Motion in a Plane

Two straight-line problems wearing a trench coat.

12 minUpdated 21 October 2026independence of axescomponent equations

Lesson at a glanceShow the key points

v = (dx/dt) î + (dy/dt) ĵ and a = (dvₓ/dt) î + (dv_y/dt) ĵ.
With constant acceleration each axis obeys its own equations: vₓ = uₓ + aₓt, x = uₓt + ½aₓt², and the same for y.
The x and y motions are independent; they are linked only by sharing the time t.

Motion in two dimensions looks complicated, but it hides a simplification: what happens along x has nothing to do with what happens along y.

Coordinates in a plane

When a particle is free to move in a plane, call it the x–y plane and pick any two perpendicular directions in it as the axes. Start the clock at a convenient moment and put the origin where the particle is at t = 0. Its position at any time is then fully specified by the pair $(x, y)$ .

Between $t$ and $t + \Delta t$ the particle moves from $(x, y)$ to $(x + \Delta x, y + \Delta y)$ , so its displacement splits into two perpendicular parts:

\Delta\vec r = \Delta x\,\hat\imath + \Delta y\,\hat\jmath \quad\Rightarrow\quad \vec v = \lim_{\Delta t\to0}\frac{\Delta\vec r}{\Delta t} = \frac{dx}{dt}\hat\imath + \frac{dy}{dt}\hat\jmath \quad (3.15)

Differentiating once more gives the acceleration, component by component:

v_x = \frac{dx}{dt},\ \ v_y = \frac{dy}{dt} \qquad a_x = \frac{dv_x}{dt},\ \ a_y = \frac{dv_y}{dt} \qquad (3.16\text{–}3.19)

The big idea: split it in two

The relations between $x$ , $v_x$ , $a_x$ are exactly the straight-line relations from earlier. The same is true for $y$ , $v_y$ , $a_y$ . So if the acceleration is constant, each axis obeys its own set of equations:

Constant acceleration in a plane(3.20, 3.21)

\begin{aligned} v_x &= u_x + a_xt & v_y &= u_y + a_yt \\ x &= u_xt + \tfrac12a_xt^2 & y &= u_yt + \tfrac12a_yt^2 \\ v_x^2 &= u_x^2 + 2a_xx & v_y^2 &= u_y^2 + 2a_yy \end{aligned}

Interactive

Two independent straight-line motions

Set the initial velocity u and constant acceleration a by dragging their arrowheads in the small panel. In the big panel the coloured beads on the axes are the particle’s x and y ‘shadows’: each obeys its own straight-line equations.

Set u and a

a is drawn at 2× scale

t

2.00s

x,\ v_x

15.0 m, 9.0 m/s

y,\ v_y

4.0 m, 4.0 m/s

|\vec v|

9.85m/s

Time2.00s

u = (6.0, 0.0) m/s · a = (1.5, 2.0) m/s²

Worked example

A slanting push

A particle starts at the origin moving at 6.0 m/s along +x. It has a constant acceleration of 2.5 m/s² directed at 53° to the x-axis. Find its velocity and position at t = 2.0 s. (Use cos 53° ≈ 0.6, sin 53° ≈ 0.8.)

What to remember

1In a plane, treat x and y as two separate straight-line motions sharing one clock.
2Resolve u and a into components, apply the 1D equations on each axis, then recombine.

Check yourself

A particle has a = 3ĵ m/s² and starts with u = 4î m/s. What is its x-velocity after 5 s?

If u and a point in different directions, the path in a plane is generally…

Keep exploring

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

WatchVectors and 2D MotionCrash Course Physics #4 ReadDisplacement and velocity vectorsOpenStax University Physics §4.1 ReadTwo-dimensional motion (unit)Khan Academy

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Motion in a Plane

Two straight-line problems wearing a trench coat.

12 minUpdated 21 October 2026independence of axescomponent equations

Lesson at a glanceShow the key points

v = (dx/dt) î + (dy/dt) ĵ and a = (dvₓ/dt) î + (dv_y/dt) ĵ.
With constant acceleration each axis obeys its own equations: vₓ = uₓ + aₓt, x = uₓt + ½aₓt², and the same for y.
The x and y motions are independent; they are linked only by sharing the time t.

Coordinates in a plane

(x, y)

Between

t

and

t + \Delta t

the particle moves from

(x, y)

(x + \Delta x, y + \Delta y)

, so its displacement splits into two perpendicular parts:

\Delta\vec r = \Delta x\,\hat\imath + \Delta y\,\hat\jmath \quad\Rightarrow\quad \vec v = \lim_{\Delta t\to0}\frac{\Delta\vec r}{\Delta t} = \frac{dx}{dt}\hat\imath + \frac{dy}{dt}\hat\jmath \quad (3.15)

Differentiating once more gives the acceleration, component by component:

v_x = \frac{dx}{dt},\ \ v_y = \frac{dy}{dt} \qquad a_x = \frac{dv_x}{dt},\ \ a_y = \frac{dv_y}{dt} \qquad (3.16\text{–}3.19)

The big idea: split it in two

The relations between

x

v_x

a_x

are exactly the straight-line relations from earlier. The same is true for

y

v_y

a_y

. So if the acceleration is constant, each axis obeys its own set of equations:

Constant acceleration in a plane(3.20, 3.21)

\begin{aligned} v_x &= u_x + a_xt & v_y &= u_y + a_yt \\ x &= u_xt + \tfrac12a_xt^2 & y &= u_yt + \tfrac12a_yt^2 \\ v_x^2 &= u_x^2 + 2a_xx & v_y^2 &= u_y^2 + 2a_yy \end{aligned}

Interactive

Two independent straight-line motions

Set u and a

a is drawn at 2× scale

t

2.00s

x,\ v_x

15.0 m, 9.0 m/s

y,\ v_y

4.0 m, 4.0 m/s

|\vec v|

9.85m/s

Time2.00s

u = (6.0, 0.0) m/s · a = (1.5, 2.0) m/s²

Worked example

A slanting push

What to remember

1In a plane, treat x and y as two separate straight-line motions sharing one clock.
2Resolve u and a into components, apply the 1D equations on each axis, then recombine.

Check yourself

A particle has a = 3ĵ m/s² and starts with u = 4î m/s. What is its x-velocity after 5 s?

If u and a point in different directions, the path in a plane is generally…

Coordinates in a plane#

The big idea: split it in two#

Check yourself

Keep exploring

Coordinates in a plane#

The big idea: split it in two#

Check yourself

Keep exploring

Coordinates in a plane

The big idea: split it in two

Coordinates in a plane

The big idea: split it in two