Kinematics Equation Calculator: SUVAT — s, u, v, a, t with Signs
Solve 1D constant-acceleration motion problems using the five SUVAT equations. Enter at least 3 of the 5 variables (s, u, v, a, t) and the solver will find the remaining values. Compare up to 3 cases side by side.
SUVAT is what physicists call the package of five equations that describe motion under constant acceleration. The acronym lists the variables: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time). Each equation drops exactly one of those, which is the practical insight you exploit when solving problems. Don't have time? Use v² = u² + 2as. Don't have final velocity? Use s = ut + ½at². Picking the right equation isn't algebra, it's bookkeeping. This SUVAT solver applies that logic for you. Enter any three values and it returns the rest, with a sign-convention check (positive means displacement and acceleration agree in direction, negative doesn't mean wrong, it means opposed).
The Five SUVAT Equations at a Glance
| Equation | Missing Variable | Best For |
|---|---|---|
| v = u + at | s | Finding final velocity or time |
| s = ½(u + v)t | a | Average-velocity approach |
| s = ut + ½at² | v | Displacement when final v unknown |
| s = vt − ½at² | u | Displacement when initial u unknown |
| v² = u² + 2as | t | Braking distance, speed checks |
Setting Up the Problem: Free Body, Frame, and What's Constant
SUVAT is a kinematics package. It describes how positions and velocities evolve, and it doesn't need a free-body diagram to apply. But you still need one to know whether the problem qualifies. The single hard requirement is that net force is constant in direction and magnitude, which means acceleration is constant. The instant a brake force ramps, a rocket throttles, or a parachute deploys, SUVAT breaks because a stops being a constant.
Pick a one-dimensional coordinate axis aligned with the motion. Forward is positive. Stick with that choice for every variable in the problem. If a car decelerates from 30 m/s to 0, then u = +30 m/s, v = 0, and a = −5 m/s² (some negative number). Don't flip signs partway through to "make it positive" because the algebra carries the geometry. What's constant: a. What changes: u becomes v, position grows, time elapses. What stays put because the problem already finished it: the choice of axis.
On Earth's surface, free fall qualifies for SUVAT because g = 9.81 m/s² is constant for any drop short of a few hundred meters. Air resistance breaks this. A bowling ball dropped from a balcony stays inside SUVAT. A piece of paper dropped from the same balcony falls through a varying-acceleration regime, and SUVAT lies to you. Diagnose first, calculate second.
Picking the Right Equation Without Memorizing All of Them
Each SUVAT equation links exactly four of the five variables. To choose the right one, identify which variable you don't have and don't need. Then pick the equation that omits it.
- Don't have s? Use v = u + at
- Don't have a? Use s = ½(u + v)t
- Don't have v? Use s = ut + ½at²
- Don't have u? Use s = vt − ½at²
- Don't have t? Use v² = u² + 2as
This single rule eliminates guesswork. If a braking problem asks for stopping distance and gives initial speed and deceleration (no time), reach for v² = u² + 2as because it skips t entirely. If a problem gives u, v, and a, and asks for t, the only equation that contains all four is v = u + at, so use it.
Two of the equations also let you solve for any unknown via simple algebra. v = u + at rearranges to t = (v − u)/a or a = (v − u)/t with no quadratic in sight. v² = u² + 2as is also linear once you square the velocities. The only one that demands the quadratic formula is s = ut + ½at² when t is unknown. Two roots come out, and the positive one is what you want unless the problem is asking when the projectile crossed a height on the way up versus the way down.
Sign Conventions and Direction Choices That Trip Most Solvers
Sign errors cause most wrong answers in kinematics. The rules are simple but unforgiving.
- Choose a positive direction (right, up, forward) and stick with it for the entire problem.
- Velocity is negative if the motion is opposite your positive direction.
- Acceleration is negative if it opposes the current velocity (deceleration), OR if it points in the negative direction.
- Displacement can be negative if the object ends up behind where it started.
Common mistake: Entering deceleration as a positive number. If a car moving at +25 m/s brakes at 5 m/s² of slowing, input a = −5 m/s². Otherwise v = u + at gives +30 m/s instead of the correct +20 m/s after one second.
For free fall with up as positive, an upward toss has u > 0, but a = −9.81 m/s² (gravity pulls down). At the peak, v = 0. On the way down, v becomes negative.
| Convention | a | Upward throw u | Downward drop u |
|---|---|---|---|
| Up = positive | −9.81 m/s² | + value | 0 |
| Down = positive | +9.81 m/s² | − value | 0 |
The "down is positive" choice is unusual but legitimate when every motion in the problem is downward (a falling object). It makes a positive and u positive and lets you forget about negative numbers. The "up is positive" convention is more common because it matches everyday vertical-axis intuition. Either works. Mixing them inside a single problem doesn't.
When the Setup Has Multiple Phases (Acceleration, Constant, Stopping)
Many real problems have two or more phases: acceleration, then constant speed, then braking. SUVAT applies inside each phase but doesn't span them. Solve each phase separately. The final values of one phase become the initial values of the next.
Example: A train accelerates from rest at 1.2 m/s² for 30 s, cruises for 2 km, then decelerates at −0.8 m/s² to a stop. Find total time and distance.
Phase 1: Acceleration
u₁ = 0, a₁ = 1.2 m/s², t₁ = 30 s
v₁ = u + at = 0 + 1.2 × 30 = 36 m/s
s₁ = ½at² = ½ × 1.2 × 30² = 540 m
Phase 2: Constant speed
v₂ = 36 m/s (constant), s₂ = 2000 m
t₂ = s/v = 2000/36 = 55.6 s
Phase 3: Deceleration
u₃ = 36 m/s, v₃ = 0, a₃ = −0.8 m/s²
t₃ = (v − u)/a = (0 − 36)/(−0.8) = 45 s
s₃ = (v² − u²)/(2a) = (0 − 1296)/(−1.6) = 810 m
Totals
Total time = 30 + 55.6 + 45 = 130.6 s
Total distance = 540 + 2000 + 810 = 3350 m
Each phase uses different equations depending on which variable is unknown. The cruise phase doesn't even need SUVAT. It's just s = vt because a = 0. A useful sanity check: the sum of phase distances should equal the area under your v-t graph. Sketch it and you'll spot phase-boundary errors fast.
Energy vs. Force Approaches: When Each One Wins
SUVAT is a force-flavored toolkit. You implicitly assume Newton's second law gave you a constant acceleration, and then you track time and position. Energy methods (½mv² + mgh = constant, or work-energy theorem W = ΔKE) skip time entirely. They give you final speed for a given displacement without needing to know how long the trip took.
For braking problems, both work. v² = u² + 2as gives stopping distance with one equation. The work-energy theorem says ½mu² = F·s, so s = mu²/(2F). Same answer, different bookkeeping. SUVAT is faster if you already have a in hand. Energy is faster if you have force and mass and don't want to compute a separately.
SUVAT wins whenever the question asks "when?" or "where at time t?" Energy can't answer those. Energy wins whenever there's a height change and you don't need to track time, or when force varies with position (a spring, a Hooke's law extension, a stretched bungee). SUVAT can't handle non-constant a. Energy doesn't care: ∫F dx replaces F·s and the calculation goes through.
For multi-phase motion with friction, energy can be cleaner because you write one balance equation across the whole trip, with friction work as a single subtracted term. SUVAT forces you to chain phases. The choice is judgement, not rule. If the problem screams "time" or "instantaneous velocity at this position," go SUVAT. If it screams "speed at the bottom of the ramp" or "compression of the spring," go energy.
Worked Example: Police Skid-Mark Forensics on a 60 m Stopping Distance
Investigators measure a 60.0 m skid mark on dry asphalt. The vehicle came to rest at the end. Skid friction tests give μ_k = 0.78 for the tire-asphalt pair. The driver claims they were traveling at 25 m/s (about 56 mph) when they hit the brakes. Was the driver telling the truth?
Step 1: deceleration from kinetic friction
On a level surface with locked wheels, the only horizontal force is friction. f = μ_k · m · g, so a = μ_k · g.
a = 0.78 × 9.81 = 7.65 m/s² of deceleration. In SUVAT: a = −7.65 m/s² with forward positive.
Step 2: solve for u using v² = u² + 2as
v = 0 (stopped), s = +60.0 m, a = −7.65 m/s².
0 = u² + 2(−7.65)(60.0) = u² − 918
u = √918 = 30.3 m/s, which is about 67.8 mph
Step 3: time to stop
t = (v − u)/a = (0 − 30.3) / (−7.65) = 3.96 s. The braking event lasted about four seconds.
The driver claimed 25 m/s. The skid mark says at least 30.3 m/s, about 21 percent higher than the claim. That gap matters because kinetic energy scales as v². The actual collision energy was (30.3/25)² ≈ 1.47 times what the driver's number would predict.
Real forensic reconstruction adds reaction-time distance (typically 0.7 to 1.5 s of pre-brake travel), grade corrections if the road wasn't level, and friction adjustments for moisture or surface contamination. The Northwestern University Center for Public Safety publishes the standard reference manuals used in U.S. accident-reconstruction courts. SUVAT is the math underneath every line of those reports.
References & Further Reading
- Halliday, Resnick & Walker (2018). Fundamentals of Physics (11th ed.), Wiley. Chapter 2 on 1-D kinematics, including a derivation of all five SUVAT relations from a constant-acceleration assumption.
- Tipler, P. A. & Mosca, G. (2007). Physics for Scientists and Engineers (6th ed.), W. H. Freeman. Chapter 2 covers SUVAT with worked problems on car-chase and free-fall scenarios.
- HyperPhysics Motion Equations. Georgia State University physics reference, with interactive sliders.
- NIST Standard gravitational acceleration g = 9.80665 m/s² (exact CGPM definition).
- Northwestern University Center for Public Safety Traffic Crash Reconstruction (CRR) curriculum. Practical application of SUVAT to skid-mark and stopping-distance forensics.
Limitations & Assumptions
- •Constant acceleration only: SUVAT equations fail when a varies (rocket thrust, air drag, spring forces, etc.).
- •One dimension: For 2D projectile motion, apply SUVAT separately to x and y components.
- •No air resistance: Real free-fall at high speed reaches terminal velocity. SUVAT doesn't model that.
- •Point mass: Rotation and extended-body effects are ignored.
Fixing Common SUVAT Mistakes
Real questions from students stuck on sign conventions, negative time, and equation selection.
What are the four equations of kinematics?
The four kinematic equations describe motion at constant acceleration. They're sometimes called SUVAT in UK textbooks, after the variables they connect: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time). The first is v = u + at. If you know the starting velocity and how long an acceleration acts, this gives the final velocity. The second, s = ut + ½at², gives displacement after time t under constant acceleration. The third, v² = u² + 2as, drops time entirely. Use it for braking-distance problems where you know speeds and acceleration but not duration. The fourth, s = ½(u + v)t, drops acceleration. It works whenever you know both velocities and time. Each equation excludes one of the five variables. Pick the one that excludes whatever you don't have. A car decelerating from 30 m/s to a stop with a = −5 m/s² needs v² = u² + 2as: 0 = 900 + 2(−5)s, so s = 90 m. These equations don't apply when acceleration changes during the motion. A rocket with variable thrust or a falling object with significant air drag needs calculus instead.
My problem doesn't give me time—is there an equation that doesn't need t?
Yes: v² = u² + 2as. It relates final velocity, initial velocity, acceleration, and displacement without time. Perfect for braking-distance problems where you know speed and deceleration but not how long it takes to stop.
I entered deceleration as a positive number and now my velocity is increasing—what went wrong?
Deceleration means acceleration opposite to velocity. If velocity is positive, braking acceleration must be negative. Entering a = +5 m/s² when the car moves at +20 m/s gives v = u + at = 20 + 5t, which increases. Use a = −5 m/s² to model slowing down. The equations don't 'know' what deceleration means—you control signs.
The solver returned negative time—does that mean I set something up wrong?
Usually, yes. Negative time indicates the scenario can't happen moving forward from your chosen t = 0. Check your sign convention: did you mix positive and negative directions? Did you enter impossible values (like reaching v = 50 m/s with a = −2 m/s²)? Occasionally both roots of a quadratic are negative, meaning the event occurred in the 'past.'
How do I apply SUVAT to a ball thrown upward?
Choose 'up' as positive. Initial velocity u > 0 (upward throw). Acceleration a = −9.81 m/s² (gravity points down). At the peak, v = 0. On the way down, v becomes negative. Displacement s is positive while above the launch point, zero when back at the start, and negative if it falls below. The equations handle all of this—just keep signs consistent.
Displacement came out negative—is that wrong?
Not necessarily. Negative displacement means the object ended behind where it started, relative to your positive direction. If you threw a ball upward and asked for displacement when it hits the ground 2 m below the release point, s = −2 m is correct (down is negative if up is positive).
When I solve for t from s = ut + ½at², I get two answers. Which do I use?
Usually the positive root represents forward time. The negative root corresponds to a hypothetical earlier time when the object would have been at the same position if the motion had started earlier. In rare edge cases (e.g., multi-phase problems), both positive roots might be meaningful. Always interpret physically.
Can I use the same equations for horizontal and vertical motion separately?
Yes—that's exactly how you solve projectile motion. Horizontal motion has a = 0 (constant speed), so s_x = u_x × t. Vertical motion uses a = −g. Time t links both: find t from one direction, then use it in the other. SUVAT works independently for each axis.
Why is stopping distance four times larger when I double my speed?
From v² = u² + 2as with v = 0, stopping distance s = u²/(2|a|). Doubling u quadruples u², so s quadruples. That's why speeding is dangerous—going 60 mph instead of 30 mph doesn't double your stopping distance, it multiplies it by four.
My answer doesn't match the textbook, but I think I did everything right. What could be off?
Common culprits: (1) Sign errors—double-check that acceleration opposes or matches velocity as intended. (2) Unit mismatch—mixing m/s with km/h or feet. (3) Rounding mid-calculation—keep extra decimals until the final answer. (4) Using the wrong equation—verify the one you chose excludes the variable you don't have.
Does SUVAT work for a rocket accelerating through variable thrust?
No. SUVAT requires constant acceleration. A rocket's thrust changes as fuel burns, so acceleration isn't constant. You'd need calculus (integrating a(t)) or numerical simulation. SUVAT is strictly for uniform acceleration: free fall, constant braking, steady engine power on a car.