Equations of Motion

Three equations that solve every straight-line, constant-acceleration problem.

16 minUpdated 21 October 2026v = u + atx = ut + ½at²v² = u² + 2axposition vs distance

Lesson at a glanceShow the key points

For constant acceleration: v = u + at, x = ut + ½at², v² = u² + 2ax.
They follow from integrating a = dv/dt and v = dx/dt.
x is the position at time t, not the distance travelled; they differ if the particle turns back.
Signs encode direction: u, v, a and x are positive or negative along the chosen axis.
Slope of v–t is a; area under v–t is displacement; displacement in the last 1 s of a t-second motion is u(1 s) + ½a(2t − 1 s)(1 s).

If the acceleration never changes, the whole future of a straight-line motion is fixed by just two numbers: where it starts moving from and how fast. Three short equations do all the work.

Setting up

Put the line of motion along the x-axis, start the clock at $t = 0$ and, usually, put the origin where the particle is at that moment. Let the acceleration $a$ be constant, the velocity at $t = 0$ be $u$ , and the velocity at time $t$ be $v$ .

Deriving the equations

Derivation

From a = dv/dt to three equations

Reveal each step to see where the equations come from. Only the calculus of the previous chapter is needed.

Equations of motion (constant a)

\begin{aligned} v &= u + at &(3.12)\\ x &= ut + \tfrac12 at^2 &(3.13)\\ v^2 &= u^2 + 2ax &(3.14)\end{aligned}

Interactive

Motion lab: one cart, three graphs

Choose an initial velocity and a constant acceleration, then run. The position, velocity and acceleration graphs share a time axis; the green area under v–t is the displacement so far.

t

0.00s

x

0.00m

v

6.00m/s

distance covered0.00m

turns around at4.00s

Initial velocity

u

6.0m/s

Acceleration

a

-1.5m/s²

Time0.00s

Signs carry direction: $u$ , $v$ , $a$ and $x$ are each positive or negative according to whether they point along +x or −x.

Worked example

A steadily accelerating trolley

A trolley moving at 3.0 m/s along +x picks up speed at a constant 0.40 m/s². (a) How far does it go in the next 5.0 s? (b) How long until it reaches 9.0 m/s? (c) How far has it gone by then?

Worked example

Displacement in the last second

A particle starts with velocity $u$ and constant acceleration $a$ . (a) Find its displacement during the last 1 second of a $t$ -second journey. (b) Evaluate for u = 4 m/s, a = 3 m/s², t = 8 s.

Worked example

Drawing a bow

An archer draws an arrow back 0.50 m. When released, the string pushes it over that 0.50 m and it leaves at 40 m/s. Assuming constant acceleration during the push, how large is it?

What to remember

1Constant a: $v = u + at$ , $x = ut + \tfrac12at^2$ , $v^2 = u^2 + 2ax$ .
2Pick the equation that skips the quantity you don’t know or don’t need.
3Signs encode direction; x is position, not distance travelled.
4Slope of v–t = a; area under v–t = displacement.

Check yourself

A car at 20 m/s brakes at a constant 5 m/s². How far does it travel before stopping?

Starting from rest with constant acceleration, a body covers distance d in the first second. How far does it go in the first 3 seconds?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadThe kinematic equationsThe Physics Classroom ReadMotion with constant accelerationOpenStax University Physics §3.4 ReadVelocity and displacement from acceleration (calculus)OpenStax University Physics §3.6 ReadConstant-acceleration motionHyperPhysics ReadOne-dimensional motion (unit)Khan Academy

PreviousAcceleration Next Free Fall

Equations of Motion

Three equations that solve every straight-line, constant-acceleration problem.

16 minUpdated 21 October 2026v = u + atx = ut + ½at²v² = u² + 2axposition vs distance

Lesson at a glanceShow the key points

For constant acceleration: v = u + at, x = ut + ½at², v² = u² + 2ax.
They follow from integrating a = dv/dt and v = dx/dt.
x is the position at time t, not the distance travelled; they differ if the particle turns back.
Signs encode direction: u, v, a and x are positive or negative along the chosen axis.
Slope of v–t is a; area under v–t is displacement; displacement in the last 1 s of a t-second motion is u(1 s) + ½a(2t − 1 s)(1 s).

Deriving the equations

Derivation

From a = dv/dt to three equations

Reveal each step to see where the equations come from. Only the calculus of the previous chapter is needed.

Equations of motion (constant a)

\begin{aligned} v &= u + at &(3.12)\\ x &= ut + \tfrac12 at^2 &(3.13)\\ v^2 &= u^2 + 2ax &(3.14)\end{aligned}

Interactive

Motion lab: one cart, three graphs

Choose an initial velocity and a constant acceleration, then run. The position, velocity and acceleration graphs share a time axis; the green area under v–t is the displacement so far.

t

0.00s

x

0.00m

v

6.00m/s

distance covered0.00m

turns around at4.00s

Initial velocity

u

6.0m/s

Acceleration

a

-1.5m/s²

Time0.00s

Signs carry direction:

u

v

a

and

x

are each positive or negative according to whether they point along +x or −x.

Worked example

A steadily accelerating trolley

A trolley moving at 3.0 m/s along +x picks up speed at a constant 0.40 m/s². (a) How far does it go in the next 5.0 s? (b) How long until it reaches 9.0 m/s? (c) How far has it gone by then?

Worked example

Displacement in the last second

A particle starts with velocity $u$ and constant acceleration $a$ . (a) Find its displacement during the last 1 second of a $t$ -second journey. (b) Evaluate for u = 4 m/s, a = 3 m/s², t = 8 s.

Worked example

Drawing a bow

An archer draws an arrow back 0.50 m. When released, the string pushes it over that 0.50 m and it leaves at 40 m/s. Assuming constant acceleration during the push, how large is it?

What to remember

1Constant a: $v = u + at$ , $x = ut + \tfrac12at^2$ , $v^2 = u^2 + 2ax$ .
2Pick the equation that skips the quantity you don’t know or don’t need.
3Signs encode direction; x is position, not distance travelled.
4Slope of v–t = a; area under v–t = displacement.

Check yourself

A car at 20 m/s brakes at a constant 5 m/s². How far does it travel before stopping?

Starting from rest with constant acceleration, a body covers distance d in the first second. How far does it go in the first 3 seconds?

Setting up#

Deriving the equations#

Check yourself

Keep exploring

Setting up#

Deriving the equations#

Check yourself

Keep exploring

Setting up

Deriving the equations

Setting up

Deriving the equations