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    v² − u² = 2as Calculator

    GCSE PhysicsAQAEdexcelOCRFoundation & Higher

    Find final velocity, initial velocity, acceleration or distance for uniform acceleration using v² − u² = 2as. Use a negative acceleration for braking.

    Quick answer

    For uniform acceleration, v² − u² = 2as. To find v, use v = √(u² + 2as). For example, from rest with a = 2.0 m/s² over 25 m: v² = 0 + 2 × 2.0 × 25 = 100, so v = 10 m/s. Slowing down gives a negative a.

    By Engr. Mehmood Kakar · Updated

    v² − u² = 2 × a × s
    Find:

    Starting from rest means u = 0.

    Use a negative value for a deceleration (slowing down). Free fall: a = 9.8 m/s².

    Simulation

    Speed after a distance

    Set the starting velocity, the acceleration (negative for braking) and the distance. The car travels the distance and the speedometer shows its final velocity.

    20 m/s
    -4 m/s²
    40 m

    With braking (negative a), a car moving at u stops in a distance s = u² ÷ (2 × deceleration). Double the speed and the braking distance is four times as long.

    Put these numbers in the calculator above

    How to use v² − u² = 2as

    For an object with uniform (constant) acceleration, final velocity, initial velocity, acceleration and distance are linked by:

    (final velocity)² − (initial velocity)² = 2 × acceleration × distance

    • v is final velocity and u is initial velocity, both in m/s
    • a is acceleration in m/s² (negative when slowing down)
    • s is distance in metres (m)

    A car that starts from rest (u = 0) and accelerates at 2.0 m/s² for 25 m reaches v² = 0² + 2 × 2.0 × 25 = 100, so v = √100 = 10 m/s. The square root at the end is the step most often forgotten.

    It is especially useful when there is no time in the question. Braking questions are common: a car slowing from 20 m/s to rest over 40 m has a = (0² − 20²) ÷ (2 × 40) = −5.0 m/s². The minus sign shows a deceleration.

    For a falling object with no air resistance, use a = 9.8 m/s². This equation is on your equation sheet.

    Common mistakes that cost marks

    1. Forgetting the square root. 2as + u² gives v², not v.
    2. Squaring after adding. Square u first, then add 2as: v² = u² + 2as.
    3. Getting the sign of a wrong. Slowing down means a is negative.
    4. Using it when acceleration isn’t constant. The equation only works for uniform acceleration.

    Practice questions

    Try each one on paper first, then reveal the worked answer.

    Q1A car starts from rest and accelerates at 2.0 m/s² over a distance of 25 m. Calculate its final velocity.[3 marks]
    1. v² = u² + 2as
    2. v² = 0² + 2 × 2.0 × 25 = 100
    3. v = √100
    4. v = 10 m/s
    Q2A cyclist brakes from 20 m/s to rest over 40 m. Calculate her acceleration.[3 marks]
    1. a = (v² − u²) ÷ 2s
    2. a = (0² − 20²) ÷ (2 × 40) = −400 ÷ 80
    3. a = −5.0 m/s² (a deceleration of 5.0 m/s²)
    Q3A ball is dropped from a height of 20 m. Ignore air resistance; a = 9.8 m/s². Calculate its speed just before it hits the ground.[3 marks]
    1. v² = u² + 2as
    2. v² = 0² + 2 × 9.8 × 20 = 392
    3. v = √392
    4. v = 19.8 m/s

    Endless practice

    A new question every time, marked instantly. Work it out on paper, then type your answer.

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    Frequently asked questions

    When do I use v² − u² = 2as?

    When an object has uniform acceleration and the question gives (or asks for) velocities, acceleration and distance, but not time.

    How do I rearrange v² − u² = 2as to find v?

    Add u² to both sides to get v² = u² + 2as, then take the square root: v = √(u² + 2as).

    What does a negative acceleration mean here?

    The object is slowing down (decelerating). Put the acceleration in as a negative number.

    Is v² − u² = 2as on the equation sheet?

    Yes. For AQA GCSE Physics and Combined Science it is printed on the equation sheet.