The Velocity Calculator (v² = u² + 2as) finds final velocity from initial velocity, acceleration and displacement, without needing to know the elapsed time at all. Enter u, a and s, and the tool returns the final velocity, taking care with the square root that this equation requires.
This is the SUVAT equation to reach for whenever a problem gives a distance traveled and an acceleration, but does not state how long the motion took, a common setup in braking-distance and free-fall-height problems.
Calculate velocity from acceleration and displacement
When time is unknown, final velocity still follows from initial velocity, acceleration and displacement, through a relation obtained by eliminating time from the linear velocity equation.
v² = u² + 2as
v = ±√(u² + 2as)Worked example: a vehicle brakes from 10 m/s at −2 m/s² over a distance of 25 m.
1. Square the initial velocity. 10² = 100.
2. Calculate the 2as term. 2 × (−2) × 25 = −100.
3. Add the two terms. 100 + (−100) = 0.
4. Take the square root. v = √0 = 0 m/s.
The vehicle comes to a complete stop exactly at that 25 m mark.
Choose the correct sign for the square root
Squaring a number always loses its original sign, so the equation v² = u² + 2as technically has two mathematical solutions, one positive and one negative. Physically, only one of them usually makes sense for a given problem, and choosing correctly depends on the direction convention already set for the problem.
If motion continues in the same direction as the initial velocity, take the positive root. If the object has reversed direction by the time it reaches the given displacement (which can happen with a ball thrown upward, for instance), the negative root may be the physically meaningful one instead. The result panel should be checked against the physical scenario, not just accepted at face value.
Recognize when no real velocity exists
If the quantity under the square root, u² + 2as, works out negative, there is no real number solution, meaning the object never actually reaches that displacement given the stated initial velocity and acceleration.
Worked example: a vehicle at 10 m/s decelerating at −2 m/s² is asked for its velocity at 30 m, beyond the 25 m stopping point found earlier.
u² + 2as = 100 + 2(−2)(30) = 100 − 120 = −20, a negative value, confirming the vehicle stops before ever reaching 30 m under this deceleration.
Understand where this equation comes from
Starting from v = u + at, solving for t gives t = (v − u)/a. Substituting that into s = ut + ½at² and simplifying algebraically eliminates t entirely, leaving v² = u² + 2as.
Because it derives from squaring a linear relationship, the equation naturally produces two mathematical roots even though most real problems only need one of them.
Apply the formula to a launch problem
A ball is launched upward at 20 m/s. Taking up as positive, gravity acts as a = −9.80665 m/s². Find its velocity after rising 15 m. 1. Square the initial velocity. 20² = 400. 2. Calculate the 2as term. 2 × (−9.80665) × 15 = −294.20.
3. Add the two terms. 400 − 294.20 = 105.80.
4. Take the square root. v = √105.80 ≈ 10.29 m/s.
Since the ball is still rising at 15 m (it has not yet reached its peak), the positive root, +10.29 m/s, is the physically correct choice here.
Work through a positive-acceleration example
A car starts at u = 5 m/s and accelerates at a = 3 m/s² over a displacement of s = 40 m. 1. Square the initial velocity. 5² = 25. 2. Calculate the 2as term. 2 × 3 × 40 = 240.
3. Add the two terms. 25 + 240 = 265.
4. Take the square root. v = √265 ≈ 16.28 m/s.
Since the car is accelerating rather than decelerating, only the positive root is physically meaningful here, matching the direction of the original motion.
Frequently asked questions
What is the formula for velocity without time?
v² = u² + 2as, then take the square root: v = ±√(u² + 2as). This works from initial velocity, acceleration and displacement alone.
Why does this equation have two possible answers?
Squaring a number removes its sign, so mathematically the equation admits both a positive and negative root. Physically, the correct sign depends on the direction of motion at the displacement in question.
What does it mean if the value under the square root is negative?
It means the object never reaches that displacement under the given initial velocity and acceleration; no real velocity exists for that combination of inputs.
How is this equation derived?
It comes from combining v = u + at and s = ut + ½at² to eliminate time algebraically, leaving a relationship between v, u, a and s directly.
Can this formula be used for free-fall height problems?
Yes. Enter the initial velocity, acceleration due to gravity, and displacement (height fallen or risen) to find velocity at that point without needing to know how much time has passed.
What if acceleration is zero?
Then the equation simplifies to v² = u², meaning v = ±u. With no acceleration, velocity magnitude never changes, so displacement alone doesn't add any new information about speed.
Is this the same equation used on the main Velocity Calculator?
Yes. This page focuses specifically on the no-time v² = u² + 2as form; the main Velocity Calculator also covers final velocity from time and average velocity in the same tool.
Summary
Final velocity without time follows v² = u² + 2as, taking the square root to solve for v, with a braking example (u = 10 m/s, a = −2 m/s², s = 25 m) giving exactly v = 0 m/s at the stopping point.
Because squaring removes sign information, the correct root, positive or negative, must be chosen based on the physical direction of motion. A negative value under the square root means the displacement in question is never reached under those conditions.