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Displacement from Initial Velocity and Acceleration

Evaluate the displacement equation using an initial velocity, constant acceleration and duration.

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Displacement s (m)
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Final velocity v (m/s)
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    The calculator uses the following formula or method.

    s = ut + ½at²

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What this tool answers

Use this page when initial velocity and constant acceleration are known. The acceleration contribution is quadratic in time, unlike the linear contribution from the initial velocity.

Formula

s = u × t + ½ × a × t².

Worked example: Six seconds of uniform acceleration

For u = 4 m/s, a = 1.5 m/s² and t = 6 s, displacement is 4 × 6 + 0.5 × 1.5 × 36 = 51 m. The two contributions are 24 m and 27 m.

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Limitations

Acceleration is assumed constant. Use compatible units and one signed direction convention. This equation is not a trajectory reconstruction for arbitrary forces.

Frequently asked questions

Why does the acceleration contribution contain time squared?

Integrating constant acceleration changes velocity linearly and displacement quadratically with time.

What remains when acceleration is zero in this equation?

The expression reduces to initial velocity multiplied by elapsed time.

Can this constant-acceleration expression represent any force history?

No. A varying acceleration needs an appropriate time-dependent model.

Sources
  • No external reference is listed for this calculator. Its formula and variable definitions are shown above.
Limitations
  • The model uses the stated units and idealized assumptions. Check applicability, tolerances, and safety requirements for real equipment.
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