2.1.6Motion and Forces

Calculating Uniform Acceleration — Edexcel GCSE Combined Science 1SC0 - Physics Revision Notes & Topical Questions

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📚 Edexcel GCSE Combined Science 1SC0 - Physics Hub / Topical Revision

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Section 1 of 8Uniform acceleration

Acceleration is uniform when it has a constant value: the velocity changes by the same amount every second. On a velocity-time graph this is a straight sloping line. Examples include a car braking steadily, a ball falling when air resistance is small, and a trolley on a smooth slope.

The symbols used are:

u: initial velocity (m/s)
v: final velocity (m/s)
a: acceleration (m/s²)
t: time (s)
s: distance (or displacement) moved in the time t (m)

Section 2 of 8The equations

There are two main relationships. The first is the definition of acceleration:

a = (v − u) ÷ t, or v = u + a t

What the symbols mean: a = acceleration (m s⁻²) · v = final velocity · u = initial velocity (m s⁻¹) · Δv = change in velocity · t = time taken (s)

The second links the velocities, acceleration and distance, with no time:

v² = u² + 2 a s (written v² − u² = 2as)

This comes from the area under a velocity-time graph. The distance is the area of a trapezium, s = ½ (u + v) t. Substituting t = (v − u) ÷ a gives s = (v² − u²) ÷ 2a, which rearranges to the equation above.

What the symbols mean: s = displacement (m) · u = initial velocity · v = final velocity (m s⁻¹) · t = time (s); (u + v)/2 is the average velocity

Use the equation that contains the three quantities you know and the one you want, and does not contain the quantity that you do not need.

Section 3 of 8Method and sign conventions

List the quantities: write what is known (u, v, a, t, s) and what is wanted. If an object starts from rest, u = 0; if it stops, v = 0.

Choose a positive direction.

If a quantity is in the opposite direction, give it a negative sign.
A braking car has negative a if the direction of travel is positive.

Choose the equation containing the three known values and the unknown one.

Rearrange, substitute with units and calculate.

Check that the answer is sensible: for example, a braking distance must be positive.

With v², remember that taking the square root gives a positive and a negative root. Decide which one is physically sensible from the direction of motion.

Section 4 of 8Free fall and projectiles in one dimension

For objects in free fall, a = g = 9.8 m/s² downwards if air resistance is ignored.

Free fall and projectiles in one dimension

If we take downwards as positive, a dropped object from rest has u = 0 and a = +9.8 m/s².

Then v² = 2 g s gives the speed after falling through a height s, and v = g t gives the speed after a time t.

For an object thrown upwards, if upwards is taken as positive, then a = −9.8 m/s²; at the highest point v = 0.

These equations apply only when the acceleration is constant. If the acceleration changes, as with a falling object that is reaching terminal velocity, they do not apply. They build on 1.1.4 (acceleration) and 1.1.6 (area under a graph).

Section 5 of 8Choosing the right equation

A quick way to choose is to write down the five symbols and cross out the one that is neither given nor wanted.

If time is not involved, use v² = u² + 2as.

If distance is not involved, use v = u + at.

If the final velocity is not given and not wanted, you can use s = ½(u + v)t.

If you have to find two unknowns, use one equation to find one quantity and then a second equation to find the other.

A worked checka motorbike moves from rest with an acceleration of 4.0 m/s² for 6.0 s.

Using v = u + at gives v = 24 m/s.

Using s = ½(u + v)t gives s = ½ × 24 × 6.0 = 72 m.

Check with v² = u² + 2as: 576 = 0 + 2 × 4.0 × 72 = 576. The three results agree, which shows that the equations are consistent.

Whenever possible, use a second route like this to check your answer,

and make sure that the sign of each quantity matches the direction you chose as positive.

Section 6 of 8Worked examples

A car accelerates uniformly from 5.0 m/s to 25 m/s over a distance of 150 m. Find the acceleration.
A train moving at 30 m/s brakes uniformly at 1.5 m/s² to rest. Find the braking distance and the time taken.
A stone is dropped from a bridge and hits the water at 14 m/s. Ignoring air resistance, how high is the bridge above the water?

Section 7 of 8Key terms

Term	Meaning
Uniform acceleration	Constant acceleration; a straight line on a velocity-time graph.
Initial velocity (u)	Velocity at the start of the time interval.
Final velocity (v)	Velocity at the end of the time interval.
Free fall	Falling under gravity alone, with a = g = 9.8 m/s².
Braking distance	Distance a vehicle travels while braking to rest.

Section 8 of 8Exam technique

Forgetting that v² = u² + 2as has no time, so it cannot be used if the time is the only thing given.

Using the equations when acceleration is not constant.

Forgetting to square the velocities, or squaring after adding: (u + 2as)².

Using the wrong sign for a when braking.

Taking u as the final velocity and v as the initial velocity.

Last updated: 20 October 2026