Uniform Acceleration Calculator v = v₀ + at: Final Velocity

About this calculator

The uniform-acceleration calculator solves $v = v_0 + at$ : given an initial velocity $v_0$ , an acceleration $a$ and a time $t$ , it finds the final velocity $v$ ; any three of the four quantities determine the fourth—how long braking takes, what acceleration reaches 100 km/h in 8 seconds, and how fast a ball left the hand when it takes 2 seconds to come to rest.

“Uniform acceleration” means the acceleration is constant: free fall ( $a = g$ ), smooth acceleration and braking of a car, a train pulling into a station, and a vertical throw can all be approximated this way. It is the most basic kinematic equation—the definition of acceleration rewritten—and the starting point of all the other kinematic formulas: the displacement formula $s = v_0 t + \tfrac12 at^2$ is on the uniform acceleration displacement calculator; where acceleration comes from ( $a = F/m$ ), its units and vector conventions are covered by Newton's second law.

What it does not do: it does not compute displacement or distance; it does not handle motion whose acceleration changes with time (falling with significant air resistance, variable-rate starts); it is one-dimensional only—curved motion and projectiles need the horizontal and vertical directions treated separately; and it does not convert units (convert km/h to m/s first).

Formula and variables

v = v_0 + a\,t \qquad\Longleftrightarrow\qquad a = \frac{v - v_0}{t} = \frac{\Delta v}{\Delta t}

Inverse forms:

t = \frac{v - v_0}{a},\qquad a = \frac{v - v_0}{t},\qquad v_0 = v - a\,t

Symbol	Meaning	Calculator field	Unit	Range and notes
$v_0$	initial velocity (at $t = 0$ )	Initial velocity v₀	m/s	any real number; default 0. A minus sign means opposite to the positive direction
$a$	acceleration (constant)	Acceleration a	m/s²	any real number; default 9.8. Positive along the velocity (speeding up), negative against it (slowing down)
$t$	elapsed time	Time t	s	$t \ge 0$ ; no default, empty on open
$v$	final velocity (at time $t$ )	Final velocity v	m/s	output; an input when solving in reverse

Unit consistency: $\text{m/s}^2 \times \text{s} = \text{m/s}$ . The default acceleration 9.8 is the textbook value for gravity; the standard value is $g_0 = 9.806\,65$ , and the discussion of which value to use is in Newton's second law.

How to use this calculator

Open the panelthis page corresponds to “Math tools → Formulas → Uniform acceleration” (/#/mathtools/formula?calculator=calculator.phys-uniform-accel). On open, “Initial velocity v₀” = 0 and “Acceleration a” = 9.8 are already filled in and “Time t” is empty; both “Time t” and “Final velocity v” then carry a “calculate” marker and show 0—the engine treats the empty $t$ as 0 and $v = 0 + 9.8 \times 0 = 0$ . This is not a result.
Find the final velocityenter the time in seconds and the final velocity updates immediately. For deceleration enter a negative $a$ .
Solve for timeclear “Time t” and enter “Final velocity v” (0 for a stop); the time field switches to “calculate” and shows $(v - v_0)/a$ .
Solve for accelerationclear “Acceleration a” and enter $v_0$ , $v$ and $t$ ; the acceleration field shows $(v - v_0)/t$ .
Solve for initial velocityclear “Initial velocity v₀” and enter $a$ , $t$ and $v$ ; the initial-velocity field shows $v - at$ .
Read the resultvalues keep at most 8 decimal places with trailing zeros removed. With $a = 0$ , solving for time is $0/0$ and the field stays empty—for constant velocity there is no unique answer to “when does it reach this speed”.
Copy the formulathe copy button at the right of the panel title puts the LaTeX v = v_0 + at on the clipboard.

Worked examples

All four examples were recomputed by the engine's phys-uniform-accel compute routine; the display convention is at most 8 decimal places with trailing zeros removed, in units m/s, m/s² and s.

Example 1: 3 seconds of free fall

Inputs: $v_0 = 0$ , $a = 9.8$ , $t = 3$ . $v = 0 + 9.8 \times 3 = 29.4$ . Interface shows: final velocity v 29.4—about 106 km/h. Every second the velocity increases by 9.8 m/s, which is what “acceleration 9.8 m/s²” means: how much the velocity changes each second. Falling from 10 storeys (about 30 m) takes roughly 2.5 s and reaches 24 m/s.

Example 2: a car accelerating to 100 km/h in 8 seconds, and then braking

Solving for acceleration: clear $a$ and enter $v_0 = 0$ , $v = 27.77777778$ (100 km/h), $t = 8$ : $a = 27.77777778 / 8 = 3.47222222$ . Interface shows: acceleration a 3.47222222. Entering it back ( $v_0 = 0$ , $a = 3.47222222$ , $t = 8$ ) makes the interface show final velocity 27.77777776—the final 6 is a rounding tail from substituting the 8-decimal acceleration back in, not an error.

Mid-braking: $v_0 = 27.77777778$ , $a = -5$ , $t = 3$ : $v = 27.78 - 15 = 12.77777778$ . Interface shows: final velocity v 12.77777778 (about 46 km/h).

Solving for the stopping time: clear $t$ and enter $v_0 = 27.77777778$ , $a = -5$ , $v = 0$ : $t = (0 - 27.78)/(-5) = 5.55555556$ . Interface shows: time t 5.55555556. A negative divided by a negative is positive—the time is of course positive. The 77.16 m travelled in that time is calculated with the displacement calculator.

Figure 1: on a v–t graph, uniformly accelerated motion is a straight line whose slope is the acceleration. Left: the car of Example 2; right: the thrown ball of Example 3—the acceleration is still −9.8 as the velocity passes through zero, and the line has no kink

Example 3: a vertical throw—acceleration is not zero when velocity is

Taking upward as positive, $a = -9.8$ . A ball reaches its highest point ( $v = 0$ ) 2 seconds after being thrown; how fast did it leave the hand? Clear $v_0$ and enter $v = 0$ , $a = -9.8$ , $t = 2$ : $v_0 = 0 - (-9.8)(2) = 19.6$ . Interface shows: initial velocity v₀ 19.6.

Now the third second: $v_0 = 19.6$ , $a = -9.8$ , $t = 3$ : $v = 19.6 - 29.4 = -9.8$ . Interface shows: final velocity v −9.8—the minus sign says it is falling. Throughout, the acceleration is always $-9.8$ , including the instant at the highest point; only the velocity changes from positive to negative. The straight line on the right of Figure 1 crosses zero continuously, without pausing.

Example 4: a train and the boundary case of constant velocity

Train: a train travelling at 20 m/s accelerates at 0.5 m/s² for 1 minute. $v_0 = 20$ , $a = 0.5$ , $t = 60$ : $v = 20 + 30 = 50$ . Interface shows: final velocity v 50 (180 km/h). A small acceleration does not mean a small velocity change—given enough time it is just as significant.

Constant velocity: $v_0 = 5$ , $a = 0$ , solving for “when does it reach $v = 5$ ” leaves the time field empty: $0/0$ is undefined, and at constant velocity the speed is 5 m/s at every instant.

The inputs of Example 1 can be entered directly into the panel to reproduce it; change the time to 1, 2, 4 and 5 and watch the final velocity grow in multiples of 9.8; change the acceleration to 1.62 (the Moon) and see that the same 3 seconds reaches only 4.86 m/s.

Principle and derivation

This formula is the definition of acceleration

Acceleration is defined as the rate of change of velocity: $a = \Delta v / \Delta t$ . With constant acceleration, $\Delta v = a\Delta t$ , and starting from $v_0$ at $t = 0$ , after $t$ seconds $v = v_0 + at$ . It needs no “derivation”—it is the definition rearranged. The substantive assertion is that the acceleration is constant, an assumption about the motion guaranteed by Newton's second law: a constant net force (such as gravity) gives constant acceleration.

The v–t graph: a straight line and its slope

Plotting velocity against time, uniformly accelerated motion is a straight line with vertical intercept $v_0$ and slope $a$ . Take any two points on the car's line on the left of Figure 1: $\Delta v / \Delta t = 13.89 / 4 = 3.47$ , matching the solved acceleration—this is exactly what the slope calculator does. A negative slope (right of Figure 1) means deceleration or acceleration in the reverse direction; the line crossing the horizontal axis means the velocity reverses. The area under the line is the displacement, which is the next page's subject.

Sign conventions: one-dimensional vectors

Velocity and acceleration are vectors, and in one dimension their direction is carried by the sign. Choose a positive direction first (usually the direction of the initial velocity, or “upward”), then: $a$ with the same sign as $v$ → speed increasing; opposite sign → speed decreasing. In Example 3 the ball rises with $v > 0$ and $a < 0$ (slowing), then passes the highest point and falls with $v < 0$ and $a < 0$ (speeding up); the acceleration never changes. “Deceleration” is not another physical quantity, merely a negative acceleration.

Galileo and free fall

That “all bodies, heavy or light, fall with the same acceleration” is Galileo's conclusion in the Two New Sciences of 1638. He used an inclined plane to “dilute” gravity and found that speed is proportional to time (not to distance). $a = g \approx 9.8\ \text{m/s}^2$ is an approximation at the Earth's surface; the Moon is 1.62 and Mars 3.72. When air resistance is significant (feathers, raindrops, parachutes) the acceleration falls with speed until it reaches zero (terminal velocity) and the motion is no longer uniformly accelerated.

Relationship to the other kinematic formulas

Besides the four quantities $v_0$ , $v$ , $a$ and $t$ there is the displacement $s$ , and among the five quantities there are four common equations, each missing one: $v = v_0 + at$ (no $s$ ), $s = v_0 t + \tfrac12 at^2$ (no $v$ ), $v^2 = v_0^2 + 2as$ (no $t$ ) and $s = \tfrac12(v_0 + v)t$ (no $a$ ). This calculator is the first; the second is on the displacement calculator; the third comes from eliminating $t$ between the first two and is the source of the kinetic energy formula $\tfrac12 mv^2$ .

Assumptions

Constant acceleration: $a$ is unchanged over the whole interval. Real vehicle acceleration varies with gear and speed, so the result is an average.
Straight-line, one-dimensional motion: velocity and acceleration are collinear, with direction carried by signs.
Drag is ignored: the free-fall examples exclude air resistance.
SI units: m/s, m/s², s; no conversion is performed.
Interface conventions: unfilled fields participate as 0 in the engine (in inverse modes the denominator falls back to 1), so the final velocity shows 0 when the panel opens.

Scope and limitations

Can calculate: any one of $v_0$ , $a$ , $t$ , $v$ from the other three, including negative values.
Approximate only: variable acceleration can only be approximated piecewise; taking $g = 9.8$ carries a value difference of about 0.1%.
Cannot calculate: displacement (use the displacement calculator); time-free relations such as $v^2 = v_0^2 + 2as$ (by hand); two-dimensional projectiles (one calculation per direction); motion whose acceleration changes with time.
Does not: convert units; solve for time when $a = 0$ .
Numerics: substituting 8-decimal results back leaves a rounding tail in the last digit (27.77777776 in Example 2).

Common pitfalls

Treating km/h as m/s: the “acceleration” returned is in (km/h)/s, numerically 3.6 times larger (note after Example 4).
Assuming zero acceleration at the highest point: the acceleration is still $-9.8$ the instant the velocity is zero (Example 3).
Forgetting the minus sign on deceleration: a braking $a$ is negative; entering it positive gives an ever-increasing speed.
Taking $v$ for a speed: $v$ is signed; the $-9.8$ in Example 3 means 9.8 m/s downward, not a “negative speed”.
Entering time in minutes: 60 s is not 1; entering $t = 1$ for the train in Example 4 gives only 20.5.
Using the average velocity as the final velocity: accelerating uniformly from 0 to 27.78 gives an average of 13.89; the final velocity is 27.78.

Typical use cases

Teaching

Making the definition of acceleration intuitive; the slope of a v–t graph; sign conventions and the vertical throw; Galileo and free fall.

Transport

Converting between 0–100 km/h times and average acceleration; stopping time for a given deceleration; the time for a train to accelerate to a target speed.

Sport and safety

Velocity gain at the start of a parachute jump (before terminal velocity); starting acceleration estimates; falling-speed estimates (with the displacement page for height).

Engineering

Designing the acceleration and deceleration times of lifts, conveyors and machine-tool feeds; finding the ramp duration for an allowed acceleration.

FAQ

Why is the final velocity 0 when the panel opens?

The time has not been filled in and the engine treats $t$ as 0. Enter a time and it is normal.

How do I calculate displacement or distance?

Use the uniform acceleration displacement calculator ( $s = v_0 t + \tfrac12 at^2$ ); or first find $t$ here and then multiply the average velocity $\tfrac12(v_0 + v)$ by $t$ .

Does a negative acceleration mean deceleration?

Only when it is opposite to the velocity. With downward positive, a falling body's acceleration is positive; with upward positive it is negative, yet the body still speeds up as it falls.

Why are the rise and fall times of a vertical throw equal?

With constant acceleration and no drag, $v$ is a straight line in $t$ , so going from $v_0$ down to 0 and from 0 down to $-v_0$ take the same time (both $v_0/g$ ).

Can it be used when air resistance is large?

No. Drag grows with speed, so the acceleration is no longer constant and the body eventually reaches terminal velocity. Raindrops, parachutes and feathers are not uniformly accelerated.

Should g be 9.8 or 9.81?

Use the value given by the problem or the specification; the two differ by 0.1%. The discussion of versions is in Newton's second law.

References and further reading

The phys-uniform-accel definition in the CalcX engine source src/data/formulasScience.ts (compute(known, solveFor) solves for v / t / a / v0); every figure on this page was recomputed by that engine.
Wikipedia, Equations of motion — Constant translational acceleration in a straight line （访问日期：2026-09-09）—the complete table of five quantities and four equations.
Wikipedia, Acceleration （访问日期：2026-09-09）—the definition, units and vector nature of acceleration.
Galileo Galilei, Dialogues Concerning Two New Sciences (1638), Third Day, translated by Crew & de Salvio （访问日期：2026-09-09）—the original account of uniformly accelerated motion and free fall.
NIST, CODATA 2018: standard acceleration of gravity （访问日期：2026-09-09）— $g_0 = 9.806\,65\ \text{m/s}^2$ .
ISO 80000-3:2019, Quantities and units — Part 3: Space and time—definitions of the quantities and units of velocity and acceleration.

Privacy

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Sources & review

Reviewed by: CalcX 编辑组
Updated: 2026-09-09

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Open Uniform Acceleration Calculator v = v₀ + at: Final Velocity in CalcX