AQA · GCSE · Combined Science

    Course version: 8464 · Version from Sept 2018

    Acceleration: v² − u² = 2as

    This equation links the four quantities of motion when time is not known.

    Checked against the AQA 8464 specification (opens in a new tab) on .

    • Point 16 of 21

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    This equation links the four quantities of motion when time is not known. The symbol v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s² and s is displacement in m. It is derived from combining a = (v − u)/t with s = ((u + v)/2)t, eliminating t. To use it, list the known values, identify the unknown, substitute into v² − u² = 2as and solve. For example, a car accelerating at 2 m/s² from rest over 25 m gives v² = 0² + 2 × 2 × 25 = 100, so v = 10 m/s. If the object decelerates, a is negative, and if it reverses direction, v may be negative. Always check units are m/s, m/s² and m before substituting.

    What earns the marks

    • State the equation as v² − u² = 2as and identify each symbol with its unit: v and u in m/s, a in m/s², s in m.
    • Substitute known values correctly, keeping signs for direction so deceleration uses a negative value.
    • Rearrange to isolate the squared unknown, for example v² = u² + 2as, before taking the square root.
    • Evaluate the arithmetic accurately, including squares and products, and give the final velocity with the correct unit and a sensible number of significant figures.
    • Interpret the physical result, such as recognising that a negative square root indicates motion in the opposite direction to the chosen positive direction.

    Common mistakes and how to put them right

    • Forgetting to take the square root after finding v², leaving an answer such as 100 m/s instead of 10 m/s.

      Put it right: Finish by square-rooting the isolated squared term.

    • Using a positive value for deceleration.

      Put it right: If the object slows down, substitute a negative acceleration so the equation reflects the change in velocity.

    • Mixing units, for example using distance in km with acceleration in m/s².

      Put it right: Convert all quantities to m, m/s and m/s² before substituting.

    Exam tips

    • Write the equation, then a substitution line, then the rearranged line, then the answer with unit; this makes method marks clear even if arithmetic slips.
    • Check whether the question gives time; if it does not, this equation is usually the intended route.
    • Keep the sign of a and v consistent with your chosen positive direction, and state that direction if the answer is negative.

    Try the lesson

    The start of the Acceleration: v² − u² = 2as lesson, made from checked teaching notes for this specification point.

    Have a go first

    A train travelling at 10 m/s accelerates steadily to 30 m/s over a distance of 400 m. Using v² − u² = 2as, what is the acceleration a of the train?

    Listen

    This equation links the four quantities of motion when time is not known. The symbol v is final velocity in m/s, u is initial velocity in m/s, a is acceleration in m/s² and s is displacement in m. It is derived from combining a = (v − u)/t with s = ((u + v)/2)t, eliminating t. To use it, list the known values, identify the unknown, substitute into v² − u² = 2as and solve. For example, a car accelerating at 2 m/s² from rest over 25 m gives v² = 0² + 2 × 2 × 25 = 100, so v = 10 m/s. If the object decelerates, a is negative, and if it reverses direction, v may be negative. Always check units are m/s, m/s² and m before substituting. A strong answer earns marks for these points. State the equation as v² − u² = 2as and identify each symbol with its unit: v and u in m/s, a in m/s², s in m. Substitute known values correctly, keeping signs for direction so deceleration uses a negative value. Rearrange to isolate the squared unknown, for example v² = u² + 2as, before taking the square root. Evaluate the arithmetic accurately, including squares and products, and give the final velocity with the correct unit and a sensible number of significant figures. A common slip: Forgetting to take the square root after finding v², leaving an answer such as 100 m/s instead of 10 m/s. Instead, finish by square-rooting the isolated squared term. Exam tip: Write the equation, then a substitution line, then the rearranged line, then the answer with unit; this makes method marks clear even if arithmetic slips.

    See it

    A top-down flowchart with seven boxes joined by arrows. The first box says list the known values and points to a box saying identify the unknown. That points to a box saying check units are m/s, m/s² and m. That points to a box saying substitute into v² − u² = 2as. That points to a box saying rearrange to isolate the squared unknown. That points to a box saying take the square root. The final box says give final velocity with unit.
    The sequence of steps for using v² − u² = 2as, from listing known values to giving the final velocity with its unit.

    Put it in order

    Put these steps in the order they happen.

    1. 1. Give final velocity with unit
    2. 2. Take the square root
    3. 3. Check units are m/s, m/s² and m
    4. 4. Rearrange to isolate the squared unknown
    5. 5. List the known values
    6. 6. Identify the unknown
    7. 7. Substitute into v² − u² = 2as

    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-down flowchart with seven boxes joined by arrows. The first box says list the known values and points to a box saying identify the unknown. That points to a box saying check units are m/s, m/s² and m. That points to a box saying substitute into v² − u² = 2as. That points to a box saying rearrange to isolate the squared unknown. That points to a box saying take the square root. The final box says give final velocity with unit.
    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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