The kinematic equations are four formulas that link displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t) for motion in a straight line with constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as and s = ½(u + v)t. Each equation leaves out one of the five quantities, so if you know any three you can find the other two.
They are often called the SUVAT equations, after the five symbols, or the equations of motion in Class 11 and JEE physics. This page covers the formulas, where they come from, how to pick the right one, and three solved problems.
The four kinematic equations
| No. | Equation | Quantity it leaves out |
|---|---|---|
| 1 | v = u + at | s (displacement) |
| 2 | s = ut + ½at² | v (final velocity) |
| 3 | v² = u² + 2as | t (time) |
| 4 | s = ½(u + v)t | a (acceleration) |
Symbols and SI units:
- s = displacement (m), measured from the starting point, with a sign
- u = initial velocity (m/s)
- v = final velocity (m/s)
- a = constant acceleration (m/s²); negative when it opposes the chosen positive direction
- t = time taken (s)
A fifth form, s = vt − ½at², leaves out u. It is less common but handy when you know the final speed and not the starting one.
When do the kinematic equations apply?
Two conditions must both hold:
- Acceleration is constant in size and direction for the whole interval you apply them to.
- Motion is along one straight line, so every vector can be written as a plus or minus number.
If the acceleration changes (a car with a varying throttle, a spring, air drag at high speed), you need calculus: v = ∫a dt and s = ∫v dt. If the motion is in two dimensions, as in projectile motion, you apply the equations separately along each axis. For the idea of motion along a line itself (position, displacement, the sign convention), see the page on rectilinear motion.
Derivation of the kinematic equations
Equation 1 from the definition of acceleration
Acceleration is the rate of change of velocity. With constant acceleration, the velocity changes from u to v in time t, so
a = (v − u) / t, which rearranges to v = u + at.
Equation 4 from the area under the v-t graph
For constant acceleration the velocity-time graph is a straight line from u (at t = 0) to v (at time t). Displacement equals the area under that line, which is a trapezium with parallel sides u and v and width t:
s = ½(u + v)t
Equation 2 by removing v
Substitute v = u + at into equation 4:
s = ½(u + u + at)t = ½(2u + at)t, so s = ut + ½at².
On the graph this is the rectangle ut plus the triangle ½ × t × at on top of it.
Equation 3 by removing t
From equation 1, t = (v − u)/a. Put this into equation 4:
s = ½(u + v)(v − u)/a = (v² − u²)/(2a), so v² = u² + 2as.
Which kinematic equation should you use?
List the three quantities the question gives you and the one it asks for. The fifth quantity is the one that does not appear, and that tells you the equation.
| Not given and not asked | Use | Typical question |
|---|---|---|
| s | v = u + at | How fast after 5 s? |
| v | s = ut + ½at² | How far in 5 s? |
| t | v² = u² + 2as | Stopping distance from a known speed |
| a | s = ½(u + v)t | Distance when start and end speeds and time are known |
| u | s = vt − ½at² | Distance covered in the last t seconds before reaching v |
Tip: always convert km/h to m/s first (divide by 3.6) and fix one direction as positive before you write any sign.
Worked examples
Example 1: stopping distance of a braking car
A car moving at 72 km/h brakes with a constant deceleration of 5 m/s². How far does it travel before stopping, and how long does it take?
- u = 72 ÷ 3.6 = 20 m/s, v = 0, a = −5 m/s². Time is not given, so use v² = u² + 2as.
- 0 = 20² + 2(−5)s, so 0 = 400 − 10s, giving s = 40 m.
- For the time, v = u + at: 0 = 20 − 5t, so t = 4 s.
- Check with equation 4: s = ½(20 + 0)(4) = 40 m. It matches.
Note that stopping distance goes with u². At 144 km/h (40 m/s) the same brakes need 1600/10 = 160 m, four times as far, not twice.
Example 2: ball thrown straight up
A ball is thrown vertically upward at 15 m/s. Take g = 9.81 m/s² and ignore air resistance. Find the maximum height and the time to return to the hand.
- Take up as positive: u = +15 m/s, a = −9.81 m/s², and at the top v = 0.
- Height: 0 = 15² − 2(9.81)h, so h = 225 ÷ 19.62 = 11.47 m.
- Time to the top: 0 = 15 − 9.81t, so t = 15 ÷ 9.81 = 1.53 s.
- Total time: set s = 0 in s = ut + ½at², giving 0 = t(15 − 4.905t), so t = 15 ÷ 4.905 = 3.06 s, exactly twice the rise time.
The acceleration is −9.81 m/s² the whole time, including at the top where the velocity is momentarily zero. That is why one set of equations covers both the rise and the fall.
Example 3: a train leaving a station
A train starts from rest and accelerates uniformly at 0.5 m/s² for 60 s. Find its final speed and the distance covered.
- u = 0, a = 0.5 m/s², t = 60 s.
- v = u + at = 0 + 0.5 × 60 = 30 m/s (108 km/h).
- s = ut + ½at² = 0 + ½ × 0.5 × 60² = 0.25 × 3600 = 900 m.
- Check: s = ½(0 + 30)(60) = 900 m.
The acceleration comes from the net force on the train, which is where Newton’s second law takes over: kinematics describes the motion, dynamics explains its cause.
Free fall: a special case
An object dropped from rest near the Earth’s surface, with air resistance neglected, falls with constant acceleration g ≈ 9.81 m/s². Put u = 0 and a = g (taking down as positive), and the equations become:
- v = gt
- h = ½gt²
- v² = 2gh
Example: a stone dropped from a 20 m roof. Time to fall: t = √(2h/g) = √(40/9.81) = √4.08 = 2.02 s. Impact speed: v = √(2 × 9.81 × 20) = √392.4 = 19.8 m/s. Check: v = gt = 9.81 × 2.02 = 19.8 m/s.
Common mistakes
- Mixing signs. Deceleration is just acceleration pointing against the motion. Choose a positive direction once and keep it.
- Using the equations when a is not constant. Split the motion into stages with constant a, or integrate.
- Confusing distance with displacement. For the ball in Example 2, displacement at landing is 0 m but distance travelled is 2 × 11.47 = 22.94 m.
- Forgetting unit conversion. 72 km/h is 20 m/s, not 72 m/s.
Linear and rotational forms
Rotation about a fixed axis with constant angular acceleration α uses the same four equations with θ, ω₀, ω and α in place of s, u, v and a: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ and θ = ½(ω₀ + ω)t. The two sets are tied together by v = ωr, explained in the relationship between linear and angular velocity.
FAQs
What are the 4 kinematic equations?
They are v = u + at, s = ut + ½at², v² = u² + 2as and s = ½(u + v)t, where u and v are initial and final velocity, a is constant acceleration, s is displacement and t is time.
Why are they called SUVAT equations?
The name comes from the five symbols they use: s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time). Each equation contains four of the five.
Can kinematic equations be used when acceleration changes?
No. They are derived assuming constant acceleration. For changing acceleration, integrate a(t) to get velocity and then displacement, or split the motion into stages where a is constant.
Which kinematic equation has no time in it?
v² = u² + 2as. It is the one to use for stopping distances and heights reached when the time is neither given nor asked.
Do the kinematic equations work for projectile motion?
Yes, if you apply them separately to the horizontal and vertical directions. Horizontally a = 0, so the velocity is constant; vertically a = −g.