Kinematics: Displacement
Displacement under constant acceleration.
Formula
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Example
From rest at 2 m/s² for 3 s: s = 0 + ½·2·3² = 9 m.
How it works
This formula gives the displacement of an object when it moves with constant acceleration: the first part, ut, is how far it would travel if it kept its starting velocity, and the second part, 1/2 at^2, adds the extra distance gained from speeding up or the reduction if slowing down. It comes from combining the ideas that velocity changes uniformly and that displacement is the area under a velocity-time graph. Because the acceleration is constant, that graph is a straight line, so the average velocity over the time interval can be found exactly.
When to use it
Use this in one-dimensional motion problems where acceleration is constant, such as cars speeding up, objects falling under gravity, or any motion where you know the initial velocity, acceleration, and time. It is especially useful when you need displacement rather than final velocity.
Worked example, step by step
A cyclist starts with an initial velocity of 4 m/s and accelerates at 1.5 m/s² for 6 s. Use s = ut + 1/2 at² s = (4 × 6) + 1/2 × 1.5 × 6² s = 24 + 0.75 × 36 s = 24 + 27 s = 51 m Final answer: the cyclist’s displacement is 51 m.
Common mistakes to avoid
- Using the formula when acceleration is not constant
- Forgetting to square the time in the 1/2at² term
- Mixing up displacement with distance or using the wrong sign for acceleration
Did You Know?
Galileo discovered that falling distance grows with the square of time by rolling balls down ramps.
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