P6.9

Acceleration and SUVAT

Indicative relationship: Edexcel International GCSE Physics · Forces and Motion
KS3 science lesson

Acceleration and SUVAT

Lesson focus: Constant-acceleration equations connect , , acceleration, time and displacement when their conditions are satisfied.

Learning objective

Select and use a = (v − u)/t and v² = u² + 2as for one-dimensional constant-acceleration motion.

Build on earlier learning

Velocity and acceleration graphs

The previous lesson derived acceleration and displacement from velocity–time graphs. This lesson expresses the same constant-acceleration model algebraically.

You will be successful when you can…

  • Define u, v, a, s and t.
  • Select an equation from known values.
  • Use consistent signs and units.
  • Check constant-acceleration assumptions.
KEY VOCABULARY

Words that carry the science

SUVAT
A label for displacement s, initial velocity u, final velocity v, acceleration a and time t.
initial velocity
Velocity u at the start of the chosen interval.
final velocity
Velocity v at the end of the chosen interval.
uniform acceleration
Constant acceleration throughout the interval.
kinematic equation
An equation relating motion quantities without directly naming the forces.
model assumption
A condition accepted so a mathematical representation can be used.
BUILD YOUR UNDERSTANDING

Choose quantities before choosing an equation

01Choose quantities before choosing an equation

Define a positive direction, list known values with units and identify the unknown. The symbols are s displacement, u , v , a acceleration and t time.

Use a = (v − u)/t when time is involved. Use v² = u² + 2as when time is absent.

02Signs carry direction information

A velocity or acceleration opposite the positive direction is negative. Displacement also carries a sign; distance does not.

For braking with forward positive, u is positive and a is negative. The term 2as then reduces v² as expected.

03SUVAT is a constant-acceleration model

These equations apply only when acceleration is constant over the chosen interval. They are not valid across changing drag, multiple stages or curved two-dimensional motion without careful separation.

A calculator result does not repair an invalid model. Check whether forces and acceleration are reasonably constant.

Five cards define s, u, v, a and t. A decision route selects acceleration equals final minus initial velocity over time when time is known, or v squared equals u squared plus 2as when time is absent.sdisplacementmuinitial velocitym s⁻¹vfinal velocitym s⁻¹aaccelerationm s⁻²ttimestime is knowna = (v − u) ÷ ttime is absentv² = u² + 2ascondition: acceleration is constant
Figure · List quantities, directions and units first. Use these equations only for a constant-acceleration interval.
Interactive model

SUVAT with a moving car

When acceleration is constant, five quantities are linked: displacement s, initial velocity u, final velocity v, acceleration a and time t. Set u, a and t, watch the car, and check that the equations agree. The animation runs at double speed.

0 m20 m40 m60 m80 ms = 96.0 mv = 10.0 m/st = 0.00 s

Controls

10 m/s
2.0 m/s²
6.0 s
v = u + at10 + (2.0 × 6.0) = 22.0 m/s
s = ut + ½at²96.0 m
Check: v² = u² + 2as484.0 = 484.0
A straight line: its gradient is the acceleration and the area under it is the displacement.012345678910010203040506070v = 22.0 m/sGradient = acceleration; area under the line = displacementTime / sVelocity / m/s
WORKED EXAMPLE

Find braking displacement

A car slows from 20 m s⁻¹ to rest at constant acceleration −5.0 m s⁻².

  1. Choose forward as positive: u = 20, v = 0, a = −5.0.
  2. Use v² = u² + 2as because time is not given.
  3. 0² = 20² + 2(−5.0)s.
  4. Rearrange: s = 400 ÷ 10 = 40 m.

How to make the reasoning visible: The braking displacement is 40 m under the constant-deceleration model.

LEARN BY DOING

Select the valid relationship

Identify the missing variable and solve only where constant acceleration is stated.

Known valuesUnknownEquationResult
u=0, v=16, t=8 saa=(v−u)/t2 m s⁻²
u=10, a=3, s=24 mvv²=u²+2as15.6 m s⁻¹
u=25, v=0, a=−6.25sv²=u²+2as50 m
changing drag throughout fallvneither directlyacceleration not constant
  1. Why is the final row unsuitable?

  2. Why take the positive square root in row two?

RETRIEVE · APPLY · CHECK

Check your understanding

Answer from memory, then check against the suggested answer.

What does s represent?

State the no-time SUVAT equation.

What condition must hold?

Apply the idea

A ball’s acceleration changes because air resistance grows. Can one SUVAT calculation model the whole fall?

Stretch and challenge

How does v² = u² + 2as agree with a velocity–time graph?

Exam Question Practice

P6.9Calculate5 marks

Exam Question Practice

Mark points shown after marking

A cyclist accelerates uniformly from 6.0 m/s at 1.4 m/s² over a distance of 35 m.

Calculate the final speed. Use v² = u² + 2as and show your working.

(5)
m/s

Total for this question: 5 marks

Command word: For , select the equation by the variables present and state the constant-acceleration assumption.

Tick when you have worked through it. It shows on My progress and is saved on this device only.