Equations of motion
The suvat variables
When motion has constant acceleration, five variables describe it:
- s = displacement (m)
- u = initial velocity (m/s)
- v = final velocity (m/s)
- a = acceleration (m/s²)
- t = time (s)
The Edexcel Combined Science equations
The specification requires only:
v = u + a t (final velocity)
v² − u² = 2 a s (final velocity squared)
(The full triple-science set adds s = ½(u+v)t and s = ut + ½at². Combined Science only requires v = u + at and v² = u² + 2as.)
Choosing the right equation
Identify the four variables you have or want, then pick the equation that uses only those four:
- If t is involved → use v = u + at.
- If t is not given and not asked → use v² − u² = 2as.
Always list out s, u, v, a, t at the start of a problem with units. Most dropped marks come from substituting the wrong number into the wrong slot.
✦Worked example— Worked example — v² − u² = 2as
A car accelerates from rest at 3 m/s² over a distance of 24 m. Find its final velocity.
s = 24, u = 0, a = 3, v = ? v² = u² + 2 a s = 0 + 2 × 3 × 24 = 144 v = √144 = 12 m/s.
Falling objects and terminal velocity
Near Earth’s surface, gravity gives all falling objects an acceleration of g ≈ 9.8 m/s² (Edexcel uses 10 m/s² in some questions — check the data).
For a falling object in air the forces are:
- Weight (downwards) — constant, equal to mg.
- Air resistance / drag (upwards) — increases with speed.
As the object speeds up, drag rises until drag = weight. The resultant force is then zero, so by Newton’s first law the object continues at a steady velocity — its terminal velocity.
A skydiver: ~120 mph in free fall, ~12 mph after parachute opens (much larger area → more drag → much smaller terminal velocity).
Edexcel exam tip
When using v² − u² = 2as, watch for negative accelerations (deceleration). If the question says "the car decelerates at 4 m/s²", use a = −4. Also: always square-root after the substitution, not before.
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