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    Home » SUVAT Equations Worksheet with Answers: GCSE and A-Level Practice (PDF)
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    SUVAT Equations Worksheet with Answers: GCSE and A-Level Practice (PDF)

    Maurice CotterellBy Maurice CotterellJune 27, 2026Updated:June 28, 2026No Comments7 Mins Read
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    SUVAT equations worksheet with answers
    SUVAT equations worksheet with answers
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    SUVAT Equations Worksheet with Answers (PDF): GCSE and A-Level Practice

    This page contains a free SUVAT equations worksheet with 10 fully worked questions, 5 at GCSE level and 5 at A-Level, plus a downloadable PDF. Every question includes a step-by-step worked answer. The worksheet covers all 5 SUVAT equations and the most common exam question types.

    Save the SUVAT equations worksheet PDF

    What Are SUVAT Equations?

    SUVAT equations are 5 formulas that describe motion under constant acceleration. The name SUVAT comes from the 5 variables: s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). Each equation uses 4 of the 5 variables, which allows a solver to find any unknown when 3 values are known.

    What Are the 5 SUVAT Equations?

    For exam-board context, AQA’s official 2026 formulae and equation sheet update explains how equation-sheet arrangements apply for GCSE Mathematics, Physics, and Combined Science.

    The 5 SUVAT equations are the standard equations of motion used in GCSE Physics, GCSE Maths, and A-Level Mechanics.

    EquationFormulaVariable Not Included
    1v = u + ats
    2s = ut + ½at²v
    3v² = u² + 2ast
    4s = ½(u + v)ta
    5s = vt − ½at²u

    Select the equation that contains your 3 known variables and 1 unknown. The variable absent from the problem matches the "Variable Not Included" column.

    How Do GCSE and A-Level SUVAT Questions Differ?

    GCSE SUVAT questions use 1 equation and 1 unknown. A-Level questions use 2 or more equations, include vertical motion under gravity, or require forming a quadratic.

    LevelTypical StructureCommon Topics
    GCSE1 equation, 1 unknownStopping distance, free fall, acceleration
    A-LevelMulti-step, quadratic, vectorsProjectiles, relative motion, nth-second formula

    GCSE SUVAT Practice Questions

    The 5 GCSE questions below test each of the 5 SUVAT equations in a direct, 1-step format.

    1. A car starts from rest and accelerates at 4 m/s² for 6 seconds. Find the final velocity.
    2. A ball rolls with an initial velocity of 8 m/s and decelerates at 2 m/s². Find the displacement after 3 seconds.
    3. A cyclist accelerates from 3 m/s to 9 m/s over a distance of 18 m. Find the acceleration.
    4. A train travels at 30 m/s and brakes with a deceleration of 5 m/s². Find the stopping distance.
    5. A stone is dropped from rest and falls for 4 seconds under gravity (g = 9.8 m/s²). Find the distance fallen.

    GCSE SUVAT Worked Answers

    Each GCSE answer below states the equation used, substitutes values, and gives the final answer with units.

    Question 1 Answer: Final Velocity

    Known: u = 0, a = 4 m/s², t = 6 s. Unknown: v. Use v = u + at.

    v = 0 + (4)(6) = 24 m/s

    Question 2 Answer: Displacement With Deceleration

    Known: u = 8 m/s, a = −2 m/s², t = 3 s. Unknown: s. Use s = ut + ½at².

    s = (8)(3) + ½(−2)(3²) = 24 − 9 = 15 m

    Question 3 Answer: Acceleration From Velocities and Distance

    Known: u = 3 m/s, v = 9 m/s, s = 18 m. Unknown: a. Use v² = u² + 2as.

    81 = 9 + 2a(18) → 72 = 36a → a = 2 m/s²

    Question 4 Answer: Stopping Distance

    Known: u = 30 m/s, v = 0, a = −5 m/s². Unknown: s. Use v² = u² + 2as.

    0 = 900 + 2(−5)s → 10s = 900 → s = 90 m

    Question 5 Answer: Free Fall Distance

    Known: u = 0, a = 9.8 m/s², t = 4 s. Unknown: s. Use s = ut + ½at².

    s = 0 + ½(9.8)(16) = 78.4 m

    A-Level SUVAT Practice Questions

    The 5 A-Level questions below require multi-step working, sign conventions, or quadratic equations.

    1. A ball is thrown vertically upward at 15 m/s from a point 2 m above the ground. Find the maximum height above the ground. (g = 9.8 m/s²)
    2. A particle decelerates uniformly from 20 m/s to rest over 8 seconds. Find (a) the deceleration and (b) the total distance traveled.
    3. A stone is projected vertically upward from the top of a 45 m cliff with initial velocity 10 m/s. Find the time for the stone to reach the ground. (g = 9.8 m/s²)
    4. A particle moves with initial velocity 5 m/s under constant acceleration. It covers 18 m during the 3rd second of motion. Find the acceleration.
    5. Particle A moves from a fixed point at 6 m/s with zero acceleration. Particle B starts from the same point 4 seconds later, with initial velocity 2 m/s and acceleration 2 m/s². Find the time after A starts when B catches A.

    A-Level SUVAT Worked Answers

    Each A-Level answer states the sign convention, identifies the equation, shows full substitution, and solves completely.

    Question 1 Answer: Maximum Height

    At maximum height, v = 0. Taking upward as positive: u = 15 m/s, a = −9.8 m/s².

    Use v² = u² + 2as → 0 = 225 − 19.6s → s = 225 ÷ 19.6 = 11.48 m above launch point.

    Total height above ground = 2 + 11.48 = 13.5 m (3 s.f.)

    Question 2 Answer: Deceleration and Distance

    (a) Use v = u + at → 0 = 20 + 8a → a = −2.5 m/s²

    (b) Use s = ½(u + v)t → s = ½(20 + 0)(8) = 80 m

    Question 3 Answer: Time to Reach Ground From Cliff

    Take upward as positive. The stone ends 45 m below the launch point, so s = −45 m.

    Use s = ut + ½at²: −45 = 10t − 4.9t² → 4.9t² − 10t − 45 = 0.

    Quadratic formula: t = (10 ± √(100 + 882)) ÷ 9.8 = (10 ± 31.34) ÷ 9.8.

    Taking the positive root: t = 41.34 ÷ 9.8 = 4.22 s (3 s.f.)

    Question 4 Answer: Acceleration Using the Nth-Second Formula

    Distance in the nth second: s_n = u + a(n − ½). Substitute n = 3, s_3 = 18, u = 5.

    18 = 5 + a(2.5) → 13 = 2.5a → a = 5.2 m/s²

    Question 5 Answer: When B Catches A

    Let t = time in seconds after A departs. B departs at t = 4, so B travels for (t − 4) seconds.

    Distance A: s_A = 6t. Distance B: s_B = 2(t − 4) + ½(2)(t − 4)².

    Set s_A = s_B: 6t = 2(t − 4) + (t − 4)² → t² − 12t + 8 = 0.

    t = (12 ± √(144 − 32)) ÷ 2 = (12 ± 10.58) ÷ 2. Taking the valid root: t = 11.3 s (3 s.f.)

    What Are the Most Common SUVAT Mistakes at A-Level?

    The 3 most frequent errors are incorrect sign convention, misidentifying displacement as distance, and applying SUVAT when acceleration is not constant.

    1. Sign errors: Set a clear positive direction before substituting. Deceleration and gravity (when downward) are negative in the upward-positive convention.
    2. Displacement vs distance: A particle returning to its starting point has s = 0, not s = total path length. The 3rd and 4th SUVAT equations use displacement only.
    3. Non-constant acceleration: SUVAT does not apply to variable acceleration problems. Those require calculus, specifically integration of the acceleration function.

    What Should You Know Before Using This Worksheet?

    SUVAT equations apply to straight-line motion under constant acceleration. GCSE questions require 1 equation and 3 known values. A-Level questions add vertical motion under gravity, quadratic equations, and multi-particle scenarios. Assigning a sign convention before every problem eliminates the most common source of error across both levels.

    Maurice Cotterell
    MEd., BSc. QTS/OCT – Physics Teacher at Davenant Foundation School
    The University of the West Indies

    My passion as a teacher and leader in education is to contribute to students’ enhanced learning by creation of a positive teaching/learning environment, where they are motivated to develop strong critical thinking, analytical and problem-solving skills.
    As acting Vice-Principal and head of Physics, I have amassed five years of experience in leading meetings, implementing strategies to maintain discipline and school ethos, liaise with stakeholders in planning for future development of the institution; successfully negotiated the implementation of various projects for student development as well as supervising effective execution of such. For the past 19 years, I have incorporated discovery learning within the classroom through practical activities, problem solving and cooperative learning and have realized that students retention in Physics and Mathematics have increased significantly.

    I am always cognizant that students have different learning styles, and as such teaching strategies during planning and delivery must take this crucial factor into consideration and be tailored to meet such differences.

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    Maurice Cotterell
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    MEd., BSc. QTS/OCT - Physics Teacher at Davenant Foundation School The University of the West Indies
    My passion as a teacher and leader in education is to contribute to students’ enhanced learning by creation of a positive teaching/learning environment, where they are motivated to develop strong critical thinking, analytical and problem-solving skills. As acting Vice-Principal and head of Physics, I have amassed five years of experience in leading meetings, implementing strategies to maintain discipline and school ethos, liaise with stakeholders in planning for future development of the institution; successfully negotiated the implementation of various projects for student development as well as supervising effective execution of such. For the past 19 years, I have incorporated discovery learning within the classroom through practical activities, problem solving and cooperative learning and have realized that students retention in Physics and Mathematics have increased significantly. I am always cognizant that students have different learning styles, and as such teaching strategies during planning and delivery must take this crucial factor into consideration and be tailored to meet such differences.

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