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Displacement from Velocity-Time

Get displacement and final velocity from kinematics

Result

42.00 m
8.00 m/s

How it is calculated

The tool uses two constant-acceleration formulas: s = u·t + ½·a·t² for displacement and v = u + a·t for final velocity. The variables are u, the initial velocity in m/s, a, the acceleration in m/s², t, the time in seconds, s, the displacement in metres, and v, the final velocity in m/s. Nothing is converted, so every input must already be in metres and seconds. The sign matters: choose one direction as positive, and a velocity or acceleration pointing the other way is negative. Time cannot be negative, and all three fields are required. Both results are rounded to two decimals.

Worked example

A car moving at 20 m/s brakes with a steady deceleration of 4 m/s², which is a = −4. We look at the state after 3 s. Displacement: 20 × 3 + ½ × (−4) × 3² = 60 − 18 = 42.00 m. Final velocity: 20 + (−4) × 3 = 8.00 m/s. After three seconds of braking the car has covered 42.00 m and is still moving forward at 8 m/s.

Try this example →

Common mistakes

  • Entering a deceleration as a positive number. Braking from 20 m/s needs a = −4; with a = 4 the tool computes a car that speeds up and returns 78.00 m.
  • Reading displacement as distance travelled. With u = 20, a = −4 and t = 8 the result is 32.00 m and −12.00 m/s: the object stopped after 5 s (50.00 m), turned back and ended up closer to the start.
  • Applying the formulas past the moment a real object would stop. A braking car does not reverse, so keep t no larger than −u/a, here 5 s.
  • Typing km/h or minutes. The formulas only work in m/s, m/s² and seconds, so convert first (72 km/h is 20 m/s).
  • Using the tool when acceleration changes during the motion. Both formulas assume a constant a for the whole time.

About this calculator

This calculator applies the two core constant-acceleration kinematics equations: displacement s = ut + ½at² and final velocity v = u + at, where u is initial velocity, a is constant acceleration, and t is elapsed time — the foundation of every projectile, free-fall, and straight-line motion problem in introductory physics.

Physics students use it to check homework and lab results, and it's the same math engineers use for quick estimates of braking distance, elevator motion, or how far a dropped object falls in a given time, whenever acceleration can reasonably be treated as constant.

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Comments (1)

  • Editorial team

    A handy check is free fall, treating down as positive and ignoring air resistance. Start from rest, use a = 9.81 and t = 2: the tool returns 19.62 m of fall and 19.62 m/s, and the matching numbers show that dropping something for two seconds covers roughly the height of a six-storey building. Change the sign convention and every sign flips with it, so decide which direction is positive before typing. What motion are you modelling?

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