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A Level Maths 9709 · Paper 4 Mechanics · 4.2 · ~5 min read

Kinematics of motion in a straight line

Distance, displacement, speed, velocity and acceleration, displacement-time and velocity-time graphs, the constant-acceleration (suvat) equations including vertical motion under gravity, and calculus for motion with variable acceleration.

Displacement vs distanceVelocity vs speeds–t and v–t graphsArea & gradientsuvat equationsVertical motionVariable accelerationDifferentiating & integrating
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Vectors vs scalars: displacement and velocity have direction (and can be negative); distance and speed don't.
  • v–t graph: the gradient is the acceleration; the area under it is the displacement.
  • suvat works only when acceleration is constant. List s, u, v, a, t, mark the three you know, and pick the equation without the one you don't want.
  • Variable acceleration: differentiate s → v → a; integrate a → v → s, using initial conditions for the constants.

By the end you can

  • Distinguish distance and displacement, speed and velocity.
  • Sketch and interpret displacement-time and velocity-time graphs.
  • Use the constant-acceleration formulae, including vertical motion under gravity (g = 10 m s⁻²).
  • Use differentiation and integration for motion with acceleration depending on time.

1Graphs

Gradient and area

t (s)v (m/s)1234-11234/3−4/34/3v = t² − 4t + 3: displacement = ∫v dt = 4/3 m, distance = ∫|v| dt = 4 mBelow the axis the particle is moving backwards.
On a v–t graph, gradient = acceleration and area = displacement.
GraphGradientArea under it
displacement–timevelocity(not used)
velocity–timeaccelerationdisplacement
  • Area below the t-axis is negative displacement. For total distance, add the sizes of all the areas.
Worked example

A train accelerates uniformly from rest to 12 m s⁻¹ in 8 s, travels at 12 m s⁻¹ for 20 s, then decelerates uniformly to rest in 6 s. Find the total distance.

  1. area = ½ × 8 × 12 + 20 × 12 + ½ × 6 × 12 M1
  2. = 48 + 240 + 36 = 324 m A1
Common mistakeReading a value off the graph as an area, or counting a trapezium as a rectangle. Sketch the shapes and label each one.

2Constant acceleration

The suvat equations

v = u + at ; s = ½(u + v)t ; s = ut + ½at² ; v² = u² + 2as ; s = vt − ½at² FORMULA LIST (MF19)
a must be constant
Worked example

A car moving at 4 m s⁻¹ accelerates at 2.5 m s⁻² for 6 s. Find its final speed and the distance covered.

  1. v = 4 + 2.5 × 6 = 19 m s⁻¹ B1
  2. s = 4 × 6 + ½ × 2.5 × 36 = 69 m M1 A1

Vertical motion

  • Choose a positive direction and stick to it. With up positive, a = −10 and a point below the start has negative s.
Worked example

A ball is thrown vertically upwards at 15 m s⁻¹ from a point 20 m above the ground. Find (a) its greatest height above the ground and (b) the time it takes to hit the ground.

  1. (a) at the top v = 0: 0 = 225 − 20s ⇒ s = 11.25 m, so 31.25 m above ground M1 A1
  2. (b) up positive: −20 = 15t − 5t² ⇒ t² − 3t − 4 = 0 M1
  3. (t − 4)(t + 1) = 0 ⇒ t = 4 s A1
Common mistakeSign errors with a = −g. Forgetting that s is negative when the ball ends below its start. Mixing up directions is common, says the 2024 Paper 4 report.

3Variable acceleration

Using calculus

v = ds/dt ; a = dv/dt = d²s/dt² ; s = ∫v dt ; v = ∫a dt
use these whenever a depends on t
  • At rest: v = 0. Changes direction: v changes sign. Maximum velocity: a = 0.
Worked example

A particle moves with v = t² − 5t + 4 m s⁻¹. Find (a) when it is at rest, (b) its displacement and (c) the total distance travelled from t = 0 to t = 5.

  1. (a) (t − 1)(t − 4) = 0 ⇒ t = 1 and t = 4 M1 A1
  2. s(t) = t³/3 − 5t²/2 + 4t: s(1) = 11/6, s(4) = −8/3, s(5) = −5/6 M1
  3. (b) displacement = −5/6 m A1
  4. (c) distance = 11/6 + 27/6 + 11/6 = 49/6 ≈ 8.17 m M1 A1
Common mistakeIntegrating straight from 0 to 5 to get the distance. If the particle turns round, split the integral at each turning time.
Linking questionAccelerations come from forces in Newton's laws of motion; the calculus is from Differentiation and Integration; and graphs of motion link to Kinematics in Physics.

Test yourself — cover the answers first

Quick check

  1. What does the area under a v–t graph give?
  2. Which suvat equation has no t?
  3. A stone is dropped from rest. Find its speed after 3 s.
  4. s = t³ − 6t². Find a when t = 2.
  5. Can you use suvat when a = 2t?

Exam-style

  1. A particle is projected vertically upwards at 25 m s⁻¹. Find the time for which it is at least 20 m above the start.
  2. v = 6t − t² for 0 ≤ t ≤ 6. Find the maximum velocity and the distance travelled.
  3. A car decelerates uniformly from 20 m s⁻¹ to 8 m s⁻¹ over 70 m. Find the deceleration and the time taken.
Show answers
  1. displacement
  2. v² = u² + 2as
  3. 30 m s⁻¹
  4. a = 6t − 12 = 0
  5. no, a must be constant
  6. 20 = 25t − 5t² ⇒ t = 1, 4; so 3 s
  7. a = 6 − 2t = 0 at t = 3: v = 9 m s⁻¹; ∫₀⁶ (6t − t²) dt = 108 − 72 = 36 m
  8. 64 = 400 − 140a ⇒ a = 2.4 m s⁻²; t = 12/2.4 = 5 s
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.