SUVAT Kinematics Solver: Solve 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.
A car brakes from 25 m/s to a stop. Your problem gives you acceleration (−5 m/s²) but not time—how do you find the stopping distance? The SUVAT equations handle exactly this situation. Each of the five formulas omits one variable, so choosing the right equation means picking the one that leaves out what you don't know. This calculator solves for any two unknowns once you supply three; below you'll find rearranged forms for every variable, a missing-variable cheat sheet, and worked examples covering multi-phase motion, free fall, and "catching-up" problems where two objects start at different times.
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 |
Which Equation Do I Use? Missing-Variable Shortcuts
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.
Rearranging Each SUVAT Formula for Any Unknown
Most textbooks present the equations with v or s on the left, but exam problems often ask for the other variable. Here are all rearrangements you might need:
From v = u + at:
u = v − at | a = (v − u)/t | t = (v − u)/a
From s = ½(u + v)t:
t = 2s/(u + v) | u = 2s/t − v | v = 2s/t − u
From s = ut + ½at²:
a = 2(s − ut)/t² | u = (s − ½at²)/t
Solving for t requires the quadratic formula—see Section 4.
From v² = u² + 2as:
s = (v² − u²)/(2a) | u = √(v² − 2as) | a = (v² − u²)/(2s)
The calculator handles these automatically, but knowing the algebra lets you verify answers and spot errors. Notice that solving for t from s = ut + ½at² yields a quadratic—you'll get two roots, and the positive one usually represents forward time.
Handling Negative Values: Sign Convention Pitfalls
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.
- Velocity is negative if motion is opposite your positive direction.
- Acceleration is negative if it opposes 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, input a = −5 m/s². Otherwise v = u + at gives +30 m/s instead of the correct +20 m/s after 1 second.
For free fall with "up" as positive: initial upward throw has u > 0, but a = −9.81 m/s² (gravity pulls down). At the peak, v = 0. On the way down, v becomes negative.
Step-by-Step: Multi-Phase Motion Problem
Many real problems have two or more phases—acceleration, then constant speed, then braking. 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
Notice that each phase uses different equations depending on which variable is unknown. The cruise phase doesn't even need SUVAT—it's just s = vt.
Free Fall Special Case (g = 9.81 m/s²)
Free-fall problems are just SUVAT with a = ±g. The sign depends on your coordinate choice:
| Convention | a | Upward throw u | Downward drop u |
|---|---|---|---|
| Up = positive | −9.81 m/s² | + value | 0 |
| Down = positive | +9.81 m/s² | − value | 0 |
At the highest point of an upward throw, v = 0. That's a key condition for finding maximum height:
h_max = u²/(2g) (derived from v² = u² − 2gh with v = 0)
Air resistance is ignored in basic SUVAT. At low speeds and short drops this is fine; at higher speeds or with light objects (paper, feathers), drag becomes significant and acceleration is no longer constant.
Braking Distance Without Knowing Time
The time-independent equation v² = u² + 2as is perfect for braking problems. When a vehicle stops, v = 0:
s = −u²/(2a) = u²/(2|a|)
Stopping distance scales with the square of speed. Double your speed and stopping distance quadruples. That's why speeding is so dangerous—going from 30 mph to 60 mph doesn't double your braking distance, it multiplies it by four.
Quick example:
u = 25 m/s, a = −5 m/s² → s = 25²/(2 × 5) = 625/10 = 62.5 m
If you also need time: t = (v − u)/a = (0 − 25)/(−5) = 5 s. Both answers come from the same three knowns.
Worked Problem: Catching Up / Collision Timing
A police car starts from rest and accelerates at 4 m/s² to chase a suspect traveling at constant 20 m/s. The suspect has a 100 m head start. When and where does the police car catch up?
Step 1: Write position equations
Police: x_p = ½at² = ½(4)t² = 2t²
Suspect: x_s = 100 + 20t (head start + constant speed)
Step 2: Set positions equal
2t² = 100 + 20t
2t² − 20t − 100 = 0
t² − 10t − 50 = 0
Step 3: Solve quadratic
t = (10 ± √(100 + 200))/2 = (10 ± √300)/2 = (10 ± 17.32)/2
t = 13.66 s (positive root)
Step 4: Find position
x = 2(13.66)² = 2 × 186.6 ≈ 373 m from police start
This problem combines SUVAT for the accelerating car with simple kinematics for the constant-speed car. The key is writing both positions as functions of the same t, then equating them.
Unit Reference: SI, Imperial, Conversions
SUVAT equations work in any consistent unit system. The most common are SI (meters, seconds) and imperial (feet, seconds). Time is always in seconds.
| Quantity | SI Unit | Imperial Unit | Conversion |
|---|---|---|---|
| Displacement (s) | m | ft | 1 m = 3.281 ft |
| Velocity (u, v) | m/s | ft/s | 1 m/s = 3.281 ft/s |
| Acceleration (a) | m/s² | ft/s² | 1 m/s² = 3.281 ft/s² |
| Time (t) | s | s | — |
| g (free fall) | 9.81 m/s² | 32.17 ft/s² | — |
Common speed conversions:
- 1 km/h = 0.2778 m/s (divide km/h by 3.6)
- 1 mph = 0.447 m/s = 1.467 ft/s
- 60 mph ≈ 26.8 m/s ≈ 88 ft/s
Limitations & Assumptions
- •Constant acceleration only: SUVAT equations fail when a varies (rocket thrust, air drag, 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.
Sources & Further Reading
- Halliday, Resnick, Walker — Fundamentals of Physics (11th ed.), Wiley 2018. Chapter 2 on 1-D kinematics.
- HyperPhysics — Motion Equations. Georgia State University physics reference.
- NIST — Standard gravitational acceleration g = 9.80665 m/s² (exact definition).
Fixing Common SUVAT Mistakes
Real questions from students stuck on sign conventions, negative time, and equation selection.
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.
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.
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.