# Uniformly accelerated motion equations

> CIE A-Level Physics · 9702 (2022-2024)
> Source: https://www.owlsprep.com/study/cie-9702-u2-uniformly-accelerated-motion-equations/

This module covers derivation and application of the kinematic (SUVAT) equations for one-dimensional motion with constant uniform acceleration. You will learn to solve common problems including free fall and stopping distances for CIE A-Level Physics.

**Prerequisites:** [Basic kinematic quantities (velocity, acceleration, displacement)](https://www.owlsprep.com/study/cie-9702-u2-basic-kinematic-quantities/); [Motion graphs for linear motion](https://www.owlsprep.com/study/cie-9702-u2-motion-graphs/)

## Learning objectives

- Derive the five uniformly accelerated motion equations from first principles
- Correctly identify known/unknown quantities and choose the right SUVAT equation
- Solve one-dimensional motion problems for constant acceleration
- Apply equations to free fall motion under gravity

## Definition and First Principles Derivation

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.

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

*Notation:* $a = \text{constant}$

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

**Derivation:** 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. $$s = \left( \frac{u + v}{2} \right) t$$

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

**Worked example:** Derive the displacement equation $s = ut + \frac{1}{2} a t^2$ from the two base equations

1. Substitute $v = u + at$ into $s = \left( \frac{u + v}{2} \right) t$:
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. $$s = ut + \frac{1}{2} a t^2$$

> **tip**
>
> Always show full derivation steps if an exam question asks for derivation, you get marks for every intermediate step.

## SUVAT Equations: Summary and Selection

> **SUVAT Mnemonic**
>
> Each letter of SUVAT names one variable, and each full equation omits exactly one variable to let you solve for an unknown when you know three other quantities.

| 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 |

1. Omitting *s*: $v = u + at$
2. Omitting *a*: $s = \left(\frac{u+v}{2}\right) t$
3. Omitting *v*: $s = ut + \frac{1}{2} a t^2$
4. Omitting *t*: $v^2 = u^2 + 2 a s$
5. Omitting *u*: $s = vt - \frac{1}{2} a t^2$

**Worked example:** A cyclist accelerates uniformly from rest at $1.5 \text{ m s}^{-2}$ for 6.0 s. Find the cyclist's displacement.

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

> **exam_tip**
>
> Always list known and unknown variables first before choosing your equation, this avoids picking the wrong equation for the problem.

## Application to General Linear Motion

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

1. Define positive direction as the direction of initial travel. Variables: $u = 20 \text{ m s}^{-1}$, $v = 0 \text{ m s}^{-1}$ (car stops), $a = -4.0 \text{ m s}^{-2}$ (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. $$s = 50 \text{ m}$$

**Check your understanding**

Test your understanding of sign conventions

1. What result would you get if you incorrectly used $a = +4.0 \text{ m s}^{-2}$ for the problem above?

   - +50 m
   - -50 m
   - 25 m
   - 100 m

   *Answer:* -50 m

   *Why:* A non-physical negative displacement tells you you got the sign of acceleration wrong. Always confirm your sign convention before starting calculations.

## Application to Free Fall Motion

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

**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 $14.7 \text{ m s}^{-1}$. Find the maximum height reached.

1. Take upwards as positive direction. Variables: $u = 14.7 \text{ m s}^{-1}$, $v = 0 \text{ m s}^{-1}$ (velocity is zero at maximum height), $a = -g = -9.81 \text{ m s}^{-2}$ (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. $$s = 11.0 \text{ m}$$

> **tip**
>
> CIE accepts $g = 9.8 \text{ m s}^{-2}$ or $10 \text{ m s}^{-2}$ if specified in the question, but always use 9.81 by default.

## Common pitfalls

- **Wrong:** Using positive acceleration for deceleration or upward motion against gravity
  - Why it fails: Acceleration is a vector, sign depends on your defined positive direction
  - Correct: Always define your positive direction first, then assign correct signs to all vector quantities
- **Wrong:** Using SUVAT equations when acceleration is not constant
  - Why it fails: SUVAT is only valid for uniform (constant) acceleration, it gives incorrect results for changing acceleration
  - Correct: Confirm acceleration is constant before using SUVAT; use integration or area under graphs for non-uniform acceleration
- **Wrong:** Confusing net displacement with total distance travelled when an object changes direction
  - Why it fails: SUVAT calculates net displacement from the starting point, not total distance moved over the full journey
  - Correct: Split motion into segments with constant direction, calculate distance for each segment, then add the distances
- **Wrong:** Assuming acceleration is zero at maximum height for vertical projection
  - Why it fails: Velocity is zero at maximum height, but gravity is still acting on the object
  - Correct: Always use $a = -g$ (for upwards positive) even at maximum height
- **Wrong:** Rounding intermediate results to 3 significant figures early
  - Why it fails: Early rounding causes avoidable rounding error in the final answer
  - Correct: Keep extra significant figures for intermediate steps, round only the final answer

## Cheatsheet

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

## 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.

- [Projectile Motion](https://www.owlsprep.com/study/cie-9702-u2-projectile-motion/)
- [Free fall](https://www.owlsprep.com/study/cie-9702-u2-free-fall/)
- [Dynamics](https://www.owlsprep.com/study/cie-9702-u3-overview/)

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