Velocity is the derivative of position, and acceleration is the derivative of velocity. To go the other way, integrate. Every kinematics question with a formula in t is a path along this chain, and the conditions in the question fix the constants.
This lesson joins the integration from calculating displacement from velocity with the differentiation you met in tangents, normals and rates. It also closes the loop on distance versus displacement.
How do the three quantities connect?
Use this chain, written from left to right:
position s → velocity v → acceleration a
Each arrow to the right means differentiate: v = ds/dt and a = dv/dt. Each arrow to the left means integrate, and add a constant that a given condition will fix.
The units help. Metres, then metres per second, then metres per second squared: each step divides by one more second.
How to set out a question
- Write the chain first and mark which quantity you are given.
- Choose the direction. From s or v to something further right, differentiate. From a or v to something further left, integrate.
- Add a constant every time you integrate.
- Use the given condition, such as “v = 2 when t = 0”, to find each constant.
- Substitute the required time last.
Worked example
A particle has acceleration a = 6t − 4 m/s². When t = 0 its velocity is 2 m/s and its position is 1 m from O. Find its velocity and position when t = 2.
Step 1, integrate a: v = 3t² − 4t + c. At t = 0, v = 2, so c = 2.
v = 3t² − 4t + 2.
Step 2, integrate v: s = t³ − 2t² + 2t + k. At t = 0, s = 1, so k = 1.
s = t³ − 2t² + 2t + 1.
Step 3, substitute t = 2:
v = 12 − 8 + 2 = 6 m/s.
s = 8 − 8 + 4 + 1 = 5 m from O.
Check by differentiating: ds/dt = 3t² − 4t + 2, which is v. And dv/dt = 6t − 4, which is a. The chain works both ways.
The mistake to watch for
A common slip is to use a constant-acceleration formula when the acceleration changes.
Mistaken answer: at t = 2, a = 8, so v = u + at = 2 + 8 × 2 = 18 m/s.
The student took the acceleration at one moment and assumed it held throughout. But a = 6t − 4 grows with time.
The correction is to integrate. The true answer is 6 m/s, which is far from 18. A quick test: if the acceleration contains t, the constant-acceleration formulas are not allowed.
Check yourself
Try these on paper, then open each answer.
1. A particle has position s = 2t³ − 9t² + 12t metres. Find its velocity and acceleration when t = 1.
Show answer
v = ds/dt = 6t² − 18t + 12. At t = 1: 6 − 18 + 12 = 0.
a = dv/dt = 12t − 18. At t = 1: 12 − 18 = −6.
v = 0 m/s and a = −6 m/s²
2. A particle has acceleration a = 4 − 2t m/s² and velocity 3 m/s when t = 0. Find its velocity when t = 3.
Show answer
v = 4t − t² + c. At t = 0, v = 3, so c = 3.
At t = 3: v = 12 − 9 + 3 = 6 m/s.
3. A particle has velocity v = 12t − 3t² m/s. For t > 0, when is it instantaneously at rest, and what is its acceleration then?
Show answer
v = 3t(4 − t) = 0, so t = 4 for t > 0.
a = dv/dt = 12 − 6t. At t = 4: a = 12 − 24 = −12 m/s².
Where this leads next
With the chain secure, try the areas and motion practice set, which mixes all five lessons. The mistake log and retest queue is a good place to record which direction you chose wrongly so that you can retest it later.
If the chain still gets tangled when a question has three parts, our teachers can trace the problem through your own working in online one-to-one Additional Mathematics tuition.