Uniformly accelerated motion equations

CIE A-Level Physics· Unit 2: Kinematics· 15 min read

1. Definition and First Principles Derivation★★☆☆☆⏱ 5 min

Uniformly accelerated motion is defined as motion where the rate of change of velocity is constant over time. This means acceleration $a$ does not change magnitude or direction throughout the motion.

📘 Definition

Uniformly accelerated motion

$a = constant$

Motion in one dimension where acceleration remains constant in magnitude and direction for the entire duration of motion

Example:

An object falling close to Earth's surface, ignoring air resistance

🔬 Derivation

Goal:

Derive the two base SUVAT equations from core definitions

Starting from:

Definitions of average acceleration and average velocity

1
Start with definition of average acceleration over time interval $t$ :
2
$a = \frac{v - u}{t}$
3
Rearrange to get the first base equation:
4
$v = u + at$
5
Average velocity for constant acceleration is the mean of initial and final velocity:
6
$v_{\text{avg}} = \frac{u + v}{2} = \frac{s}{t}$
7
Rearrange to get the second base equation for displacement:
8
$s = \left( \frac{u + v}{2} \right) t$

Result:

These two base equations can be combined to derive the three remaining SUVAT equations.

📐 Worked Example

Derive the displacement equation $s = u t + \frac{1}{2} a t^{2}$ from the two base equations

1
Substitute $v = u + a t$ into $s = (\frac{u + v}{2}) t$ :
2
$s = \left( \frac{u + (u + at)}{2} \right) t$
3
Simplify the expression inside the brackets:
4
$s = \left( \frac{2u + at}{2} \right) t = \left( u + \frac{1}{2} a t \right) t$
5
Expand to get the final derived equation:
6
$s = ut + \frac{1}{2} a t^2$

2. SUVAT Equations: Summary and Selection★★☆☆☆⏱ 3 min

Variable	Quantity Name	SI Units
u	Initial velocity	m s⁻¹
v	Final velocity	m s⁻¹
a	Constant acceleration	m s⁻²
s	Net displacement	m
t	Time interval	s

Omitting s: $v = u + a t$
Omitting a: $s = (\frac{u + v}{2}) t$
Omitting v: $s = u t + \frac{1}{2} a t^{2}$
Omitting t: $v^{2} = u^{2} + 2 a s$
Omitting u: $s = v t - \frac{1}{2} a t^{2}$

📐 Worked Example

A cyclist accelerates uniformly from rest at $1.5 m s^{- 2}$ for 6.0 s. Find the cyclist's displacement.

1
List known and unknown variables: $u = 0 m s^{- 1}$ (starts from rest), $a = 1.5 m s^{- 2}$ , $t = 6.0 s$ , unknown $s$ . We need the equation that omits $v$ .
2
Substitute values into $s = u t + \frac{1}{2} a t^{2}$ :
3
$s = (0)(6.0) + \frac{1}{2} (1.5) (6.0)^2$
4
Calculate the final result:
5
$s = 27 \text{ m}$

3. Application to General Linear Motion★★★☆☆⏱ 4 min

SUVAT equations can be applied to any one-dimensional motion with constant acceleration. The most common problems include stopping distances for vehicles, accelerating trains and sprinters, and motion down inclined planes.

📐 Worked Example

A car travels at $20 m s^{- 1}$ when brakes are applied, producing a uniform deceleration of $4.0 m s^{- 2}$ . Calculate the stopping distance.

1
Define positive direction as the direction of initial travel. Variables: $u = 20 m s^{- 1}$ , $v = 0 m s^{- 1}$ (car stops), $a = - 4.0 m s^{- 2}$ (deceleration opposes motion, so negative), unknown $s$ .
2
We know u, v, a, need s, so use the equation omitting $t$ : $v^{2} = u^{2} + 2 a s$
3
Rearrange for s and substitute values:
4
$s = \frac{v^2 - u^2}{2 a} = \frac{0^2 - (20)^2}{2 (-4.0)} = \frac{-400}{-8}$
5
Final result:
6
$s = 50 \text{ m}$

✓ Quick check

Test your understanding of sign conventions

What result would you get if you incorrectly used $a = + 4.0 m s^{- 2}$ for the problem above?
- +50 m
- -50 m
- 25 m
- 100 m
Reveal answer
1 —
A non-physical negative displacement tells you you got the sign of acceleration wrong. Always confirm your sign convention before starting calculations.

4. Application to Free Fall Motion★★★☆☆⏱ 3 min

Close to Earth's surface, all objects falling under gravity (ignoring air resistance) have a constant downward acceleration called acceleration due to gravity, $g = 9.81 m s^{- 2}$ for CIE A-Level Physics.

📘 Definition

Free fall

Motion of an object under the influence of only gravity, with constant acceleration $g$ , no air resistance

📐 Worked Example

A ball is thrown vertically upwards from ground level with initial velocity $14.7 m s^{- 1}$ . Find the maximum height reached.

1
Take upwards as positive direction. Variables: $u = 14.7 m s^{- 1}$ , $v = 0 m s^{- 1}$ (velocity is zero at maximum height), $a = - g = - 9.81 m s^{- 2}$ (gravity acts downwards), unknown $s$ (height).
2
Use the equation omitting $t$ : $v^{2} = u^{2} + 2 a s$
3
Rearrange and substitute values:
4
$s = \frac{0^2 - (14.7)^2}{2 (-9.81)} = \frac{-216.09}{-19.62}$
5
Final result (3 significant figures):
6
$s = 11.0 \text{ m}$

5. Common Pitfalls

Wrong move:

Using positive acceleration for deceleration or upward motion against gravity

Why:

Acceleration is a vector, sign depends on your defined positive direction

Correct move:

Always define your positive direction first, then assign correct signs to all vector quantities

Wrong move:

Using SUVAT equations when acceleration is not constant

Why:

SUVAT is only valid for uniform (constant) acceleration, it gives incorrect results for changing acceleration

Correct move:

Confirm acceleration is constant before using SUVAT; use integration or area under graphs for non-uniform acceleration

Wrong move:

Confusing net displacement with total distance travelled when an object changes direction

Why:

SUVAT calculates net displacement from the starting point, not total distance moved over the full journey

Correct move:

Split motion into segments with constant direction, calculate distance for each segment, then add the distances

Wrong move:

Assuming acceleration is zero at maximum height for vertical projection

Why:

Velocity is zero at maximum height, but gravity is still acting on the object

Correct move:

Always use $a = - g$ (for upwards positive) even at maximum height

Wrong move:

Rounding intermediate results to 3 significant figures early

Why:

Early rounding causes avoidable rounding error in the final answer

Correct move:

Keep extra significant figures for intermediate steps, round only the final answer

6. Quick Reference Cheatsheet

Equation	Variable Omitted
$v = u + a t$	s
$s = \frac{( u + v )}{2} t$	a
$s = u t + \frac{1}{2} a t^{2}$	v
$v^{2} = u^{2} + 2 a s$	t
$s = v t - \frac{1}{2} a t^{2}$	u
$g = 9.81 m s^{- 2}$ (free fall)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
MCQ on free fall displacement calculation
2023 · 2
Calculation of vehicle stopping distance
2024 · 1
MCQ on acceleration from motion data

Going deeper

syllabusCIE 9702 2022-2024 SyllabusFull unit 2 kinematics requirements

What's Next

Now you have mastered the derivation and application of uniformly accelerated motion (SUVAT) equations, you are ready to apply these to more complex kinematics problems. The most common next application is projectile motion, where you split motion into independent horizontal and vertical components: horizontal motion has zero acceleration so constant velocity, while vertical motion has constant acceleration due to gravity, meaning you can directly apply SUVAT here. You can also progress to studying Newton's laws of motion, where you first calculate acceleration from net force acting on an object, before using SUVAT equations to find unknown displacement, velocity or time quantities. Solid understanding of SUVAT is foundational to almost all mechanics topics in A-Level Physics.