Kinematics
Describing motion in one, two and three dimensions without asking why. 53 topics, from beginner to advanced. Open the interactive version with labs and quizzes
Kinematics describes how objects move, using position, displacement, velocity, speed and acceleration, without asking what causes the motion. It covers straight-line motion, graphs of position and velocity against time, and projectile motion under constant gravity. These ideas are the starting point for almost every other mechanics topic.
How the topics connect
Kinematics describes how things move without asking what causes the motion. It starts with position, velocity and acceleration in one dimension, then extends to projectiles and circular motion. The same graphs and equations return throughout physics, so mastering them here helps with every later topic.
Real-world uses
Estimate stopping distances, average speeds on a journey, and how long a vehicle takes to accelerate from rest to a given speed.
Analyse the path of a thrown ball, a jump or a sprint using position-time and velocity-time graphs.
Read displacement and velocity from frame-by-frame footage of a moving object to test a physical model.
Plan smooth motion profiles for a robot arm or an animated object by specifying how velocity and acceleration change over time.
Engineers use stopping distance and braking formulas to design road layouts, speed limits and crash barriers.
Coaches and players use projectile ideas to judge the launch angle and speed needed for a good throw, kick or shot.
Rest and motion
An object moves if its position changes relative to a reference frame.
Rest and motion are relative: an object is moving only compared with some reference. A passenger is at rest relative to the train but moving relative to the platform. Describing motion therefore always needs a frame and a clock.
A book on a table is at rest relative to the room but moves at about 465 m/s relative to Earth's axis at the equator.
Saying "it is moving" without naming a reference is incomplete.
Reference frames
The coordinate system and clock from which motion is described.
A reference frame is a coordinate system with an origin, axes and a clock. Positions and velocities depend on the frame, but the physical laws (in inertial frames) do not. Choosing a convenient frame, such as one with the x-axis along the slope, can simplify a problem a lot.
For a ball thrown in a moving train, the train frame sees a simple vertical throw while the ground frame sees a parabola.
Mixing coordinates from two frames in one equation is a classic source of sign errors.
Distance vs displacement
Distance is path length; displacement is net change in position.
Distance is the total path length and is never negative. Displacement is the straight-line change in position from start to end, with direction, and can be zero or negative.
Running one lap of a 400 m track: distance 400 m, displacement 0.
Average speed uses distance; average velocity uses displacement. Don't swap them.
Speed vs velocity
Speed is a scalar; velocity includes direction.
v = Δx/ΔtSpeed says how fast; velocity says how fast and which way. A car going round a circle at constant speed has changing velocity because its direction keeps changing.
A car that does 60 km/h north then 60 km/h south has the same speed but velocities of opposite sign.
Velocity can be negative (direction), speed cannot.
Average velocity
Total displacement divided by total time.
v̄ = Δx/ΔtAverage velocity is total displacement divided by total time. It ignores what happens in between, so it can be zero for a round trip. For constant acceleration it equals (u + v)/2.
Travelling 100 m east in 20 s gives 5 m/s east. A trip east and back in the same 40 s gives 0.
Average velocity is not the average of speeds unless time intervals are equal.
Instantaneous velocity
Velocity at one instant, the derivative of position.
v = dx/dtInstantaneous velocity is the limit of Δx/Δt as Δt → 0, the slope of the tangent on an x–t graph, v = dx/dt. A speedometer shows its magnitude.
If x = 3t² then v = 6t, so at t = 2 s the velocity is 12 m/s.
A velocity of zero for an instant (top of a throw) does not mean zero acceleration.
Acceleration
The rate at which velocity changes.
a = dv/dtAcceleration is the rate of change of velocity, a = dv/dt. Since velocity is a vector, a change in direction also counts as acceleration. Speeding up, slowing down and turning are all accelerations.
A car increases from 0 to 27 m/s in 6 s: a = 4.5 m/s².
Negative acceleration doesn't always mean slowing down; it depends on the sign of velocity too.
Uniform motion
Constant velocity means zero acceleration.
x = x₀ + vtIn uniform motion velocity is constant, so acceleration is zero and the x–t graph is a straight line: x = x₀ + vt. No net force acts (Newton's first law).
A train at constant 25 m/s covers 900 m in 36 s.
Real objects rarely maintain this for long because of friction.
Position–time graphs
The slope of x–t gives velocity.
On an x–t graph, the slope at any point is the velocity. A straight line means constant velocity, a curve means acceleration, and a flat line means rest. Steeper means faster.
A line rising from (0, 0) to (4 s, 20 m) has slope 5 m/s.
A hill-shaped x–t graph is not the path of the object; it's position against time.
Velocity–time graphs
Slope gives acceleration; area gives displacement.
On a v–t graph the slope is acceleration and the area between the curve and the axis is displacement (areas below the axis are negative). A horizontal line means constant velocity.
A triangle from 0 to 10 m/s over 5 s has area ½·5·10 = 25 m.
Crossing the time axis means the direction reverses, not that the object returns to the start.
Equations of motion: v = u + at
Final velocity under constant acceleration.
v = u + atFor constant acceleration a, velocity grows linearly with time: v = u + at, where u is the initial velocity. It comes from integrating a = dv/dt. Use it when you know u, a, t and want v.
A bike at 2 m/s accelerating 0.5 m/s² for 8 s reaches 6 m/s.
It only works for constant acceleration.
Equations of motion: s = ut + ½at²
Displacement under constant acceleration.
s = ut + ½at²Displacement under constant acceleration is s = ut + ½at², the area under the v–t trapezoid. The first term is what you would cover at constant u, the second is the extra from accelerating.
From rest, a = 2 m/s² for 5 s: s = ½·2·25 = 25 m.
Keep sign convention: if a opposes u, a is negative.
Equations of motion: v² = u² + 2as
Relates velocities and displacement without time.
v² = u² + 2asEliminating time between the two earlier equations gives v² = u² + 2as. It is the fastest route when time is unknown or irrelevant, and it is the kinematic form of the work–energy theorem.
Braking from 20 m/s at −5 m/s²: 0 = 400 − 10s → s = 40 m.
Taking the square root gives ±; choose the sign by physical reasoning.
Free fall
Motion under gravity alone, with acceleration about 9.8 m/s².
y = ½gt²In free fall only gravity acts. All bodies accelerate downward at about 9.8 m/s² regardless of mass, because the weight and the inertia both scale with mass. The equations of constant acceleration apply with a = g.
Dropped from rest, an object falls 4.9 m in 1 s and 19.6 m in 2 s.
Air resistance is ignored; a feather falls slowly in air but together with a hammer in a vacuum.
Acceleration due to gravity
Near Earth's surface all objects fall with the same acceleration.
g ≈ 9.8 m/s²Near Earth's surface g ≈ 9.8 m/s² downward, varying slightly with latitude and altitude (9.78 at the equator, 9.83 at the poles). It is the same for all objects in free fall. On the Moon g ≈ 1.6 and on Jupiter ≈ 24.8 m/s².
On the Moon you could jump about six times higher with the same effort.
g is an acceleration, not a force; the weight is mg.
Vertical throw upward
Rising body decelerates at g, stops at the top, then falls.
H = u²/2gA body thrown up decelerates at g, reaches zero velocity at the top, and then falls back. The motion is symmetric: it returns to the start with the same speed it left. Maximum height H = u²/2g, rise time u/g.
Thrown up at 20 m/s: top at 20 m after about 2.04 s, back at 4.08 s.
At the top the velocity is zero but the acceleration is still g downward.
Time of flight (vertical)
Up and down takes twice the rise time.
T = 2u/gFor a throw up and return to the starting height, total time T = 2u/g. If it lands at a different height solve s = ut − ½gt² as a quadratic, taking the positive root.
Thrown at 15 m/s from the ground: T = 2·15/9.8 ≈ 3.06 s.
A quadratic has two roots; one is often negative or before launch.
Relative velocity in 1D
Velocity of A as seen from B is the difference of velocities.
v_AB = v_A − v_BThe velocity of A relative to B is v_AB = v_A − v_B in the same direction convention. Overtaking and head-on motions become simple by working in the frame of one body.
Cars at 30 m/s and 20 m/s in the same direction: relative speed 10 m/s. Head-on: 50 m/s.
Define which direction is positive before you subtract.
Stopping distance
Distance covered while braking grows with the square of speed.
d = v²/2aBraking distance d = v²/(2a) grows with the square of speed: double the speed means four times the distance. Total stopping distance adds the thinking distance v·t_reaction.
At 15 m/s, 6 m/s² braking: 18.75 m. At 30 m/s: 75 m.
Wet roads reduce a and so increase d considerably.
Reaction time and braking
Total stopping distance adds thinking distance and braking distance.
Before braking begins, the car travels at constant speed during the driver's reaction time (about 0.7 s). So total stopping distance = v·t_r + v²/2a.
At 20 m/s with t_r = 0.7 s and a = 5 m/s²: 14 + 40 = 54 m.
Tired or distracted drivers have longer reaction times, which adds metres at high speed.
Non-uniform acceleration
When a varies, integrate calculus-style to get v and x.
v = ∫a dtWhen a changes with time, the constant-acceleration formulas fail. Instead integrate: v = v₀ + ∫a dt, x = x₀ + ∫v dt, or read areas on graphs.
If a = 6t, then v = 3t² and x = t³ (from rest at origin).
Never apply s = ut + ½at² to varying acceleration.
Motion graphs from calculus
Slopes and areas are derivatives and integrals of the motion graphs.
The three graphs are linked: x–t slope gives v, v–t slope gives a, and areas go the other way. At a peak of x(t), v = 0; at a peak of v(t), a = 0.
Given x = t³ − 6t², v = 3t² − 12t is zero at t = 4 s, where x is at a minimum.
Zero slope means zero of the next quantity, not zero of the quantity itself.
Terminal velocity
Falling speed levels off when drag equals weight.
mg = kvA falling body speeds up until drag equals weight, then falls at constant terminal velocity. With drag kv: v_t = mg/k. Skydivers reach about 55 m/s belly-down and slow to about 5 m/s with a parachute.
A raindrop reaches terminal velocity in a few metres, which is why rain doesn't injure you.
At terminal velocity the acceleration is zero but gravity still acts.
Two-dimensional motion
Independent horizontal and vertical motion combined.
In 2-D, horizontal and vertical motions are independent: apply the equations separately to each axis, linked only by time. Perpendicular components do not influence each other.
A ball rolling off a table keeps its horizontal speed while falling.
Don't mix x and y quantities in one equation.
Projectile motion
A launched body follows a parabola under gravity.
y = x tanθ − gx²/(2u²cos²θ)A projectile has constant horizontal velocity and constant downward acceleration g, so its path is a parabola. Split the initial velocity: ux = u cosθ, uy = u sinθ.
Launched at 20 m/s at 30°: ux = 17.3, uy = 10 m/s.
Acceleration is g everywhere, including the top, where only the vertical velocity is zero.
Time of flight
Time a projectile stays in the air.
T = 2u sinθ/gFor a launch and landing at the same height, T = 2u sinθ/g. It depends only on the vertical launch component.
At 30 m/s and 53°: T ≈ 2·24/9.8 ≈ 4.9 s.
Time is set by the vertical motion alone; horizontal speed doesn't change it.
Maximum height
Highest point reached by a projectile.
H = u² sin²θ/2gThe height at the top is H = u²sin²θ/(2g), where the vertical velocity is zero. It increases with the square of the vertical component.
u = 30 m/s at 45°: H = 900·0.5/19.6 ≈ 23 m.
At the top the speed is u cosθ, not zero.
Horizontal range
Horizontal distance covered by a projectile on level ground.
R = u² sin2θ/gOn level ground R = u² sin2θ/g = ux·T. It depends on speed squared and the sine of twice the angle.
u = 30 m/s at 45°: R = 900/9.8 ≈ 92 m.
The formula assumes launch and landing at the same level.
Complementary angles range
Launch angles θ and 90°−θ give the same range.
projectileBecause sin2θ = sin(180° − 2θ), launch angles θ and 90° − θ give the same range, with the higher angle having more height and time.
30° and 60° both reach 0.866u²/g.
They share range, not flight time or height.
Maximum range at 45°
On level ground the range is greatest at 45°.
R_max = u²/gSince sin2θ is largest (1) at θ = 45°, the maximum range is u²/g. With drag or a launch height above the target the best angle becomes smaller than 45°.
Throwing at 40 m/s: maximum range ≈ 163 m.
For a launch from a height the optimal angle is less than 45°.
Horizontal projectile
Object launched horizontally falls as in free fall while moving sideways.
projectileA body launched horizontally falls with y = ½gt² while moving at constant x = ut. The time to fall height h is √(2h/g), independent of the horizontal speed.
A ball rolled off a 1.25 m table at 4 m/s hits the floor after 0.5 s, 2 m from the base.
Fast or slow, the ball lands at the same time as one dropped from rest.
Projectile on an incline
Range along an incline depends on both launch and plane angles.
Choose axes along and perpendicular to the plane: gravity then has components g sinα and g cosα along each. Time of flight becomes T = 2u sin(θ − α)/(g cosα), and the range along the plane follows.
A ball thrown perpendicular to an incline of angle α stays in the air 2u/(g cosα).
Don't use horizontal-ground formulas on slopes.
Trajectory equation
Eliminating time gives y as a function of x.
projectileEliminating t from x = u cosθ·t and y = u sinθ·t − ½gt² gives y = x tanθ − gx²/(2u²cos²θ), a downward parabola. It lets you check if a projectile clears an obstacle at a given x.
At x = 20 m, u = 20 m/s, θ = 45°: y = 20 − 9.8·400/(2·400·0.5) ≈ 10.2 m.
Use the equation only for the path, not for timing.
Velocity at any point of projectile
Horizontal component is constant; vertical changes at g.
v = √(vx²+vy²)At any time vx = u cosθ and vy = u sinθ − gt. The speed is √(vx² + vy²) and the direction tanφ = vy/vx. Energy conservation gives speed at height y: v² = u² − 2gy.
At height 10 m launched at 20 m/s: v = √(400 − 196) ≈ 14.3 m/s.
Speed at equal heights is the same on the way up and down.
Effect of air resistance
Drag shortens range and makes the path asymmetric.
projectileDrag opposes velocity, so the range and height are smaller, the path becomes asymmetric (steeper on the way down), and the best angle drops below 45°. Large, fast and light objects are most affected.
A cricket ball might go 70 m where vacuum theory says 90 m.
Vacuum formulas overestimate real range; they are only an approximation for dense, slow objects.
Relative velocity in 2D
Vector subtraction of velocities.
v_AB = v_A − v_BIn two dimensions, v_AB = v_A − v_B is a vector subtraction. Components are subtracted separately, or a triangle of velocities is drawn.
Car A 20 m/s east and car B 15 m/s north: relative velocity of A with respect to B is (20, −15), magnitude 25 m/s.
Remember to subtract the reference body's velocity, not add.
River-boat problems
Resultant velocity of boat is vector sum of boat and current velocity.
The boat's ground velocity is the vector sum of its velocity relative to water and the river current. To cross in the shortest time, point the boat perpendicular to the bank; to land directly opposite, aim upstream so the current is cancelled: sinθ = vr/vb.
Boat speed 5 m/s, river 3 m/s, width 100 m: shortest time 20 s, drift 60 m.
Shortest time and shortest path are different strategies.
Rain and umbrella problem
Rain appears to fall at an angle depending on your own velocity.
The rain's velocity relative to you is v_rain − v_you. If you walk, the rain seems to slant toward you at angle tanθ = v_you/v_rain from the vertical, so tilt the umbrella forward.
Rain at 4 m/s vertically, you at 3 m/s: relative rain speed 5 m/s, tilt 37° forward.
The tilt depends on your speed, not the direction you face.
Uniform circular motion
Constant speed on a circle with changing direction.
v = ωrAn object in uniform circular motion moves with constant speed on a circle. Its velocity changes direction continuously, so it accelerates toward the centre.
A car circling at 10 m/s on a 50 m radius takes 31.4 s per lap.
Constant speed does not mean constant velocity.
Angular displacement and velocity
Angle swept and its rate of change.
ω = dθ/dtAngular displacement θ is in radians (arc length ÷ radius). Angular velocity ω = dθ/dt is in rad/s and is related to the linear speed by v = ωr.
A wheel turning at 120 rpm has ω = 120·2π/60 ≈ 12.6 rad/s.
Use radians in all formulas like v = ωr.
Period and frequency
Time for one revolution and revolutions per second.
T = 1/f = 2π/ωThe period T is the time for one revolution; frequency f = 1/T counts revolutions per second (hertz). ω = 2πf = 2π/T.
A fan with f = 5 Hz has T = 0.2 s and ω = 31.4 rad/s.
rpm must be converted to Hz or rad/s before use.
Centripetal acceleration
Acceleration directed to the centre of the circle.
a = v²/r = ω²rThe acceleration toward the centre is a = v²/r = ω²r. It changes direction, not speed. It is provided by a real force such as tension, gravity or friction.
At 20 m/s on a 100 m curve: a = 4 m/s².
Centripetal acceleration is a result, not an additional force on the free-body diagram.
Non-uniform circular motion
Tangential acceleration changes speed while radial changes direction.
a = √(a_r²+a_t²)If the speed changes, there is a tangential acceleration at (changing speed) as well as the radial v²/r. The total acceleration is the vector sum with magnitude √(ar² + at²).
A car speeding up in a roundabout feels pushed both inward and forward.
Don't forget at when speed varies, like a pendulum swinging.
Angular acceleration
Rate of change of angular velocity.
α = dω/dtAngular acceleration α = dω/dt is the rate of change of angular velocity and relates to tangential acceleration by at = αr. A spinning-up wheel has constant α under constant torque.
A wheel from 0 to 20 rad/s in 4 s has α = 5 rad/s².
rad/s² is not the same unit as m/s².
Rotational kinematics equations
Analogues of linear equations with θ, ω, α.
ω = ω₀ + αtWith constant α the equations mirror the linear ones: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ. Replace x→θ, v→ω, a→α.
A disc at 10 rad/s decelerates at 2 rad/s²: stops after 5 s turning 25 rad.
Use radians consistently.
Motion in polar coordinates
Radial and transverse components of velocity and acceleration.
v = ṙ r̂ + rθ̇ θ̂In polar coordinates the velocity has a radial part ṙ and a transverse part rθ̇. The acceleration has radial (r̈ − rθ̇²) and transverse (rθ̈ + 2ṙθ̇) parts; the last term is the Coriolis term.
For circular motion ṙ = 0, so the radial acceleration is −rθ̇² = −v²/r.
Unit vectors r̂ and θ̂ change direction as the particle moves.
Motion in three dimensions
Extend independent-axis analysis to x, y and z.
Treat x, y and z independently with constant-acceleration formulas, then combine as vectors. Speed is √(vx²+vy²+vz²).
A drone at (3, 4, 12) m/s has speed 13 m/s.
Add one component at a time; don't take a sum of speeds.
Parametric motion
Describe a path by x(t), y(t) and compute velocity by differentiation.
A path can be given by x(t), y(t). Velocity is (dx/dt, dy/dt) and acceleration the second derivatives. Eliminating t gives the shape of the path.
x = 2cos t, y = 2 sin t is a circle of radius 2; speed 2.
The same path can be travelled at different speeds.
Galilean transformation
Position and velocity transform between frames moving at constant velocity.
x' = x − vtBetween frames at relative speed v: x' = x − vt, t' = t, and velocities add: u' = u − v. These hold at everyday speeds but fail near the speed of light, where Lorentz transformations are needed.
A train at 20 m/s and a passenger walking 2 m/s forward: the ground sees 22 m/s.
Galilean rules assume time is universal, which relativity shows is not exactly true.
Frames: inertial vs non-inertial
Newton's laws hold in inertial frames; fictitious forces appear otherwise.
An inertial frame moves at constant velocity; Newton's laws hold in it. An accelerating or rotating frame is non-inertial, and extra fictitious forces (centrifugal, Coriolis, −ma) appear.
In a braking bus you are thrown forward: from the ground there's no force, only inertia.
Don't add fictitious forces when working in an inertial frame.
Coriolis and fictitious motion
Rotating frames show apparent deflection of moving bodies.
a = −2Ω×vIn a frame rotating at Ω, a moving body seems deflected by the Coriolis acceleration −2Ω×v and pushed outward by the centrifugal term. These explain the rotation direction of cyclones and the drift of long-range shells.
In the northern hemisphere moving air is deflected to the right, so cyclones spin counterclockwise.
Coriolis effect is negligible for sinks or baths on human scales.
Chase and meeting problems
Equate positions at the meeting time to solve for it.
Write the position of each body as a function of time and set the positions equal to find the meeting time. Be careful with different start times and directions.
A at 10 m/s starts 100 m behind B at 5 m/s: 10t = 100 + 5t → t = 20 s.
Check that the meeting time is positive and within the motion range.
Average vs instantaneous acceleration
Compare Δv/Δt over an interval with the derivative at a moment.
Average acceleration is Δv/Δt over an interval; instantaneous acceleration is the slope of the v–t graph at one moment. They coincide only if acceleration is constant.
Velocity going from 0 to 10 m/s in 5 s gives an average of 2 m/s², even if the acceleration was 4 m/s² at first and 0 later.
Don't use the constant-acceleration equations with an average of varying acceleration.
Frequently asked questions
What is the difference between speed and velocity?
Speed is how fast something moves and has no direction. Velocity is speed with a direction, so it is a vector and can be negative.
What is the difference between distance and displacement?
Distance is the total path length travelled. Displacement is the straight-line change in position from start to finish, so it can be smaller than the distance or zero for a round trip.
What does the slope of a position-time graph mean?
The slope gives the velocity at that instant. A straight line means constant velocity, and a curve means the velocity is changing.
Do I need calculus for kinematics?
Not at the start. Constant acceleration can be handled with algebra. Calculus gives the same results more generally when acceleration changes, and it links velocity and position through slopes and areas.
Why do projectiles follow a parabola?
Horizontal motion is uniform and vertical motion is uniformly accelerated under gravity. Eliminating time between the two gives a parabola, ignoring air resistance.
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