solve $v = 0$ .

This does not necessarily mean the particle stops forever. It may turn around immediately afterwards.

A velocity-time graph illustrating that a particle can be instantaneously at rest when v = 0 and then change direction.

Example

Distance between two rest positions

A particle has acceleration $a = 6 t - 12$ m s^-2. Initially its velocity is 9 m s^-1. Find the distance between the two positions where the particle is instantaneously at rest.

From the earlier result for this acceleration and initial velocity:
$v = 3t^2 - 12t + 9$
Set $v = 0$ to find the rest times:
$3t^2 - 12t + 9 = 0$
Factorise:
$3(t^2 - 4t + 3) = 0$
So the particle is at rest when:
$\quad \text{or} \quad t = 3$
To find the displacement between these times, integrate velocity from $t = 1$ to $t = 3$ :

The signed area under the velocity-time graph between the two rest times gives the displacement between the two positions.

$$
\int_1^3 (3t^2 - 12t + 9)\,dt
$$

6. Find an antiderivative:

$$
\int (3t^2 - 12t + 9)\,dt = t^3 - 6t^2 + 9t
$$

7. Evaluate between the limits:

$$
\left[t^3 - 6t^2 + 9t\right]_1^3 = 0 - 4 = -4
$$

8. The displacement is -4 m, so the distance between the two positions is 4 m.

Common Mistake

Distance is not always displacement

A negative displacement means the movement was in the negative direction. Distance is never negative, so take the magnitude when the question asks for a distance between two positions.

Total distance travelled

Definition

Distance travelled

Total distance travelled is the total length of the path followed. It ignores direction, so it cannot be negative.

On a velocity-time graph:

A velocity-time graph shows displacement as signed area and total distance as the sum of positive areas.

Displacement is the signed area under the graph.
Distance travelled is the total area, treating areas below the axis as positive.

Velocity-time graph showing positive and negative areas and the change of direction where v equals zero

To calculate total distance, split the motion at any time where $v = 0$ and add the absolute changes in displacement.

Example

Total distance when the particle changes direction

A particle has velocity $v = 12 - 2t^2$ m s^-1 for $\ge 0$ . Find the total distance travelled in the first 4 seconds.

For total distance, split the motion at the turning time where velocity is zero.

First find when the particle changes direction by solving $v = 0$ :
$12 - 2t^2 = 0$
Rearrange:
$t^2 = 6$
So the change of direction occurs at:
$\sqrt{6}$
Find the displacement function by integrating velocity:
$\int (12 - 2t^2)\,dt = 12t - \frac{2}{3}t^3 + C$
For distance travelled from $t = 0$ to $t = 4$ , calculate positions at the key times. Taking $s (0) = 0$ :
$\frac{2}{3}t^3$
At the turning time:
$s(6)=126−23(6)3=86s(\sqrt{6}) = 12\sqrt{6} - \frac{2}{3}(\sqrt{6})^3 = 8\sqrt{6}$
At $t = 4$ :
$\frac{2}{3}(4)^3 = \frac{16}{3}$
Add the two distances: from $0$ to $6\sqrt{6}$ , then from $6\sqrt{6}$ back to $4$ :
$86+(86−163)8\sqrt{6} + \left(8\sqrt{6} - \frac{16}{3}\right)$
Therefore the total distance is:
$166−163≈33.916\sqrt{6} - \frac{16}{3} \approx 33.9$

Tip

Use a mini timeline

When total distance is involved, write the key times in order: start time, any times where $v = 0$ , and the end time. Then add the distances over each interval.

Differentiating displacement

Sometimes you are given displacement directly. Then you differentiate once for velocity and twice for acceleration.

Starting from displacement, one differentiation gives velocity and a second differentiation gives acceleration.

Example

Using displacement to find velocity and acceleration

A particle travels along a line with displacement $x = t^3 - 12t^2 + 45t$ metres from a fixed point $O$ . Find its initial velocity and the time when its acceleration is zero.

Differentiate displacement to get velocity:
$\frac{dx}{dt}$
Differentiate $x = t^3 - 12t^2 + 45t$ :
$v = 3t^2 - 24t + 45$
Initial velocity means the velocity when $t = 0$ :
$v (0) = 45$
Differentiate velocity to get acceleration:
$\frac{dv}{dt} = 6t - 24$
Set acceleration equal to zero:
$6 t - 24 = 0$
Solve:
$t = 4$

Piecewise acceleration or velocity

Definition

Piecewise function

A piecewise function uses different formulae on different intervals. For example, one formula may apply for $\le t \le 3$ , and another for $t > 3$ .

For piecewise motion, be very careful about which formula applies. At the joining time, the final velocity from the first interval becomes the starting velocity for the next interval.

A piecewise motion timeline makes clear which acceleration formula applies on each interval and how the velocity carries over at the join.

Example

Piecewise acceleration

A particle starts from rest. Its acceleration is given by

\begin{cases} 3t - t^2, & 0 \le t \le 3,\\ \frac{18}{t^2}, & t > 3. \end{cases}

Find its speed when $t = 6$ .

For $\le t \le 3$ , integrate acceleration to get velocity. Since the particle starts from rest, $v = 0$ when $t = 0$ :
$\int (3t - t^2)\,dt = \frac{3}{2}t^2 - \frac{1}{3}t^3$
Find the velocity at $t = 3$ :
$\frac{3}{2}(3)^2 - \frac{1}{3}(3)^3 = \frac{9}{2}$
For $t > 3$ , use a definite integral from 3 to 6 and add it to the velocity at 3:
$\frac{9}{2} + \int_3^6 \frac{18}{t^2}\,dt$
Integrate:
$dt=−18t−1\int \frac{18}{t^2}\,dt = \int 18t^{-2}\,dt = -18t^{-1}$
Evaluate from 3 to 6:
$[−18t]36=−3−(−6)=3\left[-\frac{18}{t}\right]_3^6 = -3 - (-6) = 3$
Therefore:
$\frac{9}{2} + 3 = \frac{15}{2}$

Common Mistake

Do not mix intervals

If a formula is only valid for $t > 3$ , you cannot use it to calculate what happened before $t = 3$ . Work interval by interval.

Exam technique

In the exam

Decide what you have been given: displacement, velocity or acceleration.
Move along the chain using differentiation or integration, and include constants when integrating.
For total distance, split at every time where $v = 0$ and add positive distances, not signed displacements.

Self review

Check yourself

If you are given $a (t)$ and an initial velocity, what operation finds $v (t)$ ?
What equation do you solve to find when a particle is instantaneously at rest?
Why might the total distance travelled be different from the final displacement?

Variable Acceleration

What you'll learn

How displacement, velocity and acceleration are linked using calculus.
How to use initial conditions to find constants of integration.
How to find when a particle is instantaneously at rest or changes direction.
How to calculate total distance travelled, including when velocity becomes negative.

Why this topic needs calculus

In constant acceleration questions, you used SUVAT. In variable acceleration, acceleration is not fixed; it changes with time. That means the SUVAT formulae usually do not apply.

Instead, you use differentiation and integration.

Definition

Variable acceleration

A particle has variable acceleration when its acceleration is given as a function of time, such as $a = 4 t - 8$ , rather than as one constant value.

The big idea is:

A flow diagram showing how displacement, velocity and acceleration are connected by differentiation and integration.

\frac{ds}{dt}, \qquad a = \frac{dv}{dt}

So:

Differentiate displacement to get velocity.
Differentiate velocity to get acceleration.
Integrate acceleration to get velocity.
Integrate velocity to get displacement.

Diagram showing displacement, velocity and acceleration linked by differentiation and integration

Key Idea

The calculus chain

If you move from displacement to velocity to acceleration, you differentiate. If you move backwards from acceleration to velocity to displacement, you integrate and include a constant.

From acceleration to velocity

If you are given acceleration as a function of time, use

\frac{dv}{dt}

So to find velocity, integrate acceleration:

\int a \, dt

Because integration introduces a constant, you then use information like “initially the velocity is 6 m s^-1” to find that constant.

Example

Finding velocity from acceleration

A particle moves in a straight line. Its acceleration at time $t$ seconds is $a = 6 t - 12$ . Initially, its velocity is 9 m s^-1. Find $v$ in terms of $t$ .

Start from the relationship between acceleration and velocity:
$\frac{dv}{dt}$
Integrate $a = 6 t - 12$ with respect to $t$ :
$\int (6t - 12)\,dt$
Carry out the integration, remembering the constant:
$v = 3t^2 - 12t + C$
Use the initial condition. Initially means $t = 0$ , and the velocity is 9:
$9 = 3(0)^2 - 12(0) + C$
Hence $C = 9$ , so the velocity is:
$v = 3t^2 - 12t + 9$

Common Mistake

Forgetting the constant

When you integrate, always include $+ C$ . Many mechanics marks come from using the given initial condition correctly.

From velocity to displacement

Definition

Displacement

Displacement is the signed position of a particle from a fixed point, such as the origin $O$ . It can be positive, negative or zero depending on direction.

A number-line model of displacement showing positions to the left of the origin as negative and positions to the right as positive.

Since velocity is the rate of change of displacement,

\frac{ds}{dt}

So to find displacement, integrate velocity:

\int v \, dt

Again, you must use a given displacement value to find the constant.

Example

Finding displacement from velocity

A particle has velocity $v = 5 t - 3$ m s^-1. When $t = 4$ , its displacement from $O$ is 52 m. Find an expression for $s$ , then find its displacement when $t = 8$ .

Use $\frac{ds}{dt}$ , so integrate velocity:
$\int (5t - 3)\,dt$
Integrate term by term:
$\frac{5}{2}t^2 - 3t + C$
Use the condition $s = 52$ when $t = 4$ :
$\frac{5}{2}(4)^2 - 3(4) + C$
Simplify to find $C$ :
$52 = 40 - 12 + C$
So $C = 24$ , giving:
$\frac{5}{2}t^2 - 3t + 24$
Substitute $t = 8$ :
$\frac{5}{2}(8)^2 - 3(8) + 24 = 160$

Tip

Check your units

Velocity has units m s^-1, acceleration has units m s^-2, and displacement has units m. If your answer is a distance or displacement, it should end in metres.

Instantaneously at rest

Definition

Instantaneously at rest

A particle is instantaneously at rest when its velocity is zero at that instant. This means solve $v = 0$ .

This does not necessarily mean the particle stops forever. It may turn around immediately afterwards.

A velocity-time graph illustrating that a particle can be instantaneously at rest when v = 0 and then change direction.

Example

Distance between two rest positions

A particle has acceleration $a = 6 t - 12$ m s^-2. Initially its velocity is 9 m s^-1. Find the distance between the two positions where the particle is instantaneously at rest.

From the earlier result for this acceleration and initial velocity:
$v = 3t^2 - 12t + 9$
Set $v = 0$ to find the rest times:
$3t^2 - 12t + 9 = 0$
Factorise:
$3(t^2 - 4t + 3) = 0$
So the particle is at rest when:
$\quad \text{or} \quad t = 3$
To find the displacement between these times, integrate velocity from $t = 1$ to $t = 3$ :

The signed area under the velocity-time graph between the two rest times gives the displacement between the two positions.

$$
\int_1^3 (3t^2 - 12t + 9)\,dt
$$

6. Find an antiderivative:

$$
\int (3t^2 - 12t + 9)\,dt = t^3 - 6t^2 + 9t
$$

7. Evaluate between the limits:

$$
\left[t^3 - 6t^2 + 9t\right]_1^3 = 0 - 4 = -4
$$

8. The displacement is -4 m, so the distance between the two positions is 4 m.

Common Mistake

Distance is not always displacement

A negative displacement means the movement was in the negative direction. Distance is never negative, so take the magnitude when the question asks for a distance between two positions.

Total distance travelled

Definition

Distance travelled

Total distance travelled is the total length of the path followed. It ignores direction, so it cannot be negative.

On a velocity-time graph:

A velocity-time graph shows displacement as signed area and total distance as the sum of positive areas.

Displacement is the signed area under the graph.
Distance travelled is the total area, treating areas below the axis as positive.

Velocity-time graph showing positive and negative areas and the change of direction where v equals zero

To calculate total distance, split the motion at any time where $v = 0$ and add the absolute changes in displacement.

Example

Total distance when the particle changes direction

A particle has velocity $v = 12 - 2t^2$ m s^-1 for $\ge 0$ . Find the total distance travelled in the first 4 seconds.

For total distance, split the motion at the turning time where velocity is zero.

First find when the particle changes direction by solving $v = 0$ :
$12 - 2t^2 = 0$
Rearrange:
$t^2 = 6$
So the change of direction occurs at:
$\sqrt{6}$
Find the displacement function by integrating velocity:
$\int (12 - 2t^2)\,dt = 12t - \frac{2}{3}t^3 + C$
For distance travelled from $t = 0$ to $t = 4$ , calculate positions at the key times. Taking $s (0) = 0$ :
$\frac{2}{3}t^3$
At the turning time:
$s(6)=126−23(6)3=86s(\sqrt{6}) = 12\sqrt{6} - \frac{2}{3}(\sqrt{6})^3 = 8\sqrt{6}$
At $t = 4$ :
$\frac{2}{3}(4)^3 = \frac{16}{3}$
Add the two distances: from $0$ to $6\sqrt{6}$ , then from $6\sqrt{6}$ back to $4$ :
$86+(86−163)8\sqrt{6} + \left(8\sqrt{6} - \frac{16}{3}\right)$
Therefore the total distance is:
$166−163≈33.916\sqrt{6} - \frac{16}{3} \approx 33.9$

Tip

Use a mini timeline

When total distance is involved, write the key times in order: start time, any times where $v = 0$ , and the end time. Then add the distances over each interval.

Differentiating displacement

Sometimes you are given displacement directly. Then you differentiate once for velocity and twice for acceleration.

Starting from displacement, one differentiation gives velocity and a second differentiation gives acceleration.

Example

Using displacement to find velocity and acceleration

A particle travels along a line with displacement $x = t^3 - 12t^2 + 45t$ metres from a fixed point $O$ . Find its initial velocity and the time when its acceleration is zero.

Differentiate displacement to get velocity:
$\frac{dx}{dt}$
Differentiate $x = t^3 - 12t^2 + 45t$ :
$v = 3t^2 - 24t + 45$
Initial velocity means the velocity when $t = 0$ :
$v (0) = 45$
Differentiate velocity to get acceleration:
$\frac{dv}{dt} = 6t - 24$
Set acceleration equal to zero:
$6 t - 24 = 0$
Solve:
$t = 4$

Piecewise acceleration or velocity

Definition

Piecewise function

A piecewise function uses different formulae on different intervals. For example, one formula may apply for $\le t \le 3$ , and another for $t > 3$ .

For piecewise motion, be very careful about which formula applies. At the joining time, the final velocity from the first interval becomes the starting velocity for the next interval.

A piecewise motion timeline makes clear which acceleration formula applies on each interval and how the velocity carries over at the join.

Example

Piecewise acceleration

A particle starts from rest. Its acceleration is given by

\begin{cases} 3t - t^2, & 0 \le t \le 3,\\ \frac{18}{t^2}, & t > 3. \end{cases}

Find its speed when $t = 6$ .

For $\le t \le 3$ , integrate acceleration to get velocity. Since the particle starts from rest, $v = 0$ when $t = 0$ :
$\int (3t - t^2)\,dt = \frac{3}{2}t^2 - \frac{1}{3}t^3$
Find the velocity at $t = 3$ :
$\frac{3}{2}(3)^2 - \frac{1}{3}(3)^3 = \frac{9}{2}$
For $t > 3$ , use a definite integral from 3 to 6 and add it to the velocity at 3:
$\frac{9}{2} + \int_3^6 \frac{18}{t^2}\,dt$
Integrate:
$dt=−18t−1\int \frac{18}{t^2}\,dt = \int 18t^{-2}\,dt = -18t^{-1}$
Evaluate from 3 to 6:
$[−18t]36=−3−(−6)=3\left[-\frac{18}{t}\right]_3^6 = -3 - (-6) = 3$
Therefore:
$\frac{9}{2} + 3 = \frac{15}{2}$

Common Mistake

Do not mix intervals

If a formula is only valid for $t > 3$ , you cannot use it to calculate what happened before $t = 3$ . Work interval by interval.

Exam technique

In the exam

Decide what you have been given: displacement, velocity or acceleration.
Move along the chain using differentiation or integration, and include constants when integrating.
For total distance, split at every time where $v = 0$ and add positive distances, not signed displacements.

Self review

Check yourself

If you are given $a (t)$ and an initial velocity, what operation finds $v (t)$ ?
What equation do you solve to find when a particle is instantaneously at rest?
Why might the total distance travelled be different from the final displacement?

Variable Acceleration

What you'll learn

Why this topic needs calculus

From acceleration to velocity

From velocity to displacement

Instantaneously at rest

Total distance travelled

Differentiating displacement

Piecewise acceleration or velocity

Recap questions

You've reached the end

Variable Acceleration

What you'll learn

Why this topic needs calculus

From acceleration to velocity

From velocity to displacement

Instantaneously at rest

Total distance travelled

Differentiating displacement

Piecewise acceleration or velocity

Recap questions

You've reached the end

Variable Acceleration

What you'll learn

Why this topic needs calculus

From acceleration to velocity

From velocity to displacement

Instantaneously at rest

Total distance travelled

Differentiating displacement

Piecewise acceleration or velocity

Recap questions

You've reached the end

Variable Acceleration Revision Guide

Variable Acceleration

What you'll learn

Why this topic needs calculus

From acceleration to velocity

From velocity to displacement

Instantaneously at rest

Total distance travelled

Differentiating displacement

Piecewise acceleration or velocity

Recap questions

You've reached the end

Variable Acceleration Revision Guide