AQA · GCSE · Combined Science

    Course version: 8464 · Version from Sept 2018

    Acceleration: a = Δv / t

    This equation defines acceleration as the rate of change of velocity.

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

    • Point 3 of 21
    • 3 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    This equation defines acceleration as the rate of change of velocity. The symbol Δv means the change in velocity, found by subtracting the initial velocity from the final velocity: Δv = v − u. The symbol t is the time taken for that change, measured in seconds. Acceleration a is measured in metres per second squared, m/s². For example, a cyclist speeds up from 4 m/s to 10 m/s in 3 s, so Δv = 10 − 4 = 6 m/s and a = 6 ÷ 3 = 2 m/s². A negative value shows deceleration, meaning the object slows down. Always identify u, v and t from the question before substituting, and keep units consistent.

    What earns the marks

    • States that Δv is the change in velocity, calculated as final velocity minus initial velocity (Δv = v − u).
    • Uses the equation in the form a = Δv ÷ t, substituting the change in velocity and the time taken.
    • Keeps units consistent, using m/s for velocity and s for time, so that acceleration is in m/s².
    • Interprets a negative acceleration as deceleration, where the object is slowing down.
    • Rearranges the equation correctly when asked for Δv or t, for example Δv = a × t or t = Δv ÷ a.

    Common mistakes and how to put them right

    • Using the final velocity as Δv instead of subtracting the initial velocity.

      Put it right by calculating Δv = v − u first.

    • Mixing units, such as using time in minutes with velocity in m/s.

      Put it right by converting all values to seconds and m/s before substituting.

    • Treating a negative answer as an error.

      Put it right by recognising that a negative acceleration means the object is decelerating.

    Exam tips

    • Write down u, v and t from the question before substituting into the equation, so the change in velocity is clear.
    • Show the substitution step, for example a = (10 − 4) ÷ 3, because method marks are often available even if the final value is wrong.
    • Check the sign of your answer and add a short sentence explaining what it means, such as 'the object is decelerating'.

    Try the lesson

    The start of the Acceleration: a = Δv / t lesson, made from checked teaching notes for this specification point.

    Have a go first

    A cyclist speeds up from 4 m/s to 10 m/s in 3 s. Using a = Δv / t, what is the cyclist's acceleration?

    Listen

    This equation defines acceleration as the rate of change of velocity. The symbol Δv means the change in velocity, found by subtracting the initial velocity from the final velocity: Δv = v − u. The symbol t is the time taken for that change, measured in seconds. Acceleration a is measured in metres per second squared, m/s². For example, a cyclist speeds up from 4 m/s to 10 m/s in 3 s, so Δv = 10 − 4 = 6 m/s and a = 6 ÷ 3 = 2 m/s². A negative value shows deceleration, meaning the object slows down. Always identify u, v and t from the question before substituting, and keep units consistent. A strong answer earns marks for these points. States that Δv is the change in velocity, calculated as final velocity minus initial velocity (Δv = v − u). Uses the equation in the form a = Δv ÷ t, substituting the change in velocity and the time taken. Keeps units consistent, using m/s for velocity and s for time, so that acceleration is in m/s². Interprets a negative acceleration as deceleration, where the object is slowing down. A common slip: Using the final velocity as Δv instead of subtracting the initial velocity. Instead, calculating Δv = v − u first. Exam tip: Write down u, v and t from the question before substituting into the equation, so the change in velocity is clear.

    See it

    A flowchart running top to bottom. The first box says identify u, v and t from the question, and an arrow leads to a box saying calculate change in velocity: Δv = v − u. An arrow leads to a box saying substitute into a = Δv ÷ t, then to a box saying check units: m/s for velocity, s for time, then to a box saying acceleration in m/s². From there an arrow leads to a decision box asking is the value negative. If yes, an arrow leads to a box saying negative value shows deceleration: object slows down. If no, an arrow leads to a box saying positive value: object speeds up.
    The steps for calculating acceleration from initial velocity, final velocity and time, including interpreting a negative result as deceleration.

    Key words

    Say what each word means before you turn the card.

    Rate of change of velocity

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    Change in velocity Δv = v − u

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

    Deceleration as negative acceleration

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

    See every definition
    Rate of change of velocity
    Acceleration is defined as the rate of change of velocity.
    Change in velocity Δv = v − u
    Found by subtracting the initial velocity from the final velocity.
    Deceleration as negative acceleration
    A negative value shows deceleration, meaning the object slows down.

    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 flowchart running top to bottom. The first box says identify u, v and t from the question, and an arrow leads to a box saying calculate change in velocity: Δv = v − u. An arrow leads to a box saying substitute into a = Δv ÷ t, then to a box saying check units: m/s for velocity, s for time, then to a box saying acceleration in m/s². From there an arrow leads to a decision box asking is the value negative. If yes, an arrow leads to a box saying negative value shows deceleration: object slows down. If no, an arrow leads to a box saying positive value: object speeds up.

    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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