01 · A first check
Use u = −2 m/s, a = 3 m/s² and t = 4 s to find v.
Hint
Keep the sign of u.
Worked solution
v = −2 + 3 × 4 = 10 m/s.
Understand · explore · practise
Deriving the constant-acceleration equations
Derive all five SUVAT equations from a straight velocity graph, with signed displacement and a derivation that remains valid at zero acceleration.
Before you startSigned areas, gradients and rearranging equations.
01 / State the model before writing an equation
One straight velocity line means constant acceleration.
Use one axis and one interval.
u is initial signed velocity, v final signed velocity, a constant signed acceleration, t elapsed time and s signed displacement during that interval. The starting coordinate is not s. These equations require constant acceleration over the whole interval.
Split the signed areaExplore
Choose a constant-acceleration example lasting 4 s. A rectangle ut and a signed triangle ½at² add to displacement, even when acceleration is negative.
02 / Start from the definition of acceleration
Rearrange the slope formula.
For a nonzero elapsed time t:Worked example
a = (v − u)/t
Acceleration is the constant gradient of the velocity line.
at = v − u
Multiply by t.
v = u + at
The final velocity is the initial velocity plus its change.
02 · Rest condition
Does v = 0 imply a = 0?
Hint
Use a moment at a reversal as a counterexample.
Worked solution
No. For example, u = 6 m/s, a = −2 m/s² gives v = 0 at t = 3 s, while acceleration remains −2 m/s².
03 / Use the signed trapezium area
A linear velocity graph has mean velocity (u + v)/2.
Watch: rectangle plus signed triangle
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03 · Mean velocity
A particle has u = −4 m/s and v = 8 m/s over 6 s with constant acceleration. Find displacement.
Hint
Use the signed mean velocity.
Worked solution
s = ½(−4 + 8) × 6 = 12 m.
04 · Mean speed warning
Is ½(u + v) always the average speed?
Hint
What if velocity changes sign?
Worked solution
No. It is average velocity for constant acceleration. If the particle reverses, distance must be found by splitting at v = 0; average speed is distance divided by elapsed time.
04 / Eliminate final velocity
Substitute v = u + at into the mean-velocity formula.
s = ½(u + v)tWorked example
s = ½(u + u + at)t
Replace v.
s = ut + ½at²
Expand and simplify; this is the rectangle plus signed triangle.
05 · Formula use
Find s when u = 5 m/s, a = −2 m/s² and t = 3 s.
Hint
Both terms retain their signs.
Worked solution
s = 5 × 3 + ½ × (−2) × 9 = 6 m.
06 · Area meaning
For question 5, does the triangle term −9 m mean negative distance?
Hint
It is a correction to the rectangle in a signed-area decomposition.
Worked solution
No. It subtracts 9 m from the 15 m rectangle to give displacement 6 m. Distance requires a separate reversal check.
05 / Eliminate initial velocity instead
Use u = v − at.
s = ½(u + v)tWorked example
s = ½(v − at + v)t
Replace u.
s = vt − ½at²
Expand and simplify.
07 · Final data
A particle ends with v = 7 m/s after t = 4 s, with a = 2 m/s². Find displacement.
Hint
Use the equation containing v, a and t.
Worked solution
s = 7 × 4 − ½ × 2 × 16 = 12 m.
08 · Cross-check
Find u for question 7 and verify s with the initial-velocity formula.
Hint
u = v − at.
Worked solution
u = 7 − 8 = −1 m/s. Then s = −1 × 4 + ½ × 2 × 16 = 12 m.
06 / Eliminate time without dividing by acceleration
This derivation also covers a = 0.
Combine v − u = at with 2s = (u + v)t.Worked example
2as = (u + v)at
Multiply the area equation by a.
2as = (u + v)(v − u)
Replace at with v − u.
v² = u² + 2as
Use the difference of two squares. No division by a was needed.
09 · No time given
For u = 3 m/s, a = 2 m/s² and s = 10 m, find v².
Hint
Substitute into the squared formula.
Worked solution
v² = 9 + 40 = 49 m²/s². The algebra gives v = ±7 m/s; a physical direction argument is needed before choosing a sign.
10 · Choose a sign
If the particle in question 9 starts moving right and has positive acceleration thereafter, which v applies?
Hint
Its velocity starts positive and increases for t ≥ 0.
Worked solution
v = +7 m/s. The negative root would give a negative elapsed time in v = 3 + 2t.
07 / Check special cases and assumptions
A formula can be true without being enough to select a physical solution.
11 · Zero acceleration
Show what the equations say when a = 0.
Hint
The velocity line becomes horizontal.
Worked solution
v = u and s = ut = vt. The squared equation gives v² = u², but the model also requires v = u, not an arbitrary sign change.
12 · Zero time
What happens when t = 0?
Hint
There has been no elapsed motion in the interval.
Worked solution
v = u and s = 0. The rearranged formulae extend consistently to this boundary, though (v − u)/t cannot be evaluated at t = 0.
13 · Units
Check the units of 2as.
Hint
Acceleration times displacement.
Worked solution
(m/s²) × m = m²/s², matching the units of v² and u². The number 2 has no units.
14 · Changing acceleration
May one use these equations across an accelerate–cruise–brake journey as a single interval?
Hint
The velocity graph has different gradients.
Worked solution
Generally no. Apply the constant-acceleration equations separately to each suitable stage, or use the complete velocity-time graph.
08 / Five equations come from one graph
Remember the model and derive the form you need.
v = u + at; s = ½(u + v)t; s = ut + ½at²; s = vt − ½at²; v² = u² + 2as. These are equations for signed quantities during constant acceleration. Keep the axis, interval and units consistent, and check physical signs after solving.
Section 1 of 8 · State the model before writing an equation