The Describing Motion topic ended with two facts about a velocity–time graph: its slope is the acceleration, and the area under it is the displacement. When the acceleration is constant, the v–t graph is a straight line, and a straight line is simple enough to turn both facts into formulas. This page derives the four equations of uniformly accelerated motion from that one line, then shows how to choose the right equation for a question and how to handle the signs when a body slows down. The rest of this topic, braking, the ratio rules and free fall, is these same equations put to work.
Each of the four papers so far has had exactly one question on uniformly accelerated motion: December 2025 Q7, January 2026 Q48, March 2026 Q5 and April 2026 Q8. Two of them were the same one-line question with new numbers: a body starts from rest with a given acceleration; what is its velocity after a given time? Both are worked through below. Nearly every question in this topic has started from rest, so the from-rest forms of the equations get a section of their own. The equations also turn up as the last step of questions in other topics, so they earn marks more often than the count suggests.
The velocity after a time: v = v₀ + at
Start the clock, t = 0, at a moment when the body is moving with velocity v₀, its initial velocity. Let its acceleration be a constant, a. At a later time t its velocity is v, the final velocity: final only in the sense that it comes at the end of the interval we are looking at. Because the acceleration does not change, it equals the change in velocity divided by the time over any interval, including the whole interval from 0 to t. So a = (v − v₀)/t. Multiply both sides by t to get at = v − v₀, then add v₀ to both sides: v = v₀ + at.
Read it as a sentence. The body starts with v₀ and gains a in each of the t seconds, so by time t it has gained at. A car moving at 4 m/s that gains 3 m/s every second is moving at 4 + 3 × 5 = 19 m/s five seconds later.
On the v–t graph, v = v₀ + at is the straight line that starts at height v₀ when t = 0 and has slope a. Over a run of t, a line of slope a rises by slope × run = at, so its height at time t is v₀ + at. If you know straight lines from maths, it is y = b + kx, with v₀ as the intercept and a as the gradient.
A straight v–t line for a constant acceleration a. It starts at height v₀ when t = 0. Over a run of t its rise is slope × run = at, so its height at time t, the velocity v, is v₀ + at.
Velocity–time equation. For motion along a straight line with a constant acceleration a, the velocity at time t is v = v₀ + at, where v₀ is the velocity at t = 0. Velocities are in m/s, a in m/s² and t in s. From rest, v₀ = 0 and the equation becomes v = at.
Try it! · Check-in question
A cyclist riding along a straight path at 3 m/s speeds up with a constant acceleration of 0.5 m/s². What is her velocity 8 s later?
Her velocity starts at v₀ = 3 m/s and grows by 0.5 m/s every second for 8 s, so v = v₀ + at = 3 + 0.5 × 8 = 3 + 4 = 7 m/s, and the answer is C. A is at on its own, 0.5 × 8 = 4 m/s: the velocity she gains, without the 3 m/s she already had. B mixes in the displacement formula of the next section: ½at² = ½ × 0.5 × 8² = 16, plus 3. D adds the acceleration only once, as if she had sped up for 1 s instead of 8.
The displacement after a time: x = v₀t + ½at²
For the displacement, use the other fact from the last topic: the area under a v–t graph is the displacement. Under the straight line from t = 0 to time t, the area is a trapezium, and the dashed line at height v₀ cuts it into two shapes whose areas you already know how to find.
The rectangle at the bottom is v₀ high and t wide, so its area is v₀t. That is the displacement the body would have if it simply kept its starting velocity for the whole time.
The triangle on top has a base of t and a height of at, the rise of the line. Its area is ½ × t × at = ½at². That is the extra displacement the acceleration adds, because the body moves faster and faster as time goes on.
Add the two pieces and you have the second equation: x = v₀t + ½at².
The area under the line from 0 to t is the displacement. It splits into a rectangle, v₀ high and t wide, with area v₀t, and a triangle on top, t wide and at high, with area ½ × t × at = ½at². Together: x = v₀t + ½at².
Displacement–time equation. For motion along a straight line with a constant acceleration, the displacement in the time t after the clock starts is x = v₀t + ½at². The first term is what the starting velocity alone would give; the second is what the acceleration adds. From rest, x = ½at².
Displacement, not position. Here x is the displacement from wherever the body was at t = 0. If its position then was x₀, its position at time t is x₀ + v₀t + ½at², just as uniform motion gave x = x₀ + vt. With a = 0 the equation becomes x = v₀t, the uniform motion of the last topic.
Worked example · velocity and displacement after a time
A car moving at 4 m/s along a straight road speeds up with a constant acceleration of 2 m/s². Find its velocity and its displacement 5 s later.
Velocity: v = v₀ + at = 4 + 2 × 5 = 14 m/s.
Displacement: x = v₀t + ½at² = 4 × 5 + ½ × 2 × 5² = 20 + 25 = 45 m.
Square the time before you halve: ½ × 2 × 25 = 25 m, not ½ × 2 × 5 = 5 m. As a check, the car's velocity rose steadily from 4 m/s to 14 m/s, so it averaged 9 m/s, and 9 m/s for 5 s is 45 m.
The ½ and the square. Slips in x = v₀t + ½at² nearly always involve one of these two. Dropping the ½ gives at², twice the true extra displacement. Forgetting to square t gives ½at, which is not even a length: its unit is m/s² × s = m/s, a velocity. With the square, m/s² × s² = m, as a displacement should be. A quick look at the units catches the second slip every time.
Try it! · Check-in question
A trolley moving at 2 m/s along a straight track speeds up with a constant acceleration of 3 m/s². How far does it travel in the next 4 s?
x = v₀t + ½at² = 2 × 4 + ½ × 3 × 4² = 8 + 24 = 32 m, so the answer is A. B drops the ½: 8 + 3 × 16 = 56 m. C keeps only the triangle, ½at² = 24 m, and leaves out the 8 m the trolley would cover at its starting velocity anyway. D comes from the velocity formula, v₀ + at = 2 + 3 × 4 = 14: that is the trolley's velocity in m/s at the end of the 4 s, not a distance.
Starting from rest
Nearly every exam question in this topic so far has started from rest, so these are the forms to know by heart. Put v₀ = 0 into the two equations and both shrink: v = at and x = ½at². On the v–t graph the line now starts at the origin, the rectangle disappears, and the area under the line is a single triangle.
The two formulas use the same letters, a and t, which is exactly why they are easy to swap. Here they are side by side for a body that starts from rest with an acceleration of 2 m/s²:
Time t
Velocity v = at
Displacement x = ½at²
1 s
2 m/s
1 m
2 s
4 m/s
4 m
3 s
6 m/s
9 m
4 s
8 m/s
16 m
The velocity grows in step with the time: in twice the time, twice the velocity. The displacement grows with the square of the time: in twice the time, four times the displacement, because the later seconds are covered at higher speeds. Keep an eye on the bottom row, 8 m/s and 16 m, as the first worked example below uses exactly these numbers.
From rest. v = at gives the velocity at an instant; x = ½at² gives the displacement over the interval from t = 0. Same letters, different quantities: the velocity grows in step with t, the displacement with t².
So before you choose, read what the question asks for, and when. "Its velocity at t = 4 s" and "its velocity at the end of 5 s" name a velocity at an instant, so they want v = at. "How far does it move in 4 s" and "its displacement in the first 4 s" name a length over an interval, so they want x = ½at². As the first page of Describing Motion showed, "at the end of 5 s" is the instant t = 5 s.
Worked example · velocity after a time from rest (December 2025, Q7)
An object accelerates uniformly in a straight line from rest at a = 2 m/s² on a smooth horizontal surface. Its velocity at t = 4 s is ( ).
(A) 2 m/s (B) 16 m/s (C) 4 m/s (D) 8 m/s
Step 1: the question asks for a velocity at an instant, t = 4 s, and the object starts from rest, so v₀ = 0.
Step 2: v = v₀ + at = 0 + 2 × 4 = 8 m/s. The answer is D.
The "smooth horizontal surface" plays no part: the acceleration is given, so there is nothing to work out from forces or friction.
B, 16 m/s, is the trap. It is ½at² = ½ × 2 × 4² = 16: the number of metres the object travels in the 4 s, given a velocity unit. A student who mixes up v = at and x = ½at² finds it waiting among the options. A and C are the bare numbers in the stem, the acceleration and the time.
Worked example · the same question with new numbers (April 2026, Q8)
An object starts from rest and moves with uniformly accelerated linear motion at a = 3 m/s². Its velocity at the end of 5 s is ( )
A. 10 m/s B. 5 m/s C. 20 m/s D. 15 m/s
"At the end of 5 s" is the instant t = 5 s, and the object starts from rest, so v = at = 3 × 5 = 15 m/s. The answer is D.
B is the time, 5, with a velocity unit. A and C are round numbers on either side of the answer that no formula gives. This time the ½at² number, 37.5, is not among the options, so a student who used the wrong formula would find no match at all: a useful sign to go back and read the question again.
Exam wording. The Chinese text of April 2026 Q8 begins 物体由静止开始做匀加速直线运动. Here 由静止开始 means "starting from rest", so v₀ = 0, and 匀加速直线运动 is uniformly accelerated linear motion. Its 5 s 末, "the end of 5 s", is the instant t = 5 s. Elsewhere you will meet 初速度 for the initial velocity v₀ and 末速度 for the final velocity v.
Try it! · Check-in question
A sledge starts from rest and slides down a straight slope with a constant acceleration of 4 m/s². What are its speed at t = 3 s and the distance it has travelled by then, respectively?
The speed at the instant t = 3 s is v = at = 4 × 3 = 12 m/s. The distance over the first 3 s is x = ½at² = ½ × 4 × 3² = 18 m. In the order asked, the answer is D. A drops the ½ from the distance: 4 × 9 = 36 m. B swaps the two formulas, giving 18 for the speed and 12 for the distance. C halves the speed as well, as if v were ½at; the ½ belongs only in the displacement, where it comes from the area of a triangle.
Using the average velocity: x = ½(v₀ + v)t
The trapezium under the v–t line can also be measured in one go. The area of a trapezium is the mean of its two parallel sides times the distance between them. Here the parallel sides are the heights at the two ends, v₀ and v, and the distance between them is t. So x = ½(v₀ + v)t.
The formula also reads as average velocity × time. The velocity rises steadily from v₀ to v, so over the whole interval it averages exactly halfway between them, ½(v₀ + v). The figure shows why: the part of the trapezium above that halfway height exactly fills the gap below it. The car in the first worked example went from 4 m/s to 14 m/s in 5 s, so it averaged 9 m/s, and 9 × 5 = 45 m, as the check there found. This form needs no acceleration at all, only the two velocities and the time.
Cut off the red triangle on the right, above the dashed line, and it exactly fills the red gap on the left, because the line is straight. So the area under the line equals that of a rectangle t wide and ½(v₀ + v) high: x = ½(v₀ + v)t.
Average velocity in uniformly accelerated motion. When the acceleration is constant, the average velocity over an interval is the mean of the velocities at its start and its end, ½(v₀ + v), so the displacement is x = ½(v₀ + v)t. The acceleration does not appear.
Only for a constant acceleration. Averaging the first and last velocities works only when the velocity changes steadily. For a trip in stages at different speeds, the average velocity is the total displacement divided by the total time, as on the Speed & Velocity page: 60 km at 30 km/h followed by 60 km at 60 km/h averages 40 km/h, not 45 km/h.
Try it! · Check-in question
A lorry on a straight road speeds up uniformly from 8 m/s to 16 m/s in 10 s. How far does it travel in that time?
The velocity rises steadily from 8 m/s to 16 m/s, so it averages ½(8 + 16) = 12 m/s, and in 10 s the lorry travels 12 × 10 = 120 m. The answer is B. No acceleration was needed, but it makes a good check: a = (16 − 8) ÷ 10 = 0.8 m/s², and v₀t + ½at² = 80 + ½ × 0.8 × 10² = 80 + 40 = 120 m. A uses the starting velocity for the whole 10 s, as if the lorry never sped up. C uses the final velocity for the whole time, as if it had been moving at 16 m/s from the start. D adds the two velocities but forgets to halve them.
When no time is given: v² − v₀² = 2ax
Some questions give velocities and a distance but no time: a car speeds up from 10 m/s to 20 m/s over 75 m; what is its acceleration? With the equations so far you would have to find the time first and then use it. It is quicker to remove t from the equations once and for all.
From v = v₀ + at, the time is t = (v − v₀)/a. Put that into x = ½(v₀ + v)t:
x = ½(v + v₀) × (v − v₀)/a = (v² − v₀²)/(2a),
because (v + v₀)(v − v₀) = v² − v₀², the difference of two squares. Multiply both sides by 2a and you have the fourth equation: v² − v₀² = 2ax. From rest it becomes v² = 2ax, so v = √(2ax).
Velocity–displacement equation. v² − v₀² = 2ax links the two velocities, the acceleration and the displacement, with no time in it. From rest, v² = 2ax and v = √(2ax).
Worked example · acceleration without a time
A car on a straight road speeds up uniformly from 10 m/s to 20 m/s over a distance of 75 m. Find its acceleration.
No time is given or asked for, so use v² − v₀² = 2ax: 20² − 10² = 2 × a × 75.
400 − 100 = 150a, so 300 = 150a and a = 2 m/s².
Check: at 2 m/s² the car gains its 10 m/s in 5 s. Its average velocity is ½(10 + 20) = 15 m/s, and 15 m/s × 5 s = 75 m.
v² − v₀² is not (v − v₀)². Square each velocity first, then subtract: 20² − 10² = 400 − 100 = 300. Subtracting first gives (20 − 10)² = 100, a third of the right value. In the same way, when you solve for v, take the square root of the whole of v₀² + 2ax: √(v₀² + 2ax) is not v₀ + √(2ax).
Try it! · Check-in question
A ball starts from rest and rolls down a straight ramp with a constant acceleration of 2 m/s². How fast is it moving when it has rolled 9 m?
No time is given, and the ball starts from rest, so use v² = 2ax = 2 × 2 × 9 = 36, which gives v = √36 = 6 m/s. The answer is D. C stops at v² = 36 and forgets the square root: 36 is the square of the speed, not the speed. A multiplies the acceleration by the distance, 2 × 9 = 18, as if the distance were a time in v = at. B drops the 2 from 2ax: √(2 × 9) = √18 = 3√2.
Choosing the equation
Every question in this topic is about the same five quantities: the initial velocity v₀, the final velocity v, the acceleration a, the time t and the displacement x. Each of the four equations contains four of them and leaves one out. That is the key to choosing.
Equation
Leaves out
Use it when the question has no…
v = v₀ + at
x
displacement
x = v₀t + ½at²
v
final velocity
v² − v₀² = 2ax
t
time
x = ½(v₀ + v)t
a
acceleration
A typical question gives you three of the five and asks for a fourth. The fifth plays no part, so pick the equation that leaves it out, and the answer comes in a single step.
1
List the five. Write v₀, v, a, t and x, and fill in the three the question gives. "From rest" means v₀ = 0; "comes to rest" means v = 0.
2
Mark the one you want. Then find the one left over: neither given nor asked for.
3
Use the equation without it. Substitute, with velocities in m/s and times in s, and solve.
Worked example · choosing the equation
A train moving at 10 m/s along a straight track speeds up with a constant acceleration of 0.5 m/s² until it is moving at 20 m/s. How far does it travel while speeding up?
Given: v₀ = 10 m/s, v = 20 m/s, a = 0.5 m/s². Wanted: x. Left over: t.
The equation without t is v² − v₀² = 2ax: 20² − 10² = 2 × 0.5 × x, so 300 = 1 × x and x = 300 m.
The long way gives the same answer in two steps: t = (20 − 10) ÷ 0.5 = 20 s from the first equation, then x = ½(10 + 20) × 20 = 300 m.
Try it! · Check-in question
A skateboarder on a straight path speeds up uniformly from 2 m/s to 6 m/s over a distance of 16 m. To find the time this takes in a single step, which equation should you use?
The question gives v₀ = 2 m/s, v = 6 m/s and x = 16 m and asks for t, so the acceleration is the one left over. The equation without a is x = ½(v₀ + v)t, so the answer is C: 16 = ½(2 + 6)t = 4t, which gives t = 4 s. A and B both contain a, which is not given, so you would first have to find it from another equation, a = (6² − 2²) ÷ (2 × 16) = 1 m/s², and that takes two steps. D contains no t at all, so on its own it can never give the time; here it is the equation for finding a.
Slowing down: watch the signs
All four equations hold for a body that is slowing down, as long as the signs are right. Take the direction of motion as positive, so that v₀ is positive. As the Describing Motion topic showed, a body that is slowing down has its acceleration pointing against its motion, so a is negative. When a question says a car brakes "with a deceleration of 4 m/s²", it gives the size only; in the equations you write a = −4 m/s².
Worked example · a car braking
A car moving at 20 m/s along a straight road brakes with a constant deceleration of 4 m/s². Find its velocity and its displacement 3 s later.
Take the direction of motion as positive: v₀ = 20 m/s and a = −4 m/s².
Velocity: v = v₀ + at = 20 + (−4) × 3 = 20 − 12 = 8 m/s.
Displacement: x = v₀t + ½at² = 20 × 3 + ½ × (−4) × 3² = 60 − 18 = 42 m.
Check with the average velocity: ½(20 + 8) × 3 = 14 × 3 = 42 m. Leaving out the minus sign would give 32 m/s and 78 m: a braking car going faster than when it started, which is a clear sign that something has gone wrong.
The equations do not know that a car stops. With v₀ = 20 m/s and a = −4 m/s², v = 20 − 4t reaches zero at t = 5 s. Put in t = 8 s and the equation gives v = −12 m/s: a car reversing back down the road. A ball rolled up a smooth ramp really does turn round like that, but a braking car stops and stays stopped. So for anything that brakes, first find when it stops; the next page, Braking, Stopping & Safe Distance, makes this its first rule.
Try it! · Check-in question
A scooter moving at 10 m/s along a straight road brakes with a constant deceleration of 2 m/s². How far does it travel in the next 3 s?
Take the direction of motion as positive, so v₀ = 10 m/s and a = −2 m/s². The scooter would stop only at t = 10 ÷ 2 = 5 s, so it is still moving throughout the 3 s. Then x = v₀t + ½at² = 10 × 3 + ½ × (−2) × 3² = 30 − 9 = 21 m, so the answer is A. B treats the deceleration as positive, 30 + 9 = 39 m: how far the scooter would go if it were speeding up. C is v₀t alone, as if the brakes did nothing. D multiplies the final velocity by the time: v = 10 − 2 × 3 = 4 m/s and 4 × 3 = 12 m, which treats the scooter as moving at its slowest speed for the whole 3 s.
You will meet these equations again. In questions from other topics, x = ½at² is often the last step. December 2025 Q6 gave a 2 kg object that moves 8 m from rest in 2 s under a constant force and asked for the force: ½ × a × 2² = 8 gives a = 4 m/s², and Newton's second law, F = ma, finishes the job. December 2025 Q9 and April 2026 Q10 let a block slide from rest down a smooth 30° slope for 2 s: the slope sets the acceleration at 5 m/s², and ½ × 5 × 2² = 10 m. The forces give the acceleration; the equations on this page turn it into a distance or a speed.
Your turn! · Practice questions
Practice · Question 1
A toy car starts from rest and moves 12 m along a straight track in 4 s with a constant acceleration. What is its acceleration?
The car starts from rest, and the question gives x and t and asks for a; the final velocity is left over, so use x = ½at². Then 12 = ½ × a × 4² = 8a, so a = 1.5 m/s², and the answer is B. A divides the distance by the time, 12 ÷ 4 = 3: that is the average velocity in m/s, not the acceleration. D is the final velocity, twice that average, 6 m/s; check it with v = at = 1.5 × 4 = 6 m/s. C drops the ½ and solves 12 = a × 16.
Practice · Question 2
A motorbike moving at 6 m/s along a straight road speeds up with a constant acceleration of 2 m/s². How fast is it moving when it has covered 16 m?
No time is given or asked for, so use v² − v₀² = 2ax: v² = 6² + 2 × 2 × 16 = 36 + 64 = 100, so v = 10 m/s. The answer is A. B leaves out v₀² and uses v² = 2ax as if the motorbike started from rest: √64 = 8. C takes the square roots separately, 6 + √64 = 14, but √(36 + 64) is not √36 + √64. D puts the distance in place of the time in v = v₀ + at: 6 + 2 × 16 = 38.
Practice · Question 3
A train slows down uniformly from 30 m/s to 10 m/s while it covers 400 m of straight track. How long does this take?
The question gives v₀, v and x and asks for t, so the acceleration is left over: use x = ½(v₀ + v)t. Then 400 = ½(30 + 10)t = 20t, so t = 20 s, and the answer is C. A forgets to halve and solves 400 = 40t. B divides the distance by the final velocity, as if the train had moved at 10 m/s all the way, and D divides it by the starting velocity, 400 ÷ 30 ≈ 13.3 s, as if it had kept 30 m/s. The true average velocity, 20 m/s, lies between the two.
Practice · Question 4
A ball is rolled up a long, straight, smooth ramp at 8 m/s. Its acceleration is 2 m/s² down the ramp, the same on the way up and on the way down. Taking up the ramp as positive, what is the ball's velocity 6 s later?
With up the ramp positive, v₀ = +8 m/s and a = −2 m/s², so v = v₀ + at = 8 + (−2) × 6 = 8 − 12 = −4 m/s. The minus sign means the ball is now moving down the ramp at 4 m/s, so the answer is D. The ball stopped at t = 8 ÷ 2 = 4 s, and because its acceleration stayed the same, it then rolled back down. A forgets that the acceleration points down the ramp and adds 12 m/s instead of subtracting it. B assumes the ball stays where it stops, which is true of a braking car but not of a ball on a smooth slope. C has the right size but drops the sign, and the sign is what gives the direction.
Practice · Question 5
A car starts from rest and speeds up along a straight road with a constant acceleration of 2 m/s² for 5 s. It then keeps the velocity it has reached for a further 5 s. How far does it travel in the 10 s?
Work in two stages. Speeding up from rest: x₁ = ½at² = ½ × 2 × 5² = 25 m, and the velocity reached is v = at = 2 × 5 = 10 m/s. Then at a steady 10 m/s for 5 s: x₂ = 10 × 5 = 50 m. In total 25 + 50 = 75 m, so the answer is B. A is only the first stage and D only the second. C uses ½at² for the whole 10 s, ½ × 2 × 10² = 100 m, as if the car kept accelerating; once the acceleration changes, a new stage begins, and it needs its own equation.
Things to remember
Concept
What to know
The five quantities
Initial velocity v₀, final velocity v, constant acceleration a, time t and displacement x, in m/s, m/s², s and m.
v = v₀ + at
From the slope of the v–t line: start at v₀ and gain a every second. Leaves out x.
x = v₀t + ½at²
From the area under the v–t line: a rectangle v₀t plus a triangle ½at². Leaves out v. Square t before halving.
x = ½(v₀ + v)t
Average velocity × time; the trapezium in one go. Leaves out a. The mean of the first and last velocities is the average only when the acceleration is constant.
v² − v₀² = 2ax
v = v₀ + at and x = ½(v₀ + v)t combined, with t removed. Leaves out t. Square, then subtract: v² − v₀² is not (v − v₀)².
From rest
v = at, x = ½at², v² = 2ax. The velocity grows in step with t; the displacement grows with t², so twice the time gives four times the distance.
Instant or interval
"Velocity at t = 4 s" or "at the end of 5 s" means v = at. "How far in 4 s" means x = ½at². In December 2025 Q7 the trap, 16 m/s, was ½at², a displacement given a velocity unit.
Choosing the equation
List v₀, v, a, t and x; find the one that is neither given nor asked for; use the equation that leaves it out.
Slowing down
Take the direction of motion as positive, so a is negative: "a deceleration of 4 m/s²" means a = −4 m/s².
Braking
The equations do not know that a car stops. Find when it stops first; past that time they describe a body coming back, which a braking car does not do.
