About Kinematics of motion in a straight line in A Level Mathematics
Kinematics of motion in a straight line is one of the 8 main topics in our A Level Mathematics (Cambridge (CAIE) 9709) question bank, and about 4% of its real questions come from it. Tayyari Karo has 19 real questions on this topic from 2021 to 2026, from Paper 4. By difficulty: 10 hard, 7 medium, 2 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 7 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2026 · Paper 4 · Kinematics of motion in a straight line
A particle P travels in a straight line, starting from a point O. At time t seconds the displacement of P from O is s metres, where s = 4t³ − 39t² + 120t.
(a) Find the values of t for which P is instantaneously at rest. [3]
(b) Find the total distance travelled by P in the interval 1 ≤ t ≤ 4. [2]
(c) Find the magnitude of the acceleration of P when t = 3. [3]
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(a) v = ds/dt = 12t² − 78t + 120 = 0, so t = 2.5 and t = 4.
(b) s(1) = 85, s(2.5) = 118.75, s(4) = 112. The particle turns at t = 2.5, so distance = (118.75 − 85) + (118.75 − 112) = 33.75 + 6.75 = 40.5 m.
(c) a = dv/dt = 24t − 78; at t = 3, a = −6, so the magnitude is 6 m s⁻².
Differentiate s for v and set v = 0; split the distance at the turning point; differentiate v for the acceleration.
October/November 2025 · Paper 4 · Kinematics of motion in a straight line
A particle P starts from a point O and moves in a straight line. The velocity v m s⁻¹ of P, at time t s after leaving O, is given by v = ⅓(2t − 3)(t − 4).
(a) Find the acceleration of P when t = 2. [2]
(b) Find the total distance travelled by P in the first 3 seconds of its motion. [5]
(c) Determine whether P returns to O. [2]
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(a) v = ⅓(2t²−11t+12), so dv/dt = ⅓(4t−11). At t=2: a = ⅓(8−11) = −1 m/s².
(b) s = ∫v dt = ⅓[(2/3)t³ − (11/2)t² + 12t] (since s=0 at t=0).
P changes direction when v=0: (2t−3)(t−4)=0 → t=1.5 or t=4. Within [0,3], only t=1.5 matters.
s(1.5) = ⅓[2.25 − 12.375 + 18] = ⅓(7.875) = 2.625.
s(3) = ⅓[18 − 49.5 + 36] = ⅓(4.5) = 1.5.
Distance travelled = |2.625−0| + |1.5−2.625| = 2.625 + 1.125 = 3.75 m.
(c) s(4) = ⅓[(2/3)(64) − (11/2)(16) + 48] = ⅓(8/3) = 8/9 ≈ 0.889 ≠ 0. Since v=0 only at t=1.5 and t=4 (with v>0 for all t>4), the displacement reaches a minimum of 8/9 at t=4 and increases thereafter, so P's displacement never returns to 0. P does not return to O.
Differentiate v for the acceleration in (a); integrate v to get displacement, find where v=0 to locate turning points, and sum the absolute distances between them for (b); check the displacement at the next turning point (or beyond) for (c).
October/November 2024 · Paper 4 · Kinematics of motion in a straight line
A particle P moves in a straight line, passing through a point O with velocity 4.2 m s⁻¹. At time t s after P passes O, the acceleration, a m s⁻², of P is given by a = 0.6t − 2.7 .
Find the distance P travels between the times at which it is at instantaneous rest. [7]
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Integrate the acceleration to find velocity, using v = 4.2 when t = 0:
v = ∫(0.6t − 2.7) dt = 0.3t² − 2.7t + c, and v(0) = 4.2 gives c = 4.2
So v = 0.3t² − 2.7t + 4.2
The particle is at instantaneous rest when v = 0:
0.3t² − 2.7t + 4.2 = 0
Multiply by 10/3: t² − 9t + 14 = 0
(t − 2)(t − 7) = 0
t = 2 or t = 7
Integrate velocity to find displacement, with s(0) = 0:
s = ∫(0.3t² − 2.7t + 4.2) dt = 0.1t³ − 1.35t² + 4.2t
At t = 2: s = 0.1(8) − 1.35(4) + 4.2(2) = 0.8 − 5.4 + 8.4 = 3.8
At t = 7: s = 0.1(343) − 1.35(49) + 4.2(7) = 34.3 − 66.15 + 29.4 = −2.45 …
The full model answer is in the app.
May/June 2023 · Paper 4 · Kinematics of motion in a straight line
A particle moves in a straight line starting from rest. The displacement s m of the particle from a fixed point O on the line at time t s is given by s = t^(5/2) − (15/4)t^(3/2) + 6.
Find the value of s when the particle is again at rest. [4]
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v = ds/dt = (5/2)t^(3/2) − (45/8)t^(1/2).
Set v=0: t^(1/2)[(5/2)t − 45/8] = 0 → t=0 (start) or (5/2)t = 45/8 → t = 9/4 = 2.25 s.
At t = 9/4: t^(1/2)=3/2, t^(3/2)=27/8, t^(5/2)=243/32.
s = 243/32 − (15/4)(27/8) + 6 = 243/32 − 405/32 + 192/32 = 30/32 = 15/16
s = 15/16 (= 0.9375 m)
Differentiate s to get v, set v=0 to find the (non-zero) time, then substitute back into s.
October/November 2023 · Paper 4 · Kinematics of motion in a straight line
A particle moves in a straight line starting from a point O before coming to instantaneous rest at a point X. At time t s after leaving O, the velocity v m s⁻¹ of the particle is given by
v = 7.2t², for 0 ≤ t ≤ 2,
v = 30.6 − 0.9t, for 2 ≤ t ≤ 8,
v = 1600/t² + kt, for t ≥ 8,
where k is a constant. It is given that there is no instantaneous change in velocity at t = 8.
Find the distance OX. [9]
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At t = 8, from the second expression: v = 30.6 − 0.9(8) = 30.6 − 7.2 = 23.4
From the third expression at t = 8: v = 1600/64 + 8k = 25 + 8k
Since velocity is continuous at t = 8: 25 + 8k = 23.4
8k = −1.6
k = −0.2
So for t ≥ 8: v = 1600/t² − 0.2t
The particle comes to instantaneous rest at X when v = 0:
1600/t² − 0.2t = 0
1600 = 0.2t³
t³ = 8000
t = 20
Distance OX = ∫₀² 7.2t² dt + ∫₂⁸ (30.6 − 0.9t) dt + ∫₈²⁰ (1600/t² − 0.2t) dt
∫₀² 7.2t² dt = [2.4t³]₀² = 2.4(8) = 19.2 m
∫₂⁸ (30.6 − 0.9t) dt = [30.6t − 0.45t²]₂⁸ = (244.8 − 28.8) − (61.2 − 1.8) = 216 − 59.4 = 156.6 m
∫₈²⁰ (1600/t² − 0.2t) dt = [−1600/t − 0.1t²]₈²⁰ = (−80 − 40) − (−200 − 6.4) = −120 − (−206.4) = 86.4 m …
The full model answer is in the app.
May/June 2021 · Paper 4 · Kinematics of motion in a straight line
A particle moving in a straight line starts from rest at a point A and comes instantaneously to rest at a point B. The acceleration of the particle at time t s after leaving A is a m s⁻², where
a = 6√t − 2t.
(a) Find the value of t at point B. [3]
(b) Find the distance travelled from A to the point at which the acceleration of the particle is again zero. [5]
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(a) Integrate the acceleration to find velocity:
v = ∫(6√t − 2t) dt = 4t√t − t² + c
At t = 0 the particle starts from rest, so v = 0 gives c = 0:
v = 4t√t − t²
At B the particle is again at rest, so v = 0 (t ≠ 0):
4t√t = t² → 4√t = t → 16t = t² → t(t − 16) = 0 → t = 16
So t = 16 s at B.
(b) The acceleration is zero when a = 0 (t ≠ 0):
6√t − 2t = 0 → 6√t = 2t → 3√t = t → 3 = √t → t = 9 s
Integrate velocity to find displacement (s = 0 at t = 0, point A):
s = ∫(4t√t − t²) dt = (8/5)t²√t − (1/3)t³
At t = 9: t²√t = 81 × 3 = 243, so (8/5)(243) = 388.8; t³ = 729, so (1/3)(729) = 243
s = 388.8 − 243 = 145.8 m
Integrate the acceleration to get velocity (using v=0 at t=0 to fix the constant), solve for t at each point where the relevant quantity is zero, then integrate velocity to get displacement.
October/November 2021 · Paper 4 · Kinematics of motion in a straight line
A bus moves from rest with constant acceleration for 12 s. It then moves with constant speed for 30 s before decelerating uniformly to rest in a further 6 s. The total distance travelled is 585 m.
(a) Find the constant speed of the bus. [2]
(b) Find the magnitude of the deceleration. [1]
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(a) The area under the velocity-time graph (a trapezium) gives the total distance: (1/2)(12)V + 30V + (1/2)(6)V = 585 ⟹ 6V + 30V + 3V = 585 ⟹ 39V = 585 ⟹ V = 15 m/s
(b) Deceleration = V/6 = 15/6 = 2.5 m/s²
Split the journey into three v-t graph regions (triangle, rectangle, triangle) and use the total area to find V; then use v=u+at for the deceleration phase.
