The Velocity Calculator (u, a, t) finds final velocity from initial velocity, acceleration and elapsed time under constant acceleration. Enter u, a and t, and the tool returns the final velocity with the substituted arithmetic shown, ready for a car pulling away, a ball speeding up, or a vehicle braking to a stop.
This is the most direct of the SUVAT velocity relations, a single linear equation with no squaring or averaging involved, which makes it the natural starting point for most constant-acceleration problems.
Calculate final velocity from acceleration and time
Final velocity under constant acceleration follows from initial velocity plus the product of acceleration and time.
v = u + atWorked example: a cart starts at rest (u = 0) and accelerates at 2 m/s² for 3 seconds.
1. Multiply acceleration by time. 2 × 3 = 6.
2. Add initial velocity. 0 + 6 = 6 m/s.
A second worked example: a car pulls away at u = 5 m/s and accelerates at 2 m/s² for 3 seconds.
1. Multiply acceleration by time. 2 × 3 = 6.
2. Add initial velocity. 5 + 6 = 11 m/s.
Rearrange for the other variables
The same linear equation solves for any of the four quantities when the other three are known.
u = v − at
a = (v − u)/t
t = (v − u)/aWorked check: solve for acceleration given u = 5 m/s, v = 11 m/s, t = 3 s.
a = (11 − 5) / 3 = 6 / 3 = 2 m/s², matching the original example.
Time must not be zero when solving for acceleration, and acceleration must not be zero when solving for time (unless v already equals u, since no time would be needed for no change).
Apply the formula to braking
Deceleration fits the same equation, simply with a negative acceleration value when motion is treated as positive. Worked example: a car at 20 m/s brakes at −4 m/s² for 3 seconds. 1. Multiply acceleration by time. −4 × 3 = −12.
2. Add initial velocity. 20 + (−12) = 8 m/s.
The car has slowed to 8 m/s after 3 seconds of braking at that constant rate. If braking continued long enough for v to reach zero, the equation would show the car had stopped; continuing to apply the same negative acceleration beyond that point would (incorrectly, for a real car) predict a negative velocity, which signals the equation no longer describes the physical situation once the car has actually stopped.
Understand where this equation comes from
The v = u + at relation follows directly from the definition of constant acceleration: acceleration is the rate of change of velocity, a = Δv/t = (v − u)/t. Multiplying both sides by t and rearranging gives v = u + at directly.
Because it is a first-degree (linear) equation in each variable, it never produces two possible answers the way the squared displacement or velocity equations sometimes do.
Keep sign conventions consistent
Choose a positive direction before substituting any numbers, and keep every quantity, u, v and a, consistent with that same choice throughout the problem. A ball thrown upward with up as positive will show gravity as a negative acceleration throughout its flight, even as its velocity itself changes from positive (rising) to negative (falling).
Work through a longer time interval
A rocket sled starts at u = 10 m/s and accelerates at a = 15 m/s² for t = 6 seconds. 1. Multiply acceleration by time. 15 × 6 = 90. 2. Add initial velocity. 10 + 90 = 100 m/s.
A larger acceleration and a longer time interval compound directly in this linear equation, since the at term simply scales with both factors together, unlike the squared time term that appears in the displacement equation for the same motion.
Compare against the other SUVAT relations
The v = u + at relation is one of four standard SUVAT equations, alongside s = ut + ½at², s = ½(u + v)t, and v² = u² + 2as.
Each one links a different set of three known quantities to the fourth unknown, so choosing the right one comes down to identifying which three of the five SUVAT variables, s, u, v, a and t, a given problem actually provides.
Frequently asked questions
What is the formula for final velocity with acceleration and time?
v = u + at. For u = 5 m/s, a = 2 m/s² and t = 3 s, v = 5 + 6 = 11 m/s.
How do you find acceleration if you know u, v and t?
Rearrange the formula: a = (v − u)/t. For u = 5 m/s, v = 11 m/s and t = 3 s, a = 6/3 = 2 m/s².
How do you find time if you know u, v and a?
Rearrange the formula: t = (v − u)/a. This requires acceleration to be nonzero unless v equals u.
Does this formula work for deceleration?
Yes. Enter a negative acceleration value (relative to your chosen positive direction) and the formula works exactly the same way, showing the velocity decreasing over time.
What if I don't know the acceleration?
If you know the two end velocities and time instead, use s = ½(u + v)t or the dedicated form on the Displacement Calculator for two-velocity problems. Use v = u + at whenever acceleration is one of your three known values.
Can final velocity be negative?
Yes, if enough negative acceleration is applied over enough time to reverse the direction of motion relative to the chosen positive axis.
Is this the same equation used in the full Kinematics Calculator?
Yes. This page focuses specifically on the v = u + at relation; the Kinematics Calculator covers this equation alongside the other three SUVAT relations for a complete solve from any three known values.
Summary
Final velocity under constant acceleration follows the linear relation v = u + at, giving 11 m/s for a car starting at 5 m/s and accelerating at 2 m/s² for 3 seconds. The same equation rearranges cleanly to solve for initial velocity, acceleration or time whenever the other three quantities are known.
Because the relation is linear, it never produces the two-root ambiguity that appears in the squared-velocity or squared-time SUVAT forms.