What are not using the initial conditions?

The constants must be found from the given values; leaving c in the answer loses marks.

SingaporeAdditional MathematicsSyllabus dot point

How do we use integration to recover velocity from acceleration, and displacement from velocity?

Integrate acceleration to find velocity and integrate velocity to find displacement, using initial conditions to find the constant

A focused answer to the N(A)-Level Additional Mathematics outcome on using integration in kinematics. Integrate acceleration to get velocity and velocity to get displacement, and use initial conditions to find the constant of integration.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use integration to go backwards through the kinematics chain: from acceleration to velocity, and from velocity to displacement. Because integration introduces a constant, you use a given initial condition (such as the velocity or displacement at $t = 0$ ) to find that constant. This is the reverse of the differentiation method and relies on the integration skills from the calculus strand.

The answer

Velocity from acceleration

Since acceleration is the rate of change of velocity, integrating acceleration recovers velocity:

v = \int a\, dt

Integrate the acceleration function term by term, remembering the constant of integration.

Displacement from velocity

Likewise, integrating velocity recovers displacement:

s = \int v\, dt

So one integration takes you from acceleration to velocity, and a second takes you from velocity to displacement, exactly reversing the two differentiations.

Using initial conditions to find the constant

Each integration brings a constant $+ c$ . To find it, substitute a known value, typically given at $t = 0$ :

"velocity is $u$ when $t = 0$ " fixes the constant in the velocity function,
"displacement is $s_0$ when $t = 0$ " fixes the constant in the displacement function.

Substitute the condition, solve for $c$ , and write the function with the value of $c$ in place. Without this step the answer is incomplete.

Distance over an interval

To find the distance travelled between two times, you can use a definite integral of the velocity between those limits (taking magnitudes if the velocity changes sign). This connects directly to the area-under-a-curve idea, since distance is the area under the velocity-time graph.

Worked example

A particle has acceleration $a = 6t - 2\ \text{m s}^{-2}$ . When $t = 0$ , its velocity is $4\ \text{m s}^{-1}$ and its displacement is $0$ . Find the displacement $s$ in terms of $t$ .

Step 1: Integrate acceleration for velocity

v = \int (6t - 2)\, dt = 3t^2 - 2t + c_1

Step 2: Use the initial velocity

When $t = 0$ , $v = 4$ : $4 = 0 - 0 + c_1$ , so $c_1 = 4$ . Thus $v = 3t^2 - 2t + 4$ .

Step 3: Integrate velocity for displacement

s = \int (3t^2 - 2t + 4)\, dt = t^3 - t^2 + 4t + c_2

Step 4: Use the initial displacement

When $t = 0$ , $s = 0$ : $0 = 0 - 0 + 0 + c_2$ , so $c_2 = 0$ . Therefore:

s = t^3 - t^2 + 4t

Examples in context

Example 1. Motion under gravity. An object in free fall has constant acceleration. Integrating that constant gives a velocity that increases linearly with time, and integrating again gives the familiar displacement that grows with the square of time. The standard motion equations come straight from this double integration.

Example 2. Designing a speed profile. Engineers specify how a train should accelerate, then integrate to find the resulting speed and the distance covered, checking it stops in the right place. Integrating a chosen acceleration to predict velocity and position is exactly this kinematics method applied in design.

Try this

Q1. Given $v = 2t$ , and $s = 0$ when $t = 0$ , find $s$ . [2 marks]

Cue. $s = \int 2t\, dt = t^2 + c$ ; with $s = 0$ at $t = 0$ , $c = 0$ , so $s = t^2$ .

Q2. Given $a = 4$ , and $v = 3$ when $t = 0$ , find $v$ . [2 marks]

Cue. $v = \int 4\, dt = 4t + c$ ; $c = 3$ , so $v = 4t + 3$ .

Q3. Given $v = 3t^2$ , find the distance travelled from $t = 0$ to $t = 2$ . [3 marks]

Cue. $\displaystyle\int_0^2 3t^2\, dt = [t^3]_0^2 = 8$ metres (velocity stays positive, so distance equals displacement).

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA particle moves with velocity

v = 3t^2 - 4t

m/s. Given that its displacement is

s = 0

when

t = 0

, find

s

in terms of

t

Show worked answer →

Displacement is the integral of velocity: $s = \int (3t^2 - 4t)\, dt = t^3 - 2t^2 + c$ .

Use the initial condition $s = 0$ when $t = 0$ : $0 = 0 - 0 + c$ , so $c = 0$ .

Therefore $s = t^3 - 2t^2$ .

What markers reward: integrating the velocity, including the constant, and using the initial condition to find $c = 0$ and hence the displacement function.

Original4 marksA particle has acceleration

a = 6t

m/s squared. Its velocity is

v = 5

m/s when

t = 0

. Find

v

in terms of

t

Show worked answer →

Velocity is the integral of acceleration: $v = \int 6t\, dt = 3t^2 + c$ .

Use $v = 5$ when $t = 0$ : $5 = 0 + c$ , so $c = 5$ .

Therefore $v = 3t^2 + 5$ .

What markers reward: integrating the acceleration to $3t^2 + c$ , applying the initial velocity to find $c = 5$ , and stating the velocity function.