v = ds/dt and a = dv/dt. Going the other way, s = ∫v dt and v = ∫a dt.
Use the starting conditions (often s = 0 or a given v at t = 0) to find the constant.
'At rest' or 'instantaneously at rest' means v = 0. Maximum velocity occurs where a = 0.
Speed is the size of the velocity, so speed = 4 means v = 4 or v = −4.
Distance travelled is not always the integral of v: if v changes sign, split the integral where v = 0 and add the sizes.
Watch out: Read whether the question asks for displacement (from O, can be negative) or distance travelled (always positive).
Worked example· 3 marks
The displacement of a particle from a fixed point \(O\) at time \(t\) seconds is \(s = 2t^3 - 15t^2 + 24t\) metres.
Find the values of \(t\) for which the particle is instantaneously at rest.
M1 v = ds/dt = 6t² − 30t + 24
M1 v = 0 solved
A1 t = 1 and t = 4
v = ds/dt = 6t² − 30t + 24.
At rest: 6(t² − 5t + 4) = 0, so 6(t − 1)(t − 4) = 0.
t = 1 s and t = 4 s.
Answer: \(t = 1\) and \(t = 4\)
Kinematics questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 2 marks06064PM1no calculator
The velocity of a particle after \(t\) seconds is \(v = 2t^2 - 16t + 30\) m s⁻¹. Find its acceleration when \(t = 3\).
Show the mark scheme and worked solution
M1 a = dv/dt = 4t − 16
A1 −4 m s⁻²
a = dv/dt = 4t − 16.
At t = 3: a = 12 − 16 = −4 m s⁻².
Answer: \(-4\) m s⁻²
Question 2 · 3 marks06064PM1no calculator
A particle has acceleration \(a = 4 - 2t\) m s⁻² and initial velocity 5 m s⁻¹. Find the maximum velocity of the particle.
Show the mark scheme and worked solution
M1 v = 5 + 4t − t²
M1 a = 0 at t = 2
A1 9 m s⁻¹
Integrate: v = 4t − t² + c, and v = 5 when t = 0, so v = 5 + 4t − t².
v is greatest when dv/dt = a = 0, so t = 2.
v = 5 + 8 − 4 = 9 m s⁻¹.
Answer: \(9\) m s⁻¹
Question 3 · 4 marks06064PM1no calculator
A particle moves with velocity \(v = t^2 - 5t + 4\) m s⁻¹. Find the total distance travelled from \(t = 0\) to \(t = 4\).
Show the mark scheme and worked solution
M1 v = 0 at t = 1 (direction changes)
M1 split the integral at t = 1
A1 11/6 and 9/2
A1 19/3 m
v = (t − 1)(t − 4), so the particle turns round at t = 1.
∫ from 0 to 1 of v dt = [t³/3 − 5t²/2 + 4t] = 1/3 − 5/2 + 4 = 11/6.
∫ from 1 to 4 of v dt = (64/3 − 40 + 16) − 11/6 = −8/3 − 11/6 = −9/2, a distance of 9/2.
Total distance = 11/6 + 9/2 = 19/3 m.
Answer: \(\frac{19}{3}\) m
Question 4 · 3 marks06064PM1no calculator
A particle has velocity \(v = 10 - 2t\) m s⁻¹. Find the times at which its speed is 4 m s⁻¹.
Show the mark scheme and worked solution
M1 v = 4 or v = −4
A1 t = 3
A1 t = 7
Speed 4 means v = 4 or v = −4.
10 − 2t = 4 gives t = 3; 10 − 2t = −4 gives t = 7.
Answer: \(t = 3\) and \(t = 7\)
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