Integrate acceleration and velocity using initial conditions
Recover velocity and position with separate integration constants, using conditions at any stated time rather than assuming a zero start.
Section 4.2 · Kinematics of motion in a straight line
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Integrate powers using Paper 1 techniques.
- Substitute a time and solve for an unknown constant.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If a = 2t, which expression is the general velocity?
What you will learn
- Integrate acceleration to velocity and velocity to position.
- Use supplied conditions to determine constants and evaluate displacement over an interval.
Each integration needs its own condition
Acceleration gives a rate, so integrating a(t) produces v(t) plus a constant representing information that the rate alone cannot supply. Integrating velocity then produces s(t) with a second, independent constant. For a = 6t − 4, v = 3t² − 4t + C. If v(0) = 3, then C = 3. A further integration gives s = t³ − 2t² + 3t + D; if s(0) = 2, D = 2.
At t = 2 s, this model gives velocity 7 m s⁻¹ and position 8 m. Displacement from t = 0 is 8 − 2 = 6 m, not 8 m. The initial position is a coordinate relative to an origin and need not be zero. Differentiate the completed s(t) back to v(t) and a(t), and substitute both supplied conditions to check the recovered motion rather than only its final value.
A condition can be given away from t = 0
If a = 4 m s⁻² and v(2) = 3 m s⁻¹, integration gives v = 4t + C and 3 = 8 + C, so C = −5. A condition at t = 2 is not an instruction to set C = 3. Integrating again gives s = 2t² − 5t + D; if s(2) = 7, then D = 9. A condition at another time works by substitution in exactly the same way.
A definite integral of v over [t₁, t₂] gives s(t₂) − s(t₁), which is displacement over that interval regardless of the origin constant. It does not give total distance when v changes sign: locate and split any reversals first. When only acceleration and an initial velocity are known, absolute position cannot be fixed without a position condition, although displacement between times can still be found by integrating the recovered velocity.
Pause and explain
For v = 4t + C and v(2) = 3, what is C?
Worked example
Given a = 6t − 4 m s⁻², v(0) = 3 m s⁻¹ and s(0) = 2 m, find v(t), s(t) and displacement during the first 2 s.
Show the worked solution
- Integrate a: v = 3t² − 4t + C; the velocity condition gives C = 3.
- Integrate v: s = t³ − 2t² + 3t + D; the position condition gives D = 2.
- At 2 s, s = 8 m, so displacement is 8 − 2 = 6 m; v = 7 m s⁻¹.
Answer v = 3t² − 4t + 3; s = t³ − 2t² + 3t + 2; displacement 6 m in the first 2 s.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle has a = 2 m s⁻², v(0) = −1 m s⁻¹ and s(0) = 5 m. Find v(t), s(t), velocity at 3 s and displacement in the first 3 s.
Give me a hint
Determine one constant from the velocity condition and another from position.
Compare my reasoning
- Integrating gives v = 2t + C and C = −1, so v = 2t − 1.
- Integrating again gives s = t² − t + D and D = 5.
- At 3 s, v = 5 m s⁻¹ and s = 11 m; displacement is 11 − 5 = 6 m.
v = 2t − 1; s = t² − t + 5; v(3) = 5 m s⁻¹; displacement 6 m.
Look for these in your work
- Retain both given initial conditions.
- Subtract the initial position when reporting displacement.
Given v = 3t² − 6t m s⁻¹ and s(1) = 4 m, find s(t) and displacement from t = 1 to t = 3 s.
Give me a hint
The position condition occurs at 1 s, so evaluate the primitive there.
Compare my reasoning
- Integrating gives s = t³ − 3t² + C.
- The condition gives 4 = 1 − 3 + C, hence C = 6.
- At 3 s, s = 6 m; displacement over the specified interval is 6 − 4 = +2 m.
s = t³ − 3t² + 6; displacement +2 m from 1 to 3 s.
Look for these in your work
- Use t = 1 in the given condition.
- Report displacement without claiming it is total distance across the reversal.
For a = 4 m s⁻², v(2) = 3 m s⁻¹ and s(2) = 7 m, recover both functions and find velocity and position at t = 4 s.
Give me a hint
Both constants must satisfy conditions at the same non-zero time.
Compare my reasoning
- Integration gives v = 4t + C; 3 = 8 + C gives v = 4t − 5.
- A second integration gives s = 2t² − 5t + D; 7 = 8 − 10 + D gives D = 9.
- At 4 s, v = 11 m s⁻¹ and s = 21 m; differentiating s recovers the specified v and a.
v = 4t − 5; s = 2t² − 5t + 9; v(4) = 11 m s⁻¹; s(4) = 21 m.
Look for these in your work
- Solve for each constant from its own condition.
- Verify by differentiation and substitution at t = 2.
Common mistakes
- Keep separate constants for the two integrations.
- Substitute a non-zero-time condition instead of copying its value into a constant.
- Position and displacement are different when the initial coordinate is non-zero.
What can you explain now?
If v is known but no position condition is supplied, can absolute position be determined uniquely?
Compare with the explanation
No, an arbitrary additive position constant remains.
Displacement between two times is still determined because that constant cancels.
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Make it stick
Integrate with a fresh constant each time, impose the stated conditions and subtract positions for interval displacement.
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