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Equations of Motion

Key idea: Use the SUVAT equations of motion to solve constant-acceleration problems, with step-by-step worked examples and exam tips (O Level Physics).

  • G3 Physics topic extensions
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Learning objectives

  • Solve straight-line constant-acceleration problems with the equations of uniformly accelerated motion

1. Definition

The equations of motion (also called the SUVAT equations) are equations you can use to solve straight-line motion with constant (uniform) acceleration.

They relate these 5 quantities:

  • s = displacement (m)
  • u = initial velocity (m s⁻¹)
  • v = final velocity (m s⁻¹)
  • a = acceleration (m s⁻²)
  • t = time taken (s)

2. Key Ideas

  • The SUVAT equations work only if acceleration is constant (uniform acceleration).
  • Choose a positive direction and keep signs consistent (velocity and acceleration can be negative).
  • These equations are for one-dimensional motion (motion in a straight line).
  • Always write down the known values with units before choosing an equation.
  • Near Earth, for free-fall questions at O Level, use g ≈ 10 m s⁻² unless the question states another value.

3. Detailed Explanations

A. The four SUVAT equations

Each equation “avoids” one of the 5 quantities. So you choose the one that uses what you know.

  1. (no s)

v = u + at

  1. (no a)

s = (1/2)(u + v)t

  1. (no v)

s = ut + (1/2)at²

  1. (no t)

v² = u² + 2as

B. How to choose an equation (exam workflow)

  1. Choose a positive direction (e.g. “to the right is +” or “upwards is +”).
  2. Write down what you know: s, u, v, a, t (with units and signs).
  3. Circle the unknown.
  4. Choose an equation that includes the unknown but does not include the variable you do not have.
  5. Substitute, solve, and state the unit.

Mini-example (find acceleration):

If u = 2.0 m s⁻¹, v = 10 m s⁻¹, and t = 4.0 s, then:

a = (v-u)/t = (10-2.0)/4.0 = 2.0 m s⁻²

C. Why equation (2) makes sense (average velocity idea)

For uniform acceleration, velocity changes steadily from u to v, so the average velocity is:

v_avg = (1/2)(u + v)

Uniform acceleration: v–t graph (straight line)

A straight-line velocity–time graph for uniform acceleration from u to v. A horizontal line shows the average velocity (u+v)/2.

Scroll across the graph to read all labels.

A straight-line velocity–time graph for uniform acceleration from u to v. A horizontal line shows the average velocity (u+v)/2.A straight-line velocity–time graph for uniform acceleration from u to v. A horizontal line shows the average velocity (u+v)/2.
With constant acceleration, velocity increases linearly from u to v. The area under the line equals displacement s, so s = v_avgt = 0.5(u + v)t.
Open full-size graph
View figure data
Values for Uniform acceleration: v–t graph (straight line)
Time (s)v–t (uniform acceleration)Average velocity = (u+v)/2
026
4106

Displacement is:

s = v_avgt = (1/2)(u + v)t

A Level extension: projectiles

The SUVAT equations are for straight-line motion. For projectile motion (2D), you treat horizontal and vertical motion separately: Projectile Motion.

4. Common Mistakes

  • Using SUVAT when acceleration is not constant (e.g. a curved v–t graph).
  • Mixing up distance and displacement (sign matters for displacement).
  • Sign mistakes (e.g. taking a as positive when the object is slowing down in the positive direction).
  • Unit mistakes (e.g. using minutes instead of seconds, or km/h instead of m/s).

5. Exam Tips

  • Always write “SUVAT only if constant acceleration” before you start.
  • If an object is slowing down, acceleration is opposite in direction to velocity (so a has the opposite sign to v).
  • For “comes to rest” questions, set v = 0.
  • For free-fall at O Level, use g ≈ 10 m s⁻² unless told otherwise, and state your sign convention.
  • Do a quick reasonableness check (e.g. if you are braking, the stopping distance should be positive and v should decrease to 0).

6. Worked Examples

Modelled example 1

Accelerating from rest (find v and s)

Core

Problem

A trolley starts from rest and accelerates uniformly at 2.0 m s⁻² for 6.0 s. Find its final velocity and displacement.
Study the worked solution

  1. List the data

    Method

    Translate “starts from rest” and record the constant-acceleration quantities.

    Reason

    A data list prevents equation and sign mistakes.

    Working

    u = 0, a = 2.0 m s⁻², t = 6.0 s

  2. Find final velocity

    Method

    Use the equation without displacement.

    Reason

    v = u + at contains the known data and target v.

    Working

    v = 0 + (2.0)(6.0) = 12 m s⁻¹

  3. Find displacement

    Method

    Use the equation without final velocity.

    Reason

    s = ut + (1/2)at² uses the original data directly.

    Working

    s = 0 + (1/2)(2.0)(6.0²) = 36 m

Guided practice 2

Braking to rest (find deceleration and stopping distance)

About 6 min

Problem

Taking the cyclist’s initial direction as positive, a cyclist slows uniformly from 8.0 m s⁻¹ to rest in 4.0 s. Find acceleration and stopping displacement.

Try this before viewing the solution

Hints

Hint 1: translate comes to rest
Set v = 0 and keep u = +8.0 m s⁻¹.
Hint 2: use average velocity for distance
With constant acceleration, s = (1/2)(u + v)t.
View solution step by step

  1. Find acceleration

    Method

    Use the signed velocity change per unit time.

    Reason

    The negative result means acceleration opposes the chosen positive direction.

    Working

    a = (0-8.0)/4.0 = -2.0 m s⁻²

  2. Find displacement

    Method

    Multiply average velocity by time.

    Reason

    Velocity changes uniformly from 8.0 to 0.

    Working

    s = (1/2)(8.0 + 0)(4.0) = 16 m

Common misconception 3

Free fall (drop from rest)

Find and correct the mistake

Learner claim

A stone is dropped 45 m with air resistance ignored and g = 10 m s⁻². A learner takes downward displacement as positive but writes a = -10 m s⁻². Diagnose the signs and find the fall time.

Try this before viewing the solution

Consistent signed data when downward is positive

View solution step by step

  1. State the convention

    Method

    Take downward as positive.

    Reason

    Both the motion and gravity point in that direction.

    Working

    u = 0, s = +45 m, a = +10 m s⁻²

  2. Select the equation

    Method

    Use the equation without final velocity.

    Reason

    u,s,a are known and t is required.

    Working

    45 = 0 + (1/2)(10)t²

  3. Solve

    Method

    Choose the non-negative elapsed time.

    Reason

    Time taken is a duration.

    Working

    t² = 9 ⇒ t = 3.0 s

Examiner practice 4

Find acceleration without time (use v² = u² + 2as)

4 marks

Examination question

A car speeds up uniformly from 5.0 m s⁻¹ to 25 m s⁻¹ over 60 m. Find its acceleration. [4 marks]

Try this before viewing the solution

View solution step by step

  1. List the data

    1 mark

    Method

    Record u,v,s and note that time is absent.

    Reason

    The missing quantity determines the efficient equation.

    Working

    u = 5.0, v = 25, s = 60 in SI units

  2. Select equation

    1 mark

    Method

    Use v² = u² + 2as.

    Reason

    It contains the target a but not t.

    Working

    v² = u² + 2as

  3. Rearrange and substitute

    1 mark

    Method

    Make a the subject before evaluating.

    Reason

    This exposes the squared velocities clearly.

    Working

    a = (25²-5.0²)/2(60)

  4. Calculate

    1 mark

    Method

    The acceleration is positive.

    Reason

    The car speeds up in the positive direction.

    Working

    a = 5.0 m s⁻²

Challenge 5

Find time and displacement (given u, v, and a)

Minimal support

Independent transfer

A motorcycle speeds up uniformly from 4.0 m s⁻¹ to 16 m s⁻¹ at 3.0 m s⁻². Find the time and displacement, then name a graph check for the displacement.

Try this before viewing the solution

Hints

Hint 1: find the shared time first
Orient with v = u + at; the resulting t can feed the average-velocity relation.
View solution step by step

  1. Find time

    Method

    Rearrange the velocity equation.

    Reason

    u,v,a are known.

    Working

    t = (16-4.0)/3.0 = 4.0 s

  2. Find displacement

    Method

    Use average velocity for constant acceleration.

    Reason

    The straight-line velocity change has mean (1/2)(u + v).

    Working

    s = (1/2)(4.0 + 16)(4.0) = 40 m

  3. Check graphically

    Method

    Use the area under the straight velocity–time line.

    Reason

    The trapezium area is displacement.

    Working

    s = (1/2)(4.0 + 16)(4.0) = 40 m

7. Mind Stretchers

Mind stretcher 1: Split the motion into stagesExtension

A bus starts from rest and accelerates uniformly at 1.5 m s⁻² for 8.0 s. It then continues at constant speed for 20 s.

Find the total displacement.

Show Answer

Stage 1 (accelerating):

v = u + at = 0 + (1.5)(8.0) = 12 m s⁻¹

s₁ = (1/2)at² = (1/2)(1.5)(8.0²) = 48 m

Stage 2 (constant speed at 12 m s⁻¹ for 20 s):

s₂ = vt = (12)(20) = 240 m

Total displacement:

s = s₁ + s₂ = 48 + 240 = 288 m

Mind stretcher 2: Thrown upwards (choose signs carefully)Extension

A ball is thrown vertically upwards with speed 20 m s⁻¹. Ignore air resistance and use g = 10 m s⁻².

Find:

  1. the maximum height reached
  2. the total time taken to return to the starting point
Show Answer

Take upward as positive. Then u = +20 m s⁻¹, a = -10 m s⁻².

At the highest point, v = 0.

Maximum height:

v² = u² + 2as ⇒ 0 = 20² + 2(-10)s ⇒ s = 20 m

Time to reach the top:

v = u + at ⇒ 0 = 20 - 10t ⇒ t = 2.0 s

Going up and coming back down takes twice the time:

tₜₒₜₐₗ = 2(2.0) = 4.0 s

Continue with the next resource in this course.

Course and syllabus information
Course
G3 Physics topic extensions
Syllabus scope
Beyond the syllabus
Edition
G3 Physics topic extensions