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Final Velocity Calculator

Find final velocity and distance with kinematics.

Result

25.00 m/s
150.00 m

How it is calculated

Two equations for constant acceleration are applied: v = u + a × t for the final velocity and s = u × t + ½ × a × t² for the displacement. u is the initial velocity in m/s, a the acceleration in m/s² and t the time in seconds; the answers are in m/s and metres, to two decimals.

Velocity and acceleration can be negative, which means the opposite direction, so a braking vehicle uses a negative a. Time cannot be negative; a negative value gives no result. The formulas assume the acceleration stays constant for the whole time. The distance is displacement along the line of motion: if the object slows, stops and reverses during the interval, the figure is the net change in position, not the total path length travelled.

Worked example

An object starts at 5 m/s and accelerates at 2 m/s² for 10 s. Final velocity: 5 + 2 × 10 = 25.00 m/s. Distance: 5 × 10 + ½ × 2 × 10² = 50 + 100 = 150.00 m. For a ball thrown upward at 20 m/s with a = −9.81 for 3 s, the final velocity is −9.43 m/s: it is already coming back down.

Try this example →

Common mistakes

  • Entering km/h for velocity. The labels are m/s and m; divide km/h by 3.6 first.
  • Entering a deceleration as a positive number. Slowing down in the direction of motion needs a negative acceleration, otherwise the object speeds up.
  • Reading a negative final velocity as an error. It just means the object is moving in the opposite direction to the initial one.
  • Using the distance as total path when the object reverses. It gives net displacement, so a ball going up and down can show a small number.
  • Applying it when acceleration changes during the time, as with a vehicle changing gears. The equations need constant acceleration.

Sources

About this calculator

This calculator applies the first two equations of uniformly accelerated motion: v = u + a·t for final velocity, and s = u·t + ½·a·t² for distance travelled, where u is initial velocity, a is constant acceleration, and t is elapsed time. Both come directly from integrating a constant acceleration over time, and they're the backbone of classical kinematics whenever acceleration doesn't change during the motion.

Physics students use them to solve free-fall, braking-distance, and projectile-launch problems, while automotive and aerospace engineers use the same equations to estimate stopping distances, acceleration test results, and launch trajectories. Anywhere motion can be treated as constant acceleration over a known time interval — a car merging onto a highway, a rocket in its first seconds of ascent — this pair of equations gives the answer without needing calculus at solve time.

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Comments (1)

  • Editorial team

    For braking problems, keep the time at or below the stopping time, which is initial speed divided by the deceleration. The formulas do not know a car stops: start at 20 m/s with a = −5 and the true stop takes 4 s and 40 m, but entering 6 s returns −10 m/s and only 30 m, as if the car reversed. What scenario are you modelling, braking, launch or free fall?

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