AQA · GCSE · Physics

    Course version: 8463 · Version from Sept 2016

    Acceleration: v ² − u ² = 2 a s

    This equation links the change in the square of velocity to acceleration and distance for motion in a straight line with constant acceleration.

    Checked against the AQA 8463 specification (opens in a new tab) on .

    • Point 3 of 23
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    This equation links the change in the square of velocity to acceleration and distance for motion in a straight line with constant acceleration. The symbol v is final velocity in metres per second (m/s), u is initial velocity in m/s, a is acceleration in m/s² and s is distance in metres (m). The term v² − u² is the difference between the squares of the final and initial velocities, not the square of the difference. To use it, list the known quantities with units, rearrange for the unknown, substitute values and solve.

    For example: A car accelerating uniformly from rest (u = 0 m/s) over 50 m with a = 2 m/s² has v² = 0² + 2 × 2 × 50 = 200 m²/s², so v = √200 ≈ 14 m/s.

    What earns the marks

    • States that v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s² and s is distance in m.
    • Recognises that the equation applies to uniform acceleration in a straight line.
    • Calculates v² − u² correctly by squaring each velocity before subtracting, rather than squaring the difference between the velocities.
    • Rearranges the equation correctly for the required quantity, for example v = √(u² + 2 a s) or s = (v² − u²) ÷ (2 a).
    • Substitutes values with consistent units and evaluates the square root when finding v, giving the final velocity with the correct unit and a sensible magnitude.
    • Interprets a negative value of v² − u² as deceleration, indicating that the final speed is less than the initial speed.

    Common mistakes and how to put them right

    • Squaring the difference between the velocities instead of subtracting the squares.

      Put it right by squaring u and v separately before subtracting.

    • Forgetting to take the square root after finding v².

      Put it right by applying the square root to the value of v² to obtain v.

    • Confusing the symbol 's' with speed.

      Put it right by remembering that 's' stands for distance in this kinematic equation.

    Exam tips

    • Write down u, v, a and s with their units before substituting, so it is clear which quantity is unknown.
    • Show the rearrangement step explicitly, especially when the unknown is v or s, to make the method clear.
    • Check that the final answer has the correct unit and a realistic size for the context described.

    Try the lesson

    The start of the Acceleration: v ² − u ² = 2 a s lesson, made from checked teaching notes for this specification point.

    Have a go first

    A ball is dropped from rest and falls with an acceleration of 10 m/s² over a distance of 5 m. Using v ² − u ² = 2 a s, what is the final velocity v of the ball?

    Listen

    This equation links the change in the square of velocity to acceleration and distance for motion in a straight line with constant acceleration. The symbol v is final velocity in metres per second (m/s), u is initial velocity in m/s, a is acceleration in m/s² and s is distance in metres (m). The term v² − u² is the difference between the squares of the final and initial velocities, not the square of the difference. To use it, list the known quantities with units, rearrange for the unknown, substitute values and solve. For example, a car accelerating uniformly from rest (u = 0 m/s) over 50 m with a = 2 m/s² has v² = 0² + 2 × 2 × 50 = 200 m²/s², so v = √200 ≈ 14 m/s. A strong answer earns marks for these points. States that v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s² and s is distance in m. Recognises that the equation applies to uniform acceleration in a straight line. Calculates v² − u² correctly by squaring each velocity before subtracting, rather than squaring the difference between the velocities. Rearranges the equation correctly for the required quantity, for example v = √(u² + 2 a s) or s = (v² − u²) ÷ (2 a). A common slip: Squaring the difference between the velocities instead of subtracting the squares. Instead, squaring u and v separately before subtracting. Exam tip: Write down u, v, a and s with their units before substituting, so it is clear which quantity is unknown.

    See it

    A top-to-bottom flowchart with six steps. The first box says list known quantities with units, and an arrow leads to a box saying rearrange for the unknown. An arrow leads to a box saying substitute values, then to a box saying solve. From solve, an arrow leads to a decision box asking whether the unknown is v. If yes, an arrow leads to a box saying take the square root to find v. If no, an arrow leads to a box saying final answer with correct unit. The square-root box also leads to that final box.
    The steps for using v² − u² = 2 a s, including taking the square root when the unknown is v.

    Key words

    Say what each word means before you turn the card.

    Uniform acceleration in a straight line

    Try answering aloud or in your head before turning the card.

    Relationship between velocity, acceleration and distance

    Try answering aloud or in your head before turning the card.

    Rearranging and substituting into a kinematic equation

    Try answering aloud or in your head before turning the card.

    Using standard units and interpreting signs

    Try answering aloud or in your head before turning the card.

    See every definition
    Uniform acceleration in a straight line
    The motion to which the equation linking the change in the square of velocity to acceleration and distance applies.
    Relationship between velocity, acceleration and distance
    The change in the square of velocity is linked to acceleration and distance by v ² − u ² = 2 a s.
    Rearranging and substituting into a kinematic equation
    List the known quantities with units, rearrange for the unknown, substitute values and solve.
    Using standard units and interpreting signs
    Velocity is in metres per second, acceleration in m/s² and distance in metres, and a negative value of v² − u² indicates deceleration.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A top-to-bottom flowchart with six steps. The first box says list known quantities with units, and an arrow leads to a box saying rearrange for the unknown. An arrow leads to a box saying substitute values, then to a box saying solve. From solve, an arrow leads to a decision box asking whether the unknown is v. If yes, an arrow leads to a box saying take the square root to find v. If no, an arrow leads to a box saying final answer with correct unit. The square-root box also leads to that final box.

    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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