Kinematics: v²
Velocity–displacement relation under constant acceleration.
Formula
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Example
From rest, a = 2 m/s² over 9 m: v² = 0 + 2·2·9 = 36, so v = 6 m/s.
How it works
This equation links speed, starting speed, acceleration, and distance when an object moves with constant acceleration. It comes from combining the ideas that acceleration changes velocity steadily over time and that displacement is the area under a velocity-time graph, so time can be eliminated from the motion equations. It is especially useful because it lets you find the final velocity without needing to know how long the motion took.
When to use it
Use this formula for straight-line motion with constant acceleration, such as a car speeding up, a ball falling freely, or any object slowing down at a steady rate. It is ideal when you know the initial speed, acceleration, and distance moved, but not the time.
Worked example, step by step
A cyclist travels 50 m with an initial speed of 4 m/s and a constant acceleration of 0.6 m/s². Use v² = u² + 2as v² = 4² + 2(0.6)(50) v² = 16 + 60 v² = 76 v = √76 v ≈ 8.72 m/s Final answer: the cyclist's speed after 50 m is about 8.7 m/s.
Common mistakes to avoid
- Using the formula when acceleration is not constant
- Forgetting that s is displacement, not total path length
- Squaring the final answer incorrectly or forgetting to take the square root at the end
Did You Know?
This equation lets you find final speed without ever knowing the time — handy for braking-distance problems.
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