The kinematic equations are four formulas that link displacement, initial velocity, final velocity, acceleration and time for an object moving in a straight line with constant acceleration: v = v₀ + at, Δx = v₀t + ½at², v² = v₀² + 2aΔx and Δx = ½(v₀ + v)t. To pick one, find the variable the problem neither gives nor asks for, and use the equation that leaves it out.
This guide shows where the equations come from, how to choose between them, how signs work in free fall, and how to handle a problem where the motion changes halfway (speed up, then brake), which is where most students get stuck.
The Kinematic Equations and What Each Letter Means
v = v₀ + atΔx = v₀t + ½at²v² = v₀² + 2aΔxΔx = ½(v₀ + v)tValid only while the acceleration
a stays constant, for motion along one straight line.Every problem uses the same five quantities. Each equation contains four of them, so if you know any three you can always find the other two. US textbooks such as OpenStax College Physics write them with v₀ and Δx. UK A level courses write the same equations with the letters s, u, v, a and t, which is why students there call them the SUVAT equations (see the Pearson Edexcel mechanics teaching guide).
| Quantity | US symbol | SUVAT symbol | SI unit |
|---|---|---|---|
| Displacement (change in position, with a sign) | Δx (or Δy) | s | m |
| Initial velocity | v₀ | u | m/s |
| Final velocity | v | v | m/s |
| Acceleration (constant) | a | a | m/s² |
| Time interval | t | t | s |
UK formula sheets also list a fifth equation, s = vt − ½at² (in US letters, Δx = vt − ½at²). It is the one without the initial velocity. You can always reach the same answer with the other four, but it saves a step when u is the unknown you don't care about.
Where the Equations Come From
You don't have to memorise four unrelated formulas. Two facts produce all of them.
First, constant acceleration means velocity changes by the same amount every second: a = (v − v₀) / t. Rearranged, that is v = v₀ + at.
Second, when velocity rises at a steady rate, the average velocity is exactly halfway between the start and end values, (v₀ + v) / 2. Displacement is average velocity times time, so Δx = ½(v₀ + v)t. On a velocity-time graph this is the area of the trapezoid under the straight line.
Substitute v = v₀ + at into that second equation and you get Δx = ½(2v₀ + at)t = v₀t + ½at². Eliminate t instead (solve the first equation for t and substitute) and you get v² = v₀² + 2aΔx. If you forget an equation in a test, you can rebuild it in a few lines from these two ideas.
How to Choose the Right Kinematic Equation
Write down the five letters, fill in the three the problem gives you, and mark the one it asks for. The fifth letter is the missing variable. Use the equation that does not contain it, and you will have exactly one unknown.
| Missing (not given, not asked) | Use |
|---|---|
Displacement Δx | v = v₀ + at |
Final velocity v | Δx = v₀t + ½at² |
Time t | v² = v₀² + 2aΔx |
Acceleration a | Δx = ½(v₀ + v)t |
Initial velocity v₀ | Δx = vt − ½at² |
- Sketch and choose a positive direction
Draw the path and an arrow for "positive". Every velocity, displacement and acceleration gets a sign relative to that arrow.
- Translate the words into numbers
"Starts from rest" means
v₀ = 0. "Comes to a stop" meansv = 0. "Dropped" meansv₀ = 0anda = −9.8 m/s²with up as positive. - List the five variables
Three known, one wanted, one missing. If you have fewer than three knowns, the problem has more than one stage or hides a value in the words.
- Pick the equation and solve with units
Rearrange before substituting, carry units through, and keep the significant figures the data supports.
In the examples below the data has three significant figures, so the answers do too. Where g = 9.8 m/s² appears, treat it as exact, the usual classroom convention.
A car starts from rest and accelerates at 3.00 m/s² for 6.00 s. How far does it travel?
Known: v₀ = 0, a = 3.00 m/s², t = 6.00 s. Wanted: Δx. Missing: v.
The equation without v is Δx = v₀t + ½at².
Δx = 0 + ½(3.00 m/s²)(6.00 s)² = ½(3.00 m/s²)(36.0 s²) = 54.0 m. The s² cancels, leaving metres.
Δx = 54.0 mA car moving at 25.0 m/s brakes steadily and stops in 50.0 m. Find its acceleration and the time it takes to stop.
Known: v₀ = 25.0 m/s, v = 0, Δx = 50.0 m. Wanted: a. Missing: t, so use v² = v₀² + 2aΔx.
0 = (25.0 m/s)² + 2a(50.0 m), so a = −625 m²/s² ÷ 100 m = −6.25 m/s².
The minus sign means the acceleration points opposite to the motion: the car is slowing down. Some books call this a deceleration of 6.25 m/s², but in the equations it is always entered as a negative acceleration. Now the time: t = (v − v₀) / a = (0 − 25.0 m/s) / (−6.25 m/s²) = 4.00 s.
a = −6.25 m/s², t = 4.00 sFree Fall and the Sign of g
An object moving only under gravity (ignoring air resistance) has a constant acceleration of about 9.8 m/s² downward. The defined standard value is 9.80665 m/s² (NIST); this guide uses g = 9.8 m/s², as most courses do.
Gravity does not have a fixed sign. The sign comes from your chosen positive direction. With up as positive, a = −9.8 m/s² for the whole flight: on the way up, on the way down, and at the very top. Velocity changes sign at the top (positive going up, zero at the peak, negative coming down), but acceleration never does.
A ball is thrown straight up at 14.7 m/s. How high does it rise, and when does it land back in the thrower's hand?
Up is positive, so v₀ = +14.7 m/s and a = −9.8 m/s². At the top the ball is momentarily still, so v = 0.
Height (time missing): 0 = (14.7)² + 2(−9.8)Δy, so Δy = 216.09 ÷ 19.6 = 11.025 m ≈ 11.0 m.
Time to the top (displacement missing): 0 = 14.7 − 9.8t, so t = 1.50 s.
Back at the hand, Δy = 0: 0 = 14.7t − 4.9t² = t(14.7 − 4.9t), giving t = 0 (the throw) or t = 3.00 s. Then v = 14.7 − 9.8(3.00) = −14.7 m/s.
11.0 m; back in the hand after 3.00 s, moving at 14.7 m/s downward.The trip up and the trip down take the same time, and the ball returns at the same speed it left with. That symmetry holds whenever the launch and landing heights are equal, and it makes a handy check.
A stone is thrown straight up at 4.90 m/s from the edge of a cliff 29.4 m above the water. How long until it hits the water, and how fast is it moving then?
Up is positive and the origin is the cliff edge, so the water is at Δy = −29.4 m. Known: v₀ = 4.90 m/s, a = −9.8 m/s², Δy = −29.4 m. Missing: v.
−29.4 = 4.90t − 4.9t². Move everything to one side: 4.9t² − 4.90t − 29.4 = 0. Divide by 4.9: t² − t − 6 = 0.
Factor: (t − 3)(t + 2) = 0, so t = 3.00 s or t = −2 s. A negative time is before the throw, so reject it. (For messier numbers, use the quadratic formula.)
Speed: v = 4.90 − 9.8(3.00) = −24.5 m/s. Check with the time-free equation: v² = 4.90² + 2(−9.8)(−29.4) = 24.01 + 576.24 = 600.25, and √600.25 = 24.5.
t = 3.00 s; it hits the water at 24.5 m/s downward.Writing Δy = +29.4 m in Example 4 because "the cliff is 29.4 m high". The stone ends up below where it started, so with up as positive its displacement is negative. With the wrong sign the equation becomes t² − t + 6 = 0, which has no real solutions at all. When a quadratic in t has no real roots, check your signs before anything else.
Two-Phase Problems: Speed Up, Then Brake
The equations need one constant acceleration. When the acceleration changes partway through, as when a car speeds up and then brakes, split the motion at the moment it changes. Solve each phase with its own acceleration. The link between them is the velocity: the final velocity of phase 1 is the initial velocity of phase 2.
A car starts from rest at one traffic light and accelerates at 2.50 m/s² for 8.00 s. The driver then brakes at 4.00 m/s² and stops exactly at the next light. How far apart are the lights, and how long does the trip take?
Phase 1 (v₀ = 0, a = 2.50 m/s², t = 8.00 s). Speed at the end: v = 0 + 2.50(8.00) = 20.0 m/s. Distance: Δx₁ = ½(2.50)(8.00)² = 80.0 m.
Phase 2 starts at v₀ = 20.0 m/s, ends at v = 0, with a = −4.00 m/s² (braking, opposite to the motion).
Distance (time missing): 0 = (20.0)² + 2(−4.00)Δx₂, so Δx₂ = 400 ÷ 8.00 = 50.0 m. Time: t₂ = (0 − 20.0) / (−4.00) = 5.00 s.
Totals: 80.0 + 50.0 = 130 m and 8.00 + 5.00 = 13.0 s.
130 m apart; the trip takes 13.0 s.Notice that the shared speed of 20.0 m/s is never given. You have to earn it from phase 1 before phase 2 has enough knowns. Plugging the whole 13-second trip into one equation with an "average" acceleration gives a wrong answer, because no single constant acceleration describes this motion.
Draw the velocity-time graph for Example 5: a straight line up from 0 to 20.0 m/s over 8 s, then a straight line down to 0 over 5 s. The total displacement is the area of that triangle, ½ × 13.0 s × 20.0 m/s = 130 m. Same answer, no equations needed, which is why the graph is the fastest check for multi-phase problems.
When You Cannot Use the Kinematic Equations
The formulas are exact only under their assumptions. Stop and rethink if any of these apply:
- The acceleration changes (a car that eases off the gas, a falling skydiver with air resistance, a spring). If it changes in jumps, split the motion into phases. If it changes smoothly, you need calculus: velocity is the integral of acceleration, as in the fundamental theorem of calculus.
- The motion is not along one line. A thrown ball moving sideways and up is two problems at once. Split the velocity into horizontal and vertical components with sine and cosine, then apply the equations to each direction separately.
- The question asks for distance, not displacement, and the object turns around. The equations give displacement. In Example 3 the ball's displacement after 3.00 s is zero, but it travelled
2 × 11.025 m ≈ 22.1 m(up to the top and back down).
How to Check Your Answer
Physics answers can be checked in three quick ways, and each catches a different kind of mistake.
Units. Carry units through the substitution. A displacement must come out in m and a velocity in m/s. If you get m²/s, an equation was misremembered.
A second equation. Once you have all five variables, plug them into an equation you didn't use. In Example 2, Δx = ½(25.0 + 0)(4.00) = 50.0 m, which matches the given distance.
Size and sign. A car's braking acceleration is a few m/s², not hundreds. A falling object's velocity should be negative if up is positive. And under constant acceleration the average speed must sit between the start and end speeds.
If you want to compare your working against a full solution, the physics solver page explains how Solver AI on iPhone and iPad turns a photo of a kinematics problem into step-by-step working. Use it after you have tried the problem yourself, to find the step where your signs or numbers went off track.
Practice Problems
Use g = 9.8 m/s². Take up as positive for the falling ball and the direction of motion as positive for the vehicles. Answers follow each problem.
- A cyclist slows steadily from
12.0 m/sto4.00 m/sin4.00 s. Find the acceleration and the distance covered. Answer:a = −2.00 m/s²,Δx = ½(12.0 + 4.00)(4.00) = 32.0 m. - A ball is dropped from rest from a height of
44.1 m. How long does it fall, and how fast is it going at the bottom? Answer:−44.1 = −4.9t², sot = 3.00 s; speed29.4 m/s. - A train moving at
30.0 m/sbrakes at1.50 m/s². What is its stopping distance, and how long does it take to stop? Answer:Δx = 900 ÷ 3.00 = 300 m,t = 20.0 s. - A rocket sled starts from rest, accelerates at
5.00 m/s²for4.00 s, coasts at constant speed for6.00 s, then brakes to rest in25.0 m. Find the braking acceleration, the total distance and the total time. Answer: top speed20.0 m/s; distances40.0 m + 120 m + 25.0 m = 185 m; brakinga = −400 ÷ 50.0 = −8.00 m/s²lasting2.50 s; total time12.5 s.
Choose a positive direction, list the five variables, and use the equation without the missing one. If the acceleration changes, split the motion where it changes and pass the velocity from one phase to the next. Then check with a second equation or the velocity-time graph.