Constant-acceleration equations

Key idea: Constant-acceleration equations link signed displacement, velocity and time. At the top of a vertical throw, velocity is zero but acceleration remains downward.

Reading settings

GCE A-Level H2 Physics 2027

On this page

When an object’s acceleration stays constant, five quantities describe its motion over a time interval, and four equations link them. This lesson derives the equations from the definitions of velocity and acceleration, shows how to pick the right one, and deals with the two cases that most often cause errors: a turning point, and two answers for the time.

Predict motion at a turning point

Check your understanding 1

A ball is thrown vertically upwards near Earth’s surface. Neglect air resistance and take upward as positive. At the highest point, what are its velocity and acceleration?

Velocity

Unit: m s⁻¹

Acceleration at the top

0 m/s²+9.81 m/s²−9.81 m/s²

Show answer

At the top v = 0, but a = -9.81 m s⁻². Gravity continues to change the velocity, so the ball begins to move down.

The five quantities

Choose a positive direction. Then, over one interval of constant acceleration:

Symbol	Meaning	SI unit
u	velocity at the start	m s⁻¹
v	velocity at the end	m s⁻¹
a	constant acceleration	m s⁻²
t	time taken	s
s	displacement	m

u, v, a and s are vectors and carry signs. s is a displacement, not a distance travelled.

Deriving the equations

Acceleration is the rate of change of velocity. With constant acceleration,

a = (v-u)/t ⇒ v = u + at

The velocity–time graph is then a straight line from u to v. Displacement is the area under it, a trapezium, so the average velocity is halfway between u and v:

s = (u + v)/2 t

This still holds if the line crosses zero, because the trapezium formula counts area below the axis as negative. Substituting v = u + at gives

s = ut + (1/2)at²

To remove t, multiply s = (1/2)(u + v)t by 2a and use at = v-u:

2as = (u + v)(v-u) = v²-u² ⇒ v² = u² + 2as

Squaring loses the sign of v. If you take a square root later, use the motion to decide the sign.

Check your understanding 2

Cover the derivation above. Starting from the definitions of velocity and acceleration, derive v = u + at, s = (1/2)(u + v)t, s = ut + (1/2)at² and v² = u² + 2as. State the condition on acceleration. Work on paper; you can record the key steps below without transcribing every line.

Your derivation or key steps

Show answer

For straight-line motion with constant acceleration, a = (v-u)/t, so v = u + at. Velocity varies linearly, making its average (u + v)/2. Signed velocity–time area then gives s = (u + v)t/2, even if velocity changes sign. Substituting v = u + at gives s = ut + at²/2. Finally 2as = (u + v)at = (u + v)(v-u) = v²-u². Compare your work for the constant acceleration condition, the signed-area argument, substitution and elimination of time; a formula list alone misses the derivation.

Check your understanding 3

A car stays at rest for 3.0 s, then accelerates uniformly to 12 m s⁻¹ in the next 1.0 s. A learner averages the initial and final velocities and obtains 6.0 m s⁻¹ for the whole interval. Find the actual average velocity in metres per second and choose the error in the endpoint method.

Show the steps Optional working

Step 1: Total displacement

Unit: m

Whole-interval average velocity

Unit: m s⁻¹

Why the endpoint average fails

The three seconds at rest should be excluded from the total timeAcceleration is not constant throughout the four secondsThe final velocity should be negative

Show answer

The velocity–time area is (1/2)(1.0)(12) = 6.0 m, so the average velocity is 6.0/4.0 = 1.5 m s⁻¹. The velocity does not change uniformly over all four seconds: the three seconds at rest lower the average. Averaging the endpoint velocities works for a single straight velocity–time line.

Choosing an equation

Each equation leaves out one quantity. List what you know, with signs, and choose the equation that leaves out the quantity you neither know nor need.

Equation	Leaves out
v = u + at	s
s = (1/2)(u + v)t	a
s = ut + (1/2)at²	v
v² = u² + 2as	t

Problem

A train moving at 25 m s⁻¹ brakes uniformly to rest in 200 m. Find its acceleration and the time it takes to stop.

Show full solution

List the known values
Method
Take the direction of motion as positive.
Reason
Uniform braking means constant acceleration, so the equations apply.
Working
Answer
u = +25 m s⁻¹, v = 0, s = +200 m; a and t unknown.
Find a without t
Reason
v² = u² + 2as leaves out the unknown time.
Working
Answer
0 = 25² + 2a(200), so a = -1.5625 m s⁻², about -1.6 m s⁻².
Find t
Working
Answer
t = (v-u)/a = (0-25)/(-1.5625) = 16 s

Quick check

The negative acceleration opposes the motion, as braking should. Check: average velocity 12.5 m s⁻¹ for 16 s gives 200 m.

Check your understanding 4

An object starts with velocity -3.0 m s⁻¹ and has constant acceleration -2.0 m s⁻². Find its velocity and displacement after 4.0 s. Choose an equation for displacement that does not need final velocity. Enter signed velocity and displacement, then decide how speed changes.

Displacement without final velocity

v = u + ats = ut + ½at²s = (u + v)t/2

Velocity after 4.0 s

Unit: m s⁻¹

Displacement after 4.0 s

Unit: m

Speed change

Stays constantIncreasesDecreases

Show answer

v = -3.0 + (-2.0)(4.0) = -11 m s⁻¹ and s = (-3.0)(4.0) + (1/2)(-2.0)(4.0)² = -28 m. The speed increases from 3.0 to 11 m s⁻¹, because velocity and acceleration point the same way.

Free fall

Near the Earth’s surface, a body falling with no air resistance has a constant acceleration g = 9.81 m s⁻² downward, whether it is moving up, down or is momentarily at rest. With upward as positive, a = -9.81 m s⁻² for the whole flight. At the top of a throw v = 0 for an instant, but the acceleration is still 9.81 m s⁻² downward, so the body does not stay there.

Check your understanding 5

A ball is dropped from rest from the top of a 45.0 m cliff. Neglecting air resistance, take downward as positive. Enter the initial velocity and signed displacement, then find the time to the bottom and the speed just before landing. Give time and speed to three significant figures.

Initial velocity

Unit: m s⁻¹

Signed displacement

Unit: m

Landing time

Unit: s

Show the steps Optional working

Step 1: Landing time

Unit: s

Landing speed

Unit: m s⁻¹

Show answer

Take downward as positive: u = 0, a = +9.81 m s⁻², s = +45.0 m. From s = (1/2)at², t = square root of (2(45.0)/9.81) = 3.03 s. From v² = 2as, v = square root of 2(9.81)(45.0) = 29.7 m s⁻¹.

Two times at one position

An object that turns back can pass a position on both sides of its turning point. An equation for time can then have two positive roots, both describing the motion. A root before the stated start or after the model stops applying must be rejected for that physical interval.

Modelled position–time curve with a horizontal reference at 30 m and markers at the two intersections. The curve rises to 100 m at 10 s, then falls to the starting position at 20 s. — For x = (20 m s⁻¹)t − (1 m s⁻²)t², x = 30 m at about 1.63 s and 18.4 s. The turning point is at 10 s, where the position is 100 m and velocity is zero.

Figure data: Two times at the same position
Series	Time (s)	Position (m)
Model position	0.0000	0.00
Model position	0.1000	1.99
Model position	0.2000	3.96
Model position	0.3000	5.91
Model position	0.4000	7.84
Model position	0.5000	9.75
Model position	0.6000	11.64
Model position	0.7000	13.51
Model position	0.8000	15.36
Model position	0.9000	17.19
Model position	1.0000	19.00
Model position	1.1000	20.79
Model position	1.2000	22.56
Model position	1.3000	24.31
Model position	1.4000	26.04
Model position	1.5000	27.75
Model position	1.6000	29.44
Model position	1.6334	30.00
Model position	1.7000	31.11
Model position	1.8000	32.76
Model position	1.9000	34.39
Model position	2.0000	36.00
Model position	2.1000	37.59
Model position	2.2000	39.16
Model position	2.3000	40.71
Model position	2.4000	42.24
Model position	2.5000	43.75
Model position	2.6000	45.24
Model position	2.7000	46.71
Model position	2.8000	48.16
Model position	2.9000	49.59
Model position	3.0000	51.00
Model position	3.1000	52.39
Model position	3.2000	53.76
Model position	3.3000	55.11
Model position	3.4000	56.44
Model position	3.5000	57.75
Model position	3.6000	59.04
Model position	3.7000	60.31
Model position	3.8000	61.56
Model position	3.9000	62.79
Model position	4.0000	64.00
Model position	4.1000	65.19
Model position	4.2000	66.36
Model position	4.3000	67.51
Model position	4.4000	68.64
Model position	4.5000	69.75
Model position	4.6000	70.84
Model position	4.7000	71.91
Model position	4.8000	72.96
Model position	4.9000	73.99
Model position	5.0000	75.00
Model position	5.1000	75.99
Model position	5.2000	76.96
Model position	5.3000	77.91
Model position	5.4000	78.84
Model position	5.5000	79.75
Model position	5.6000	80.64
Model position	5.7000	81.51
Model position	5.8000	82.36
Model position	5.9000	83.19
Model position	6.0000	84.00
Model position	6.1000	84.79
Model position	6.2000	85.56
Model position	6.3000	86.31
Model position	6.4000	87.04
Model position	6.5000	87.75
Model position	6.6000	88.44
Model position	6.7000	89.11
Model position	6.8000	89.76
Model position	6.9000	90.39
Model position	7.0000	91.00
Model position	7.1000	91.59
Model position	7.2000	92.16
Model position	7.3000	92.71
Model position	7.4000	93.24
Model position	7.5000	93.75
Model position	7.6000	94.24
Model position	7.7000	94.71
Model position	7.8000	95.16
Model position	7.9000	95.59
Model position	8.0000	96.00
Model position	8.1000	96.39
Model position	8.2000	96.76
Model position	8.3000	97.11
Model position	8.4000	97.44
Model position	8.5000	97.75
Model position	8.6000	98.04
Model position	8.7000	98.31
Model position	8.8000	98.56
Model position	8.9000	98.79
Model position	9.0000	99.00
Model position	9.1000	99.19
Model position	9.2000	99.36
Model position	9.3000	99.51
Model position	9.4000	99.64
Model position	9.5000	99.75
Model position	9.6000	99.84
Model position	9.7000	99.91
Model position	9.8000	99.96
Model position	9.9000	99.99
Model position	10.0000	100.00
Model position	10.1000	99.99
Model position	10.2000	99.96
Model position	10.3000	99.91
Model position	10.4000	99.84
Model position	10.5000	99.75
Model position	10.6000	99.64
Model position	10.7000	99.51
Model position	10.8000	99.36
Model position	10.9000	99.19
Model position	11.0000	99.00
Model position	11.1000	98.79
Model position	11.2000	98.56
Model position	11.3000	98.31
Model position	11.4000	98.04
Model position	11.5000	97.75
Model position	11.6000	97.44
Model position	11.7000	97.11
Model position	11.8000	96.76
Model position	11.9000	96.39
Model position	12.0000	96.00
Model position	12.1000	95.59
Model position	12.2000	95.16
Model position	12.3000	94.71
Model position	12.4000	94.24
Model position	12.5000	93.75
Model position	12.6000	93.24
Model position	12.7000	92.71
Model position	12.8000	92.16
Model position	12.9000	91.59
Model position	13.0000	91.00
Model position	13.1000	90.39
Model position	13.2000	89.76
Model position	13.3000	89.11
Model position	13.4000	88.44
Model position	13.5000	87.75
Model position	13.6000	87.04
Model position	13.7000	86.31
Model position	13.8000	85.56
Model position	13.9000	84.79
Model position	14.0000	84.00
Model position	14.1000	83.19
Model position	14.2000	82.36
Model position	14.3000	81.51
Model position	14.4000	80.64
Model position	14.5000	79.75
Model position	14.6000	78.84
Model position	14.7000	77.91
Model position	14.8000	76.96
Model position	14.9000	75.99
Model position	15.0000	75.00
Model position	15.1000	73.99
Model position	15.2000	72.96
Model position	15.3000	71.91
Model position	15.4000	70.84
Model position	15.5000	69.75
Model position	15.6000	68.64
Model position	15.7000	67.51
Model position	15.8000	66.36
Model position	15.9000	65.19
Model position	16.0000	64.00
Model position	16.1000	62.79
Model position	16.2000	61.56
Model position	16.3000	60.31
Model position	16.4000	59.04
Model position	16.5000	57.75
Model position	16.6000	56.44
Model position	16.7000	55.11
Model position	16.8000	53.76
Model position	16.9000	52.39
Model position	17.0000	51.00
Model position	17.1000	49.59
Model position	17.2000	48.16
Model position	17.3000	46.71
Model position	17.4000	45.24
Model position	17.5000	43.75
Model position	17.6000	42.24
Model position	17.7000	40.71
Model position	17.8000	39.16
Model position	17.9000	37.59
Model position	18.0000	36.00
Model position	18.1000	34.39
Model position	18.2000	32.76
Model position	18.3000	31.11
Model position	18.3666	30.00
Model position	18.4000	29.44
Model position	18.5000	27.75
Model position	18.6000	26.04
Model position	18.7000	24.31
Model position	18.8000	22.56
Model position	18.9000	20.79
Model position	19.0000	19.00
Model position	19.1000	17.19
Model position	19.2000	15.36
Model position	19.3000	13.51
Model position	19.4000	11.64
Model position	19.5000	9.75
Model position	19.6000	7.84
Model position	19.7000	5.91
Model position	19.8000	3.96
Model position	19.9000	1.99
Model position	20.0000	0.00
Reference position: 30 m	0.0000	30.00
Reference position: 30 m	20.0000	30.00
Passages through 30 m	1.6334	30.00
Passages through 30 m	18.3666	30.00
Turning point	10.0000	100.00

Figure data: Two times at the same position
Series	Time (s)	Position (m)
Model position	0.0000	0.00
Model position	0.1000	1.99
Model position	0.2000	3.96
Model position	0.3000	5.91
Model position	0.4000	7.84
Model position	0.5000	9.75
Model position	0.6000	11.64
Model position	0.7000	13.51
Model position	0.8000	15.36
Model position	0.9000	17.19
Model position	1.0000	19.00
Model position	1.1000	20.79
Model position	1.2000	22.56
Model position	1.3000	24.31
Model position	1.4000	26.04
Model position	1.5000	27.75
Model position	1.6000	29.44
Model position	1.6334	30.00
Model position	1.7000	31.11
Model position	1.8000	32.76
Model position	1.9000	34.39
Model position	2.0000	36.00
Model position	2.1000	37.59
Model position	2.2000	39.16
Model position	2.3000	40.71
Model position	2.4000	42.24
Model position	2.5000	43.75
Model position	2.6000	45.24
Model position	2.7000	46.71
Model position	2.8000	48.16
Model position	2.9000	49.59
Model position	3.0000	51.00
Model position	3.1000	52.39
Model position	3.2000	53.76
Model position	3.3000	55.11
Model position	3.4000	56.44
Model position	3.5000	57.75
Model position	3.6000	59.04
Model position	3.7000	60.31
Model position	3.8000	61.56
Model position	3.9000	62.79
Model position	4.0000	64.00
Model position	4.1000	65.19
Model position	4.2000	66.36
Model position	4.3000	67.51
Model position	4.4000	68.64
Model position	4.5000	69.75
Model position	4.6000	70.84
Model position	4.7000	71.91
Model position	4.8000	72.96
Model position	4.9000	73.99
Model position	5.0000	75.00
Model position	5.1000	75.99
Model position	5.2000	76.96
Model position	5.3000	77.91
Model position	5.4000	78.84
Model position	5.5000	79.75
Model position	5.6000	80.64
Model position	5.7000	81.51
Model position	5.8000	82.36
Model position	5.9000	83.19
Model position	6.0000	84.00
Model position	6.1000	84.79
Model position	6.2000	85.56
Model position	6.3000	86.31
Model position	6.4000	87.04
Model position	6.5000	87.75
Model position	6.6000	88.44
Model position	6.7000	89.11
Model position	6.8000	89.76
Model position	6.9000	90.39
Model position	7.0000	91.00
Model position	7.1000	91.59
Model position	7.2000	92.16
Model position	7.3000	92.71
Model position	7.4000	93.24
Model position	7.5000	93.75
Model position	7.6000	94.24
Model position	7.7000	94.71
Model position	7.8000	95.16
Model position	7.9000	95.59
Model position	8.0000	96.00
Model position	8.1000	96.39
Model position	8.2000	96.76
Model position	8.3000	97.11
Model position	8.4000	97.44
Model position	8.5000	97.75
Model position	8.6000	98.04
Model position	8.7000	98.31
Model position	8.8000	98.56
Model position	8.9000	98.79
Model position	9.0000	99.00
Model position	9.1000	99.19
Model position	9.2000	99.36
Model position	9.3000	99.51
Model position	9.4000	99.64
Model position	9.5000	99.75
Model position	9.6000	99.84
Model position	9.7000	99.91
Model position	9.8000	99.96
Model position	9.9000	99.99
Model position	10.0000	100.00
Model position	10.1000	99.99
Model position	10.2000	99.96
Model position	10.3000	99.91
Model position	10.4000	99.84
Model position	10.5000	99.75
Model position	10.6000	99.64
Model position	10.7000	99.51
Model position	10.8000	99.36
Model position	10.9000	99.19
Model position	11.0000	99.00
Model position	11.1000	98.79
Model position	11.2000	98.56
Model position	11.3000	98.31
Model position	11.4000	98.04
Model position	11.5000	97.75
Model position	11.6000	97.44
Model position	11.7000	97.11
Model position	11.8000	96.76
Model position	11.9000	96.39
Model position	12.0000	96.00
Model position	12.1000	95.59
Model position	12.2000	95.16
Model position	12.3000	94.71
Model position	12.4000	94.24
Model position	12.5000	93.75
Model position	12.6000	93.24
Model position	12.7000	92.71
Model position	12.8000	92.16
Model position	12.9000	91.59
Model position	13.0000	91.00
Model position	13.1000	90.39
Model position	13.2000	89.76
Model position	13.3000	89.11
Model position	13.4000	88.44
Model position	13.5000	87.75
Model position	13.6000	87.04
Model position	13.7000	86.31
Model position	13.8000	85.56
Model position	13.9000	84.79
Model position	14.0000	84.00
Model position	14.1000	83.19
Model position	14.2000	82.36
Model position	14.3000	81.51
Model position	14.4000	80.64
Model position	14.5000	79.75
Model position	14.6000	78.84
Model position	14.7000	77.91
Model position	14.8000	76.96
Model position	14.9000	75.99
Model position	15.0000	75.00
Model position	15.1000	73.99
Model position	15.2000	72.96
Model position	15.3000	71.91
Model position	15.4000	70.84
Model position	15.5000	69.75
Model position	15.6000	68.64
Model position	15.7000	67.51
Model position	15.8000	66.36
Model position	15.9000	65.19
Model position	16.0000	64.00
Model position	16.1000	62.79
Model position	16.2000	61.56
Model position	16.3000	60.31
Model position	16.4000	59.04
Model position	16.5000	57.75
Model position	16.6000	56.44
Model position	16.7000	55.11
Model position	16.8000	53.76
Model position	16.9000	52.39
Model position	17.0000	51.00
Model position	17.1000	49.59
Model position	17.2000	48.16
Model position	17.3000	46.71
Model position	17.4000	45.24
Model position	17.5000	43.75
Model position	17.6000	42.24
Model position	17.7000	40.71
Model position	17.8000	39.16
Model position	17.9000	37.59
Model position	18.0000	36.00
Model position	18.1000	34.39
Model position	18.2000	32.76
Model position	18.3000	31.11
Model position	18.3666	30.00
Model position	18.4000	29.44
Model position	18.5000	27.75
Model position	18.6000	26.04
Model position	18.7000	24.31
Model position	18.8000	22.56
Model position	18.9000	20.79
Model position	19.0000	19.00
Model position	19.1000	17.19
Model position	19.2000	15.36
Model position	19.3000	13.51
Model position	19.4000	11.64
Model position	19.5000	9.75
Model position	19.6000	7.84
Model position	19.7000	5.91
Model position	19.8000	3.96
Model position	19.9000	1.99
Model position	20.0000	0.00
Reference position: 30 m	0.0000	30.00
Reference position: 30 m	20.0000	30.00
Passages through 30 m	1.6334	30.00
Passages through 30 m	18.3666	30.00
Turning point	10.0000	100.00

This graph shows x = 20t-t², in SI units. The object reaches x = 30 m on the way out, turns at t = 10 s, and passes x = 30 m again on the way back. The next problem uses different numbers; the same rising-then-falling shape explains its two times.

Examination question

A particle has initial velocity 24.0 m s⁻¹ and constant acceleration -2.00 m s⁻². Find both times at which its displacement from the start is 44.0 m, and explain why there are two. [5 marks]

Find both times

Choose the equation you must solve and enter both times to three significant figures. The explanation is a separate comparison task below.

Equation for time

44.0 = 24.0t + 1.00t²44.0 = 24.0t44.0 = 24.0t − 1.00t²

Earlier time

Unit: s

Later time

Unit: s

Show solution step by step

Form the equation
2 marks
Method
Use s = ut + (1/2)at² with signed values.
Reason
The acceleration is constant and the time is unknown.
Working
Answer
44.0 = 24.0t-1.00t², so t²-24.0t + 44.0 = 0.
Solve
2 marks
Method
Use the quadratic formula.
Reason
A quadratic can have two roots.
Working
Answer
t = (24.0± square root of (576-176))/2 = 24.0±20.0/2 = 2.00 s or 22.0 s
Explain
1 mark
Working
Answer
The velocity v = 24.0-2.00t is zero at t = 12.0 s, between the two times. The particle passes 44.0 m on the way out, turns round, and passes it again on the way back.

Self-mark with the mark scheme

Tick each mark point that your answer includes.

Self-mark the equation, the two roots and the explanation.

Mark points

2 marks: Forms the correct quadratic from signed values.
2 marks: Obtains both positive times, 2.00 s and 22.0 s.
1 mark: Explains the outward and return passages through the same position.

Check your understanding 6

For the 24.0 m s⁻¹ particle above, explain why both 2.00 s and 22.0 s are valid. Find the turning time and the signed velocity at each passage. Relate them to the position–time curve; work on paper if you prefer.

Your interpretation with turning time and signed velocities

Show answer

The velocity is v = 24.0-2.00t, so the turning time is 12.0 s. At 2.00 s, v = +20.0 m s⁻¹; at 22.0 s, v = -20.0 m s⁻¹. Both times are after the start and straddle the turning point: the particle crosses s = 44.0 m going out and again returning. Compare your answer for the turning time, both signed velocities and the two passages. The plotted x = 20t-t² model has different numbers but the same rising-then-falling shape.

Check your understanding 7

A ball is thrown vertically downward at 5.00 m s⁻¹ from a balcony 20.0 m above the ground. Take upward as positive, neglect air resistance and use g = 9.81 m s⁻². Find its landing time and signed velocity just before landing. Show your equation and explain the meaning of any discarded time root. Give the results to three significant figures; work on paper if you prefer.

Your complete signed-motion solution

Show answer

Here u = -5.00 m s⁻¹, a = -9.81 m s⁻² and s = -20.0 m. Thus -20.0 = -5.00t-4.905t², giving t = 1.572922… s or -2.592290… s. The negative root extrapolates the model before release; it is not a landing after the throw. The landing time is 1.57 s and v = -5.00-9.81(1.572922…) = -20.4 m s⁻¹, downward. Check independently with v² = u² + 2as = 417.4 m² s⁻², choosing the negative square root because the ball moves downward throughout. Compare signs, the nonzero initial velocity, both roots and their interpretation, guard digits and the final direction.

Common mistakes

Using the equations over an interval in which the acceleration changes. Split the motion into stages and carry the final velocity of one stage into the next.
Treating s as a distance when the object turns back.
Saying the acceleration is zero at the top of a throw.
Rejecting the larger root of a quadratic without checking whether it describes the return journey.

Spotted an error? Report a correction

Syllabus and review details

GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30

Constant-acceleration equations

Guided lesson

Guided steps

Reading settings

Predict motion at a turning point

Check your understanding 1

The five quantities

Deriving the equations

Check your understanding 2

Check your understanding 3

Choosing an equation

A train braking to rest

List the known values

Find a without t

Find t

Check your understanding 4

Free fall

Check your understanding 5

Two times at one position

Two times for one displacement

Form the equation

Solve

Explain

Check your understanding 6

Check your understanding 7

Common mistakes