The five kinematic equations for constant acceleration are ( v = u + at ), ( s = ut + frac12at^2 ), ( v^2 = u^2 + 2as ), ( s = frac12(u + v)t ) and ( s = vt – frac12at^2 ), where s is displacement, u initial velocity, v final velocity, a acceleration and t time. Choose the equation that leaves out the one quantity you neither know nor need.
The kinematic equations describe motion in a straight line when the acceleration is constant. They connect five quantities, often called SUVAT:
| Symbol | Quantity | SI unit |
|---|---|---|
| ( s ) | displacement | m |
| ( u ) | initial velocity | m/s |
| ( v ) | final velocity | m/s |
| ( a ) | acceleration (constant) | m/s² |
| ( t ) | time | s |
If you know any three of them, the equations give you the other two. Some textbooks write ( v_0 ) for ( u ), ( Delta x ) for ( s ) and ( v ) for the final velocity; the equations are the same.
The five kinematic equations
$$v = u + at$$
$$s = ut + tfrac12at^2$$
$$v^2 = u^2 + 2as$$
$$s = tfrac12(u + v)t$$
$$s = vt - tfrac12at^2$$
Each one is missing exactly one variable. That’s the key to choosing:
| Equation | Missing variable | Use it when |
|---|---|---|
| ( v = u + at ) | ( s ) | distance isn’t given or asked for |
| ( s = ut + frac12at^2 ) | ( v ) | final velocity isn’t involved |
| ( v^2 = u^2 + 2as ) | ( t ) | time isn’t given or asked for |
| ( s = frac12(u + v)t ) | ( a ) | acceleration isn’t involved |
| ( s = vt – frac12at^2 ) | ( u ) | initial velocity isn’t involved |
How to solve a kinematics problem
- Choose a positive direction and write it down. Everything pointing the other way gets a minus sign.
- List the five variables and fill in the three you know. Mark the one you want.
- Pick the equation that doesn’t contain the fifth variable (the one you neither know nor need).
- Solve, then check the sign and size of the answer make physical sense.
Where the equations come from
The kinematic equations are calculus with the integrals already done. Acceleration is the derivative of velocity, and velocity is the derivative of position:
$$a = frac{dv}{dt}, qquad v = frac{ds}{dt}$$
Step 1. With constant ( a ), integrate: ( v(t) = int a,dt = at + C ). At ( t = 0 ), ( v = u ), so ( C = u ) and
$$v = u + at$$
Step 2. Integrate again: ( s(t) = int (u + at),dt = ut + tfrac12at^2 + C ). Measuring displacement from the start, ( s(0) = 0 ), so
$$s = ut + tfrac12at^2$$
Step 3. Since velocity changes linearly, the average velocity is the average of the start and end: ( frac{u + v}{2} ). Multiply by time to get ( s = frac12(u + v)t ).
Step 4. From step 1, ( t = frac{v – u}{a} ). Substitute into step 3:
$$s = frac{(u + v)(v - u)}{2a} = frac{v^2 - u^2}{2a} quadRightarrowquad v^2 = u^2 + 2as$$
Step 5. Substitute ( u = v – at ) into step 2 to get ( s = vt – frac12at^2 ).
The Fundamental Theorem of Calculus is what makes “displacement = integral of velocity” work. When the acceleration is not constant, you go back to integrating directly, which the position, velocity and acceleration calculator does for any ( s(t) ).
Example 1: free fall
A ball is dropped from rest off a bridge and hits the water 3 s later. How high is the bridge, and how fast is the ball moving at impact?
Take down as positive. Then ( u = 0 ), ( a = 9.81 ) m/s², ( t = 3 ) s.
$$s = 0 + tfrac12(9.81)(3)^2 = 44.145 text{ m}$$
$$v = 0 + 9.81(3) = 29.43 text{ m/s}$$
The bridge is about 44 m high and the ball hits at about 29 m/s (106 km/h).
Example 2: braking distance
A car travelling at 25 m/s (90 km/h) brakes with a constant deceleration of 5 m/s². How far does it travel before stopping?
Known: ( u = 25 ), ( v = 0 ), ( a = -5 ). Time isn’t needed, so use ( v^2 = u^2 + 2as ):
$$0 = 625 + 2(-5)s quadRightarrowquad s = frac{625}{10} = 62.5 text{ m}$$
Notice that braking distance grows with the square of speed. At 50 m/s the same car would need 250 m, four times as far.
Example 3: thrown upward
A ball is thrown straight up at 15 m/s. How high does it go, and when does it come back down to the hand?
Take up as positive, so ( a = -9.81 ). At the top, ( v = 0 ):
$$0 = 15^2 + 2(-9.81)s quadRightarrowquad s = frac{225}{19.62} approx 11.47 text{ m}$$
For the return, displacement is zero: ( 0 = 15t – 4.905t^2 = t(15 – 4.905t) ), so ( t = 0 ) or ( t = frac{15}{4.905} approx 3.06 ) s. The ball is back in the hand after about 3.06 s, and by symmetry it reaches the top at half that time, 1.53 s.
Example 4: finding acceleration
A sprinter goes from rest to 10 m/s over 20 m. Using ( v^2 = u^2 + 2as ): ( 100 = 0 + 40a ), so ( a = 2.5 ) m/s². Then ( v = u + at ) gives ( t = frac{10}{2.5} = 4 ) s.
Kinematic equations calculator
Enter any three values and leave two blank. The calculator below picks the right equations and shows which ones it used.
For a full-page version with more examples, open the kinematic equations calculator. For two-dimensional motion, see projectile motion.
Common mistakes
- Using them when acceleration changes. A rocket burning fuel or a car with a varying push doesn’t have constant ( a ). The equations then give wrong answers.
- Inconsistent signs. If up is positive, gravity is ( -9.81 ). Mixing conventions mid-problem is the most common error.
- Mixing units. Convert km/h to m/s (divide by 3.6) before plugging in.
- Distance vs displacement. ( s ) is displacement. A ball thrown up and caught has ( s = 0 ) even though it travelled 23 m in total.
- Ignoring the second root. Quadratic equations in t can give two times. Both may be meaningful, such as passing the same height on the way up and down.
Practice problems
- A train accelerates from 10 m/s to 30 m/s in 8 s. Find its acceleration and the distance covered.
- A stone is dropped down a well and hits the water after 2.5 s. How deep is the well? (Use ( g = 9.81 ) m/s².)
- A cyclist at 8 m/s brakes to rest over 16 m. Find the deceleration and the stopping time.
- A ball is thrown up at 20 m/s. What is its maximum height?
- A car starts from rest with ( a = 3 ) m/s². How long does it take to cover 150 m?
Answers: (1) ( a = 2.5 ) m/s², ( s = 160 ) m; (2) about 30.66 m; (3) ( a = -2 ) m/s², ( t = 4 ) s; (4) about 20.39 m; (5) ( t = 10 ) s.
FAQ
What are the 4 kinematic equations?
Many courses list four: ( v = u + at ), ( s = ut + frac12at^2 ), ( v^2 = u^2 + 2as ) and ( s = frac12(u + v)t ). The fifth, ( s = vt – frac12at^2 ), is less common but completes the set.
When can I use the kinematic equations?
Only when acceleration is constant and the motion is along a straight line. Projectile motion uses them separately in the horizontal and vertical directions.
What is the difference between kinematics and dynamics?
Kinematics describes motion (position, velocity, acceleration) without asking why. Dynamics explains it using forces and Newton’s laws, such as ( F = ma ).
Why is g negative in some problems?
Gravity points down. If you choose up as the positive direction, then ( a = -9.81 ) m/s². If you choose down as positive, it’s ( +9.81 ).
How are the kinematic equations related to calculus?
They come from integrating constant acceleration once to get velocity and twice to get position. Calculus also handles the cases they can’t, where acceleration varies with time.
Further reading
- 3.4 Motion with Constant Acceleration (OpenStax University Physics Volume 1) — the kinematic equations with derivations and worked problems.
- Equations of motion (Wikipedia) — SUVAT equations, their calculus derivation and extensions to rotation.
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