What is constant acceleration in the context of Edexcel International A Levels Mechanics 1?

In Edexcel International A Levels Mechanics 1, constant acceleration refers to a situation where an object's acceleration remains unchanged over time. This means that the velocity of the object changes at a steady rate. It is a fundamental concept in kinematics and is often analyzed using equations of motion.

How do you calculate the final velocity of an object under constant acceleration?

To calculate the final velocity of an object under constant acceleration, you can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time elapsed. This formula is part of the equations of motion covered in Mechanics 1.

What are the equations of motion for constant acceleration?

The equations of motion for constant acceleration, often referred to as SUVAT equations, include: 1) v = u + at, 2) s = ut + 0.5at², 3) v² = u² + 2as, 4) s = vt - 0.5at², and 5) s = 0.5(u + v)t. These equations are essential for solving problems involving constant acceleration in Mechanics 1.

How can you determine the displacement of an object experiencing constant acceleration?

The displacement of an object experiencing constant acceleration can be determined using the equation s = ut + 0.5at², where s is the displacement, u is the initial velocity, a is the constant acceleration, and t is the time. This formula is crucial for solving kinematic problems in the Edexcel International A Levels curriculum.

What are some real-world examples of constant acceleration?

Real-world examples of constant acceleration include a car accelerating uniformly from rest, an object in free fall under gravity (ignoring air resistance), and a train increasing its speed at a steady rate. Understanding these examples helps students relate theoretical concepts from Mechanics 1 to practical situations.

Why is "Using $a = 9.8$ for vertical motion without declaring direction" a common mistake in Constant Acceleration?

Why it happens: g is written as a positive number, but the **direction** is downward. How to avoid it: State 'upward positive, so a = -9.8\,m s^-2' OR 'downward positive, so a = +9.8\,m s^-2'. Mark schemes credit any consistent convention. Source: WME01/01 examiner reports

Why is "Computing 'average velocity' using total distance" a common mistake in Constant Acceleration?

Why it happens: Both are 'rates'; students conflate them. How to avoid it: Average **velocity** uses **displacement** (final - initial position). Average **speed** uses **distance** (total path length). For motion that returns to start, average velocity is zero but average speed is not.

Why is "Forgetting the time offset when two particles start at different moments" a common mistake in Constant Acceleration?

Why it happens: Students use the same t for both particles. How to avoid it: Pick one reference clock; the particle that started t earlier has time (t + t) in its suvat equations. Source: WME01/01 examiner reports — flagged annually

Why is "Picking a suvat equation that requires a quantity you don't have" a common mistake in Constant Acceleration?

Why it happens: Memorising the equations without thinking about which variable is missing. How to avoid it: List the five quantities: u, v, a, s, t. Identify which is missing and which is asked. Choose the equation that uses the four available. For greatest height, t is missing — use v^2 = u^2 + 2as.

Why is "Computing displacement as the area under a displacement-time graph" a common mistake in Constant Acceleration?

Why it happens: Confusion with velocity-time graphs. How to avoid it: Memorise: on a v-t graph, **area = displacement**; on an s-t graph, the curve itself is the displacement (no area calculation needed).

Mechanics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Constant Acceleration

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Constant Acceleration Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

The five suvat equations connect displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ and time $t$ for motion with constant acceleration in a straight line. Apply to horizontal and vertical motion (with $g = 9.8\,\text{m s}^{-2}$ ), multi-stage problems, and the interpretation of velocity-time graphs.

At a glance

Five suvat equations: $v = u + at$ , $s = ut + \tfrac{1}{2}at^{2}$ , $v^{2} = u^{2} + 2as$ , $s = \tfrac{1}{2}(u+v)t$ , $s = vt - \tfrac{1}{2}at^{2}$ .
Each equation excludes one of $\{s, u, v, a, t\}$ . Identify the missing variable; pick the equation that uses the others.
Vertical motion: $a = -g = -9.8\,\text{m s}^{-2}$ with upward positive. State convention.
Velocity-time graph: gradient = acceleration; area under = displacement (signed).
Displacement-time graph: gradient = velocity; horizontal segment = stationary.
Multi-stage motion: solve each stage separately; carry final $v$ of one stage as initial $u$ of next.
Edexcel uses $g = 9.8\,\text{m s}^{-2}$ unless stated otherwise (NOT $10$ ).

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M1 2.1 — Use the suvat equations for motion in a straight line with constant acceleration.
M1 2.2 — Solve problems involving vertical motion under gravity (with $g = 9.8\,\text{m s}^{-2}$ ).
M1 2.3 — Interpret and construct velocity-time and displacement-time graphs.
M1 2.4 — Solve multi-stage motion problems (e.g. accelerate, cruise, decelerate).

The five suvat equations

Each equation omits one variable. Pick by elimination.

Five quantities describe straight-line motion with constant acceleration: displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ , time $t$ . Any FOUR determine the fifth.

Equation	Excludes
$v = u + at$	$s$
$s = \tfrac{1}{2}(u + v)t$	$a$
$s = ut + \tfrac{1}{2}at^{2}$	$v$
$v^{2} = u^{2} + 2as$	$t$
$s = vt - \tfrac{1}{2}at^{2}$	$u$

Strategy:

List the five variables.
Identify which are GIVEN and which is REQUESTED.
Identify the MISSING variable (the one neither given nor asked).
Pick the equation that does NOT involve the missing variable.

Example. Car accelerates from rest at $a$ , travels $200$ m, reaching $20$ m s $^{-1}$ . Find $a$ .

Given: $u = 0$ , $s = 200$ , $v = 20$ .
Asked: $a$ .
Missing: $t$ .
Equation: $v^{2} = u^{2} + 2as$ .
$400 = 0 + 400 a \Rightarrow a = 1\,\text{m s}^{-2}$ .

Why this works: avoiding $t$ saves a step. If you'd used $v = u + at$ , you'd need $t$ first.

Five variables: $s, u, v, a, t$ .
Each equation excludes one.
Identify missing variable; pick equation accordingly.
Cross-check with a second equation.

Vertical motion under gravity

Same suvat, but with $a = \pm g$ . Declare the sign convention.

Near the Earth's surface, an object in free fall (no air resistance) has acceleration $g = 9.8\,\text{m s}^{-2}$ vertically downward — the Edexcel convention. (CIE often uses $g = 10$ ; do NOT confuse them.)

Sign convention. Always declare: 'taking upward as positive, $a = -9.8$ ' OR 'taking downward as positive, $a = +9.8$ '. Mark schemes credit consistency, not a particular choice.

Standard scenarios:

Ball thrown straight up. Find max height, time of flight, speed at peak.
Ball thrown up from a height. Find time to hit ground, impact speed.
Ball dropped (free fall). $u = 0$ , find $v$ or $s$ at given $t$ .
Ball projected downward from a height. $u < 0$ (downward), find time to hit ground.

Greatest height. At the peak, $v = 0$ . Use $v^{2} = u^{2} + 2as$ :

$0 = u^{2} + 2(-g) s_{\max} \Rightarrow s_{\max} = \dfrac{u^{2}}{2g}.$

Time to peak. Use $v = u + at$ with $v = 0$ :

$t_{\text{peak}} = \dfrac{u}{g}.$

Time of flight (return to launch level). Use $s = ut + \tfrac{1}{2}at^{2}$ with $s = 0$ :

$0 = ut - \tfrac{1}{2}g t^{2} \Rightarrow t = \dfrac{2 u}{g} = 2 t_{\text{peak}}.$

Worked example. Ball thrown up from $1.5$ m above ground at $14$ m/s. Find time to hit ground.

Origin at launch; upward positive; $u = 14$ , $a = -9.8$ , $s = -1.5$ (ground is below).
$-1.5 = 14 t - 4.9 t^{2}$ .
$4.9 t^{2} - 14 t - 1.5 = 0$ .
$t = (14 + \sqrt{225.4})/9.8 \approx 2.96$ s.

$g = 9.8\,\text{m s}^{-2}$ (Edexcel convention).
Declare sign convention BEFORE writing equations.
At peak, $v = 0$ .
Greatest height $= u^{2}/(2g)$ above launch.
Time of flight (return to launch level) $= 2u/g$ .

Velocity-time and displacement-time graphs

Gradient = derivative; area = integral.

Velocity-time graph (often the most useful in M1):

Gradient = acceleration. A steep upward slope means high positive acceleration; a downward slope means deceleration.
Area under the graph = displacement (signed — areas below the $t$ -axis are negative).
A horizontal segment means constant velocity.
A straight line means constant acceleration.

Displacement-time graph:

Gradient = velocity. A horizontal segment means stationary (zero velocity).
A straight line means constant velocity.
A curve means changing velocity (acceleration).

Multi-stage motion example (train problem):

Stage 1: accelerate from rest at $0.4\,\text{m s}^{-2}$ for $50$ s. Final speed $V = 0.4 \cdot 50 = 20\,\text{m s}^{-1}$ .
Stage 2: constant $20\,\text{m s}^{-1}$ for $200$ s.
Stage 3: decelerate to rest in $80$ s. Deceleration $20/80 = 0.25\,\text{m s}^{-2}$ .

The velocity-time graph is a trapezium: rises from $(0,0)$ to $(50, 20)$ , flat to $(250, 20)$ , down to $(330, 0)$ .

Total distance = area of trapezium:

$s = \tfrac{1}{2}(\text{top edge} + \text{bottom edge}) \cdot \text{height} = \tfrac{1}{2}(200 + 330)(20) = 5300\,\text{m}.$

(Equivalent: sum of three triangle/rectangle areas.)

Average speed vs average velocity on a graph:

Average speed = (total area, all positive) / (total time).
Average velocity = (signed area) / (total time).

If the particle returns to its start, signed area = 0, so average velocity = 0 but average speed > 0.

$v$ - $t$ graph: gradient = $a$ , area = $s$ .
$s$ - $t$ graph: gradient = $v$ .
Horizontal segment on $v$ - $t$ = constant velocity.
Trapezium area for accel-cruise-decel motion.
Average speed $\ne$ average velocity when direction changes.

Multi-stage and two-particle problems

Treat each stage / particle independently; link via shared quantities.

Multi-stage motion: split the journey into segments where acceleration is constant. Apply suvat to each segment. The final $v$ of segment $i$ is the initial $u$ of segment $i+1$ .

Example — train accelerates from rest at $0.4\,\text{m s}^{-2}$ for $50$ s, then cruises at constant speed for $200$ s, then decelerates uniformly to rest in $80$ s.

Stage 1: $u = 0$ , $a = 0.4$ , $t = 50$ . $v = 20$ . $s_1 = \tfrac{1}{2}(0 + 20)(50) = 500$ m.
Stage 2: $v = 20$ (constant), $t = 200$ . $s_2 = 20 \cdot 200 = 4000$ m.
Stage 3: $u = 20$ , $v = 0$ , $t = 80$ . $a = -0.25$ . $s_3 = \tfrac{1}{2}(20+0)(80) = 800$ m.
Total $= 5300$ m.

Two particles meeting / overtaking: pick one reference clock. The particle that started $\Delta t$ later has elapsed time $(t - \Delta t)$ from start.

Example — Car $A$ starts from rest at $O$ with $a = 2$ m/s². Five seconds later, $B$ passes $O$ at $25$ m/s constant. When does $A$ catch up?

Let $t$ = time since $B$ passed $O$ . Then $A$ has been moving for $(t + 5)$ s.
Position of $A$ : $s_A = \tfrac{1}{2}(2)(t+5)^{2} = (t+5)^{2}$ .
Position of $B$ : $s_B = 25 t$ .
Meet: $(t+5)^{2} = 25 t \Rightarrow t^{2} - 15 t + 25 = 0 \Rightarrow t = (15 \pm 5\sqrt{5})/2$ .
Two roots: $t \approx 1.91$ (B overtakes A) and $t \approx 13.09$ (A overtakes B back).

Multi-stage: solve each segment separately.
Carry final $v$ of one stage as initial $u$ of next.
Two-particle: pick one reference clock; offset by start time.
Two meetings often: check both roots of the quadratic.

How it’s examined

Constant acceleration appears in every WME01/01 paper, usually as Q2 or Q3 ( $6$ - $10$ marks) plus another in a later question. Expect a multi-stage problem with a velocity-time graph to sketch ( $2$ - $4$ B-marks for the graph alone), and a vertical-motion problem with a quadratic in $t$ . Examiners specifically reward (i) declaration of sign convention, (ii) clear listing of $u, v, a, s, t$ before choosing an equation, (iii) cross-checking with a second suvat. The most common error is forgetting that 'time of flight' for a ball thrown up and returning to launch is $2 u/g$ , not $u/g$ .

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Constant Acceleration

Step-by-step solutions to past-paper-style questions on constant acceleration, written exactly the way a tutor would explain them at the board.

1Single-stage suvat — finding final velocity and time (6 marks, M1)
Core• Adapted from WME01/01 January 2024 Q2• suvat, kinematics, single stage
Question
A car accelerates uniformly from rest along a straight horizontal road. After travelling $200\,\text{m}$ its speed is $20\,\text{m s}^{-1}$ . Find (a) the acceleration, (b) the time taken.
Step-by-step solution
1. Step 1
  Draw a quick motion diagram with the car moving in one direction. List the known quantities. Take the direction of motion as positive.
  $u = 0, \quad v = 20, \quad s = 200, \quad a = ?, \quad t = ?$
2. Step 2
  For (a), the unknowns are $a$ and $t$ ; we have $u$ , $v$ , $s$ . Use $v^{2} = u^{2} + 2as$ to eliminate $t$ .
  $20^{2} = 0^{2} + 2a(200) \Rightarrow 400 = 400a \Rightarrow a = 1\,\text{m s}^{-2}$
3. Step 3
  For (b), use $v = u + at$ with the known $a$ .
  $20 = 0 + 1 \cdot t \Rightarrow t = 20\,\text{s}$
4. Step 4
  Cross-check using $s = \tfrac{1}{2}(u + v)t = \tfrac{1}{2}(0 + 20)(20) = 200\,\text{m}$ . Matches.
Answer
(a) Acceleration $a = 1\,\text{m s}^{-2}$ . (b) Time $t = 20\,\text{s}$ .
Examiner tip
M1 for selecting the correct suvat equation. A1 for $a = 1$ . M1 for substituting into $v = u + at$ . A1 for $t = 20$ . B1 for cross-checking. Examiner reports note that candidates who don't list known quantities at the top of the page are more likely to mix up $u$ and $v$ .
2Vertical motion under gravity — ball thrown up (8 marks, M1)
Extended• Adapted from WME01/01 June 2024 Q3• suvat, vertical, gravity
Question
A ball is thrown vertically upward from a point $1.5\,\text{m}$ above the ground with an initial speed of $14\,\text{m s}^{-1}$ . Taking $g = 9.8\,\text{m s}^{-2}$ , find (a) the greatest height above the ground, (b) the total time before the ball hits the ground.
Step-by-step solution
1. Step 1
  Draw a vertical motion diagram. Set the launch point as origin, upward positive. Then $a = -9.8\,\text{m s}^{-2}$ . State the convention clearly.
2. Step 2
  (a) Greatest height above launch: at the top, $v = 0$ . Use $v^{2} = u^{2} + 2as$ .
  $0 = 14^{2} + 2(-9.8) s \Rightarrow s = \dfrac{196}{19.6} = 10\,\text{m}$
3. Step 3
  Greatest height above the ground = $10 + 1.5 = 11.5\,\text{m}$ .
4. Step 4
  (b) Ball hits ground when displacement from launch $s = -1.5\,\text{m}$ . Use $s = ut + \tfrac{1}{2}at^{2}$ .
  $-1.5 = 14t - 4.9 t^{2} \Rightarrow 4.9 t^{2} - 14 t - 1.5 = 0$
5. Step 5
  Apply the quadratic formula.
  $t = \dfrac{14 \pm \sqrt{196 + 4 \cdot 4.9 \cdot 1.5}}{9.8} = \dfrac{14 \pm \sqrt{225.4}}{9.8}$
6. Step 6
  $\sqrt{225.4} \approx 15.01$ . The positive root: $t = (14 + 15.01)/9.8 \approx 2.96\,\text{s}$ .
Answer
(a) Greatest height $= 11.5\,\text{m}$ above the ground. (b) Time to hit ground $\approx 2.96\,\text{s}$ .
Examiner tip
B1 for clear sign convention (upward positive). M1 for $v^{2} = u^{2} + 2as$ with $v = 0$ at peak. A1 for $s = 10$ above launch. B1 for adding $1.5\,\text{m}$ . M1 for forming a quadratic with $s = -1.5$ . A1 for the quadratic. M1A1 for solving. ECF applies — most candidates lose marks by treating the ball as starting at ground level and forgetting the extra $1.5\,\text{m}$ .
3Train: accelerate, cruise, decelerate (10 marks, M1)
Extended• Adapted from WME01/01 January 2024 Q5• suvat, multi-stage, velocity-time graph
Question
A train accelerates uniformly from rest at $0.4\,\text{m s}^{-2}$ for $50\,\text{s}$ , then travels at constant speed for $200\,\text{s}$ , then decelerates uniformly to rest in a further $80\,\text{s}$ . (a) Sketch the velocity-time graph. (b) Find the total distance travelled.
Step-by-step solution
1. Step 1
  Stage 1: acceleration from rest. Final speed $V = u + at = 0 + 0.4(50) = 20\,\text{m s}^{-1}$ .
2. Step 2
  Stage 2: constant speed $20\,\text{m s}^{-1}$ for $200\,\text{s}$ .
3. Step 3
  Stage 3: decelerate from $20$ to $0$ in $80\,\text{s}$ . Deceleration $= 20/80 = 0.25\,\text{m s}^{-2}$ .
4. Step 4
  (a) Sketch: $v$ rises linearly from $(0,0)$ to $(50, 20)$ , flat to $(250, 20)$ , then linearly down to $(330, 0)$ . Trapezium shape.
5. Step 5
  (b) Total distance = area under graph (trapezium). The two parallel sides are the top edge ( $200\,\text{s}$ ) and bottom edge ( $330\,\text{s}$ ); height is $20\,\text{m s}^{-1}$ .
  $s = \tfrac{1}{2}(200 + 330)(20) = \tfrac{1}{2}(530)(20) = 5300\,\text{m}$
6. Step 6
  Verify by stage: Stage 1: $s_1 = \tfrac{1}{2}(0+20)(50) = 500$ . Stage 2: $s_2 = 20 \cdot 200 = 4000$ . Stage 3: $s_3 = \tfrac{1}{2}(20+0)(80) = 800$ . Total = $500 + 4000 + 800 = 5300$ . Matches.
Answer
(a) Trapezium graph as described. (b) Total distance $= 5300\,\text{m}$ .
Examiner tip
B1 for stage 1 endpoint. B1 for stage 2 (horizontal segment). B1 for stage 3 endpoint. B1 for axes labelled with units. M1 for area-of-trapezium method OR sum of three areas. A1 for $5300\,\text{m}$ . The trapezium shortcut $s = \tfrac{1}{2}(\text{top} + \text{bottom})(\text{height})$ is the marker-preferred route and saves a minute.
4Two particles starting at different times (10 marks, M1)
Challenge• Adapted from WME01/01 June 2023 Q5• suvat, relative motion, simultaneous equations
Question
A car $A$ starts from rest at point $O$ and accelerates at $2\,\text{m s}^{-2}$ along a straight road. Five seconds later, a car $B$ passes $O$ travelling in the same direction at $25\,\text{m s}^{-1}$ with constant velocity. (a) Find the time after $B$ passes $O$ when $A$ and $B$ are level. (b) Find the distance from $O$ at which this occurs.
Step-by-step solution
1. Step 1
  Let $t$ = time since $B$ passes $O$ . Then $A$ has been moving for $(t + 5)$ seconds.
2. Step 2
  Position of $A$ at time $t$ (measured from $O$ ):
  $s_A = \tfrac{1}{2}(2)(t+5)^{2} = (t+5)^{2}$
3. Step 3
  Position of $B$ at time $t$ (measured from $O$ ):
  $s_B = 25 t$
4. Step 4
  Set $s_A = s_B$ :
  $(t+5)^{2} = 25 t \Rightarrow t^{2} + 10 t + 25 = 25 t \Rightarrow t^{2} - 15 t + 25 = 0$
5. Step 5
  Quadratic formula:
  $t = \dfrac{15 \pm \sqrt{225 - 100}}{2} = \dfrac{15 \pm \sqrt{125}}{2} = \dfrac{15 \pm 5\sqrt{5}}{2}$
6. Step 6
  Numerically: $t \approx (15 + 11.18)/2 \approx 13.09$ or $t \approx (15 - 11.18)/2 \approx 1.91$ . Both physical — at $t \approx 1.91$ s, $B$ overtakes $A$ ; at $t \approx 13.09$ s, $A$ overtakes $B$ (since $A$ is still accelerating).
7. Step 7
  Distance from $O$ at $t = 1.91$ : $s_B = 25 \cdot 1.91 \approx 47.8\,\text{m}$ . At $t = 13.09$ : $s_B = 25 \cdot 13.09 \approx 327\,\text{m}$ .
Answer
(a) $t = \tfrac{15 - 5\sqrt{5}}{2} \approx 1.91$ s (first meeting) and $t = \tfrac{15 + 5\sqrt{5}}{2} \approx 13.09$ s (second meeting). (b) Distances $\approx 47.8\,\text{m}$ and $\approx 327\,\text{m}$ .
Examiner tip
M1 for setting up $s_A$ with $(t+5)^{2}$ — many candidates forget the time offset. M1 for $s_B = 25t$ . M1 for $s_A = s_B$ . A1 for the quadratic. M1A1 for solving (both roots). B1 for distances. Some questions ask only for the FIRST meeting — check carefully. Reading 'when level' physically often means just the first meeting.
5Reading information from a displacement-time graph (6 marks, M1)
Extended• Adapted from WME01/01 January 2023 Q3• suvat, displacement-time graph, interpretation
Question
A particle moves along a straight line. Its displacement-time graph from $t = 0$ to $t = 12$ is straight from $(0, 0)$ to $(4, 16)$ , horizontal from $(4, 16)$ to $(7, 16)$ , and straight from $(7, 16)$ to $(12, 0)$ . (a) Describe the motion in each phase. (b) Find the average speed and average velocity over $0 \le t \le 12$ .
Step-by-step solution
1. Step 1
  On a displacement-time graph, the gradient equals velocity. Read each segment.
2. Step 2
  Phase 1 ( $0 \le t \le 4$ ): straight line from $(0,0)$ to $(4,16)$ . Gradient $= 16/4 = 4\,\text{m s}^{-1}$ , so the particle moves with constant velocity $4\,\text{m s}^{-1}$ in the positive direction.
3. Step 3
  Phase 2 ( $4 \le t \le 7$ ): horizontal segment, gradient $= 0$ . The particle is stationary at displacement $16\,\text{m}$ .
4. Step 4
  Phase 3 ( $7 \le t \le 12$ ): straight line from $(7,16)$ to $(12,0)$ . Gradient $= (0 - 16)/(12 - 7) = -3.2\,\text{m s}^{-1}$ . Constant velocity $-3.2\,\text{m s}^{-1}$ (i.e. $3.2$ m/s back toward the origin).
5. Step 5
  (b) Total distance = $16 + 0 + 16 = 32\,\text{m}$ . Total time $= 12\,\text{s}$ . Average speed = $32/12 \approx 2.67\,\text{m s}^{-1}$ .
6. Step 6
  Total displacement = final position minus initial = $0 - 0 = 0\,\text{m}$ . Average velocity = $0/12 = 0\,\text{m s}^{-1}$ .
Answer
(a) Phase 1: constant velocity $4\,\text{m s}^{-1}$ . Phase 2: stationary at $16\,\text{m}$ . Phase 3: constant velocity $-3.2\,\text{m s}^{-1}$ . (b) Average speed $\approx 2.67\,\text{m s}^{-1}$ ; average velocity $= 0$ .
Examiner tip
B1 each for the three phases described correctly (velocity with direction). M1 for average speed (distance/time). A1. B1 for average velocity $= 0$ (since the particle returns to start). Mark schemes are strict here: 'average speed' uses distance, 'average velocity' uses displacement — confusing them costs both A-marks.

Key Formulae — Constant Acceleration

The formulae you need to memorise for constant acceleration on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

$v = u + at$
$v = u + at$
$u$
initial velocity (m s $^{-1}$ )
$v$
final velocity (m s $^{-1}$ )
$a$
constant acceleration (m s $^{-2}$ )
$t$
time (s)
When to use
Use when displacement is not needed. Connects three of the five suvat quantities. In the IAL formula booklet under 'Kinematics'.
Example
Car decelerates from $20$ to rest at $0.25\,\text{m s}^{-2}$ : $0 = 20 + (-0.25)t \Rightarrow t = 80$ s.
$s = \tfrac{1}{2}(u+v)t$
$s = \tfrac{1}{2}(u + v)t$
$s$
displacement (m)
$u, v$
initial and final velocities
$t$
time
When to use
Use when acceleration $a$ is not known but you have $u$ , $v$ , $t$ . Equivalent to the area of a trapezium on a velocity-time graph. In the formula booklet.
Example
Train decelerates from $20$ to $0$ in $80$ s: $s = \tfrac{1}{2}(20+0)(80) = 800$ m.
$s = ut + \tfrac{1}{2}at^{2}$
$s = ut + \tfrac{1}{2}a t^{2}$
$s$
displacement (m)
$u$
initial velocity
$a$
acceleration
$t$
time
When to use
Use when final velocity $v$ is not known. Most common for vertical motion (set $u$ , find $s$ at given $t$ , or solve for $t$ given $s$ ). In the formula booklet.
Example
Ball thrown up at $u = 14$ , $a = -9.8$ , find $s$ at $t = 1$ s: $s = 14(1) - 4.9 = 9.1\,\text{m}$ .
$v^{2} = u^{2} + 2as$
$v^{2} = u^{2} + 2as$
$u, v$
initial and final velocities
$a$
acceleration
$s$
displacement
When to use
Use when time $t$ is not needed. Especially useful for greatest-height problems (set $v = 0$ ). In the formula booklet.
Example
Ball thrown up at $14$ , find max height: $0 = 196 - 19.6 s \Rightarrow s = 10\,\text{m}$ .
$s = vt - \tfrac{1}{2}at^{2}$
$s = vt - \tfrac{1}{2}a t^{2}$
$s$
displacement (m)
$v$
final velocity
$a$
acceleration
$t$
time
When to use
Less common; use when $u$ is unknown but $v$ is. Can save algebra in symmetric problems. In the formula booklet.
Example
Particle finishes at $v = 10$ after $4$ s with $a = 2$ : $s = 10(4) - \tfrac{1}{2}(2)(16) = 40 - 16 = 24$ m.
Acceleration due to gravity (Edexcel convention)
$g = 9.8\,\text{m s}^{-2}$
$g$
magnitude of gravitational acceleration, directed vertically downward
When to use
Edexcel IAL uses $g = 9.8$ throughout unless a question states otherwise. (CIE often uses $g = 10$ ; do not confuse the two.) NOT in the formula booklet — must be remembered.
Example
Weight of a $5$ kg mass: $W = mg = 5 \times 9.8 = 49\,\text{N}$ .

Key Definitions and Keywords — Constant Acceleration

Definitions to memorise and the exact keywords mark schemes credit for constant acceleration answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Acceleration
Examiner keyword
The rate of change of velocity with respect to time. SI unit: m s $^{-2}$ . A vector quantity; on a velocity-time graph, the gradient equals the acceleration.
Deceleration / retardation
Examiner keyword
Acceleration in the direction opposite to the velocity. A particle moving in the positive direction with deceleration $0.25\,\text{m s}^{-2}$ has $a = -0.25$ in the chosen sign convention.
Distance vs displacement
Examiner keyword
Distance is a scalar — the total length of the path travelled. Displacement is a vector — the straight-line position change from start to end. Equal only when motion is one-way along a line; differ when the particle reverses direction.
Speed vs velocity
Examiner keyword
Speed = magnitude of velocity (scalar). Velocity has both magnitude and direction. Average speed = total distance / total time; average velocity = total displacement / total time.
Velocity-time graph
Examiner keyword
A graph of $v$ against $t$ . Gradient = acceleration. Area under the graph (between the curve and the $t$ -axis) = displacement (signed). For multi-stage motion, the graph is piecewise linear.
Displacement-time graph
A graph of displacement $s$ against time $t$ . Gradient = velocity. A horizontal segment means stationary; a steep gradient means high speed; a negative gradient means moving back toward the origin.

Common Mistakes and Misconceptions — Constant Acceleration

The traps other students keep falling into on constant acceleration questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Using $a = 9.8$ for vertical motion without declaring direction
WME01/01 examiner reports
Why it happens
$g$ is written as a positive number, but the direction is downward.
How to avoid it
State 'upward positive, so $a = -9.8\,\text{m s}^{-2}$ ' OR 'downward positive, so $a = +9.8\,\text{m s}^{-2}$ '. Mark schemes credit any consistent convention.
✕Computing 'average velocity' using total distance
Why it happens
Both are 'rates'; students conflate them.
How to avoid it
Average velocity uses displacement (final $-$ initial position). Average speed uses distance (total path length). For motion that returns to start, average velocity is zero but average speed is not.
✕Forgetting the time offset when two particles start at different moments
WME01/01 examiner reports — flagged annually
Why it happens
Students use the same $t$ for both particles.
How to avoid it
Pick one reference clock; the particle that started $\Delta t$ earlier has time $(t + \Delta t)$ in its suvat equations.
✕Picking a suvat equation that requires a quantity you don't have
Why it happens
Memorising the equations without thinking about which variable is missing.
How to avoid it
List the five quantities: $u$ , $v$ , $a$ , $s$ , $t$ . Identify which is missing and which is asked. Choose the equation that uses the four available. For greatest height, $t$ is missing — use $v^{2} = u^{2} + 2as$ .
✕Computing displacement as the area under a displacement-time graph
Why it happens
Confusion with velocity-time graphs.
How to avoid it
Memorise: on a $v$ - $t$ graph, area = displacement; on an $s$ - $t$ graph, the curve itself is the displacement (no area calculation needed).

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Constant Acceleration — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Constant Acceleration

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Single-stage suvat — finding final velocity and time (6 marks, M1)

2Vertical motion under gravity — ball thrown up (8 marks, M1)

3Train: accelerate, cruise, decelerate (10 marks, M1)

4Two particles starting at different times (10 marks, M1)

5Reading information from a displacement-time graph (6 marks, M1)

v=u+atv = u + atv=u+at

s=12(u+v)ts = \tfrac{1}{2}(u+v)ts=21​(u+v)t

s=ut+12at2s = ut + \tfrac{1}{2}at^{2}s=ut+21​at2

v2=u2+2asv^{2} = u^{2} + 2asv2=u2+2as

s=vt−12at2s = vt - \tfrac{1}{2}at^{2}s=vt−21​at2

Acceleration due to gravity (Edexcel convention)

Acceleration

Deceleration / retardation

Distance vs displacement

Speed vs velocity

Velocity-time graph

Displacement-time graph

✕Using a=9.8a = 9.8a=9.8 for vertical motion without declaring direction

✕Computing 'average velocity' using total distance

✕Forgetting the time offset when two particles start at different moments

✕Picking a suvat equation that requires a quantity you don't have

✕Computing displacement as the area under a displacement-time graph

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Constant Acceleration — frequently asked questions

$v = u + at$

$s = \tfrac{1}{2}(u+v)t$

$s = ut + \tfrac{1}{2}at^{2}$

$v^{2} = u^{2} + 2as$

$s = vt - \tfrac{1}{2}at^{2}$

✕Using $a = 9.8$ for vertical motion without declaring direction