2D Kinematics

In one dimension, position, velocity, and acceleration are each described by one signed number. In two dimensions, each requires two components. This dimensional scaling is a property of vector quantities, unlike scalars (e.g. mass and time).

Most of first-year physics relies heavily on the vector formalism, so I suggest building a strong foundation with the concept. If it’s completely new to you, check out the Vectors page.

The hedgehog returns, now running a figure eight across a field. We again use vectors to describe its motion: position $\vec{r}$ points from the origin to the hedgehog, velocity $\vec{v}$ points the way it is running, and acceleration $\vec{a}$ points the way its velocity is changing. The definitions are the same as in one dimension, but it’s now clear why we need both a magnitude and direction to describe each quantity.

At 0.0 s the hedgehog is at r = ⟨0.0, 0.0⟩ m, moving at 3.6 m/s, holding its speed and heading straight.

position rvelocity vacceleration adecompose a

Component Motion

For simplicity, let’s again consider the case of constant acceleration, with the clock starting at $t = 0$ . As in 1D kinematics, a zero subscript marks a value at $t = 0$ :

\vec{v}(t) = \vec{v}_0 + \vec{a}t,

Components Hide

\begin{aligned} v_x(t) &= v_{0x} + a_xt,\\ v_y(t) &= v_{0y} + a_yt, \end{aligned}

\vec{r}(t) = \vec{r}_0 + \vec{v}_0t + \frac{1}{2}\vec{a}t^2.

Components Hide

\begin{aligned} x(t) &= x_0 + v_{0x}t + \frac{1}{2}a_xt^2,\\ y(t) &= y_0 + v_{0y}t + \frac{1}{2}a_yt^2. \end{aligned}

This notation is very concise, but we often like to consider the $x-$ and $y-$ directions independently. For a baseball flying through the air, these could represent its motion along the horizontal and vertical directions. Click either vector equation to split it into its $x$ and $y$ components.

Each set of components is effectively just a duplicate set of 1D kinematic equations. Note that these are all functions of a common variable $t$ . The shared time is what lets us combine the separate one-dimensional calculations into a two-dimensional path.

You get to choose the coordinate system, but the splitting only works if you follow a few rules:

The axes must be perpendicular. Only then is a component along $x$ unaffected by what happens along $y$ , so the two sets of equations stay independent. With slanted axes, each component would mix in part of the other direction.
Pick the origin and orientation that make the problem easiest. The axes don’t have to be horizontal and vertical; for a block on a ramp, tilting $x$ along the slope is often simpler. By convention, the positive $y$ -axis sits $90^\circ$ counterclockwise from the positive $x$ -axis.
Keep the same axes for the whole problem. Every vector ( $\vec{r}$ , $\vec{v}$ , and $\vec{a}$ ) must be broken into components along the same fixed axes, and those axes can’t move or rotate partway through. The signs of your components follow from which directions you called positive.

Projectile Motion

Projectile problems often describe the initial velocity $\vec{v}_0$ using a launch speed $v_0 = |\vec{v}_0|$ and angle $\theta$ “from the horizontal”. This gives us a natural choice of coordinate system, and we can begin by finding the two components of the initial velocity vector:

v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta.

Angle θ (°)35Speed v₀ (m/s)34.0

v_0x = v₀ cos θ = 27.9 m/sv_0y = v₀ sin θ = 19.5 m/s

A simple projectile has no horizontal acceleration and a constant downward acceleration from gravity:

\vec{a} = \langle 0, -g \rangle, \qquad g \approx 9.8\,\text{m/s}^2.

If the projectile starts at the origin, the component equations become

\begin{aligned} x &= (v_0\cos\theta) t,\\ y &= (v_0\sin\theta) t - \frac{1}{2}gt^2, \end{aligned}

\begin{aligned} v_x &= v_0\cos\theta,\\ v_y &= v_0\sin\theta - gt. \end{aligned}

The horizontal component of velocity stays constant while the vertical component changes steadily due to gravity. A projectile effectively moves sideways at a steady rate while its height rises, pauses, and then falls.

Why is the projectile path a parabola?

Start with the horizontal equation:

x = v_{0x}t.

If $v_{0x} \neq 0$ , then

t = \frac{x}{v_{0x}}.

Substitute that time into the vertical equation:

y = v_{0y}\left(\frac{x}{v_{0x}}\right) - \frac{1}{2}g\left(\frac{x}{v_{0x}}\right)^2.

With $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$ , this can be written as

y = (\tan\theta)x - \frac{g}{2v_0^2\cos^2\theta}x^2.

That is a quadratic function of $x$ , so the trajectory is a parabola.

Projectile Launcher

Aim, launch, and compare the predicted range with the actual path. Add drag to see where the ideal equations stop being exact.

To test these equations on a real ball, try the Photogate Launch lab: time a ball rolling off a table, predict where it lands, and check.

Constant $g$ is an approximation that holds when a throw is short compared to Earth’s size. For what happens when it isn’t, see Newton’s Cannonball.

Acceleration in 2D

For small intervals of time $\Delta t$ ,

\Delta \vec{v} \approx \vec{a}\Delta t

Diagram Hide

\vec{v}_i + \vec{a}\Delta t \approx \vec{v}_f

Place the tail of a Δt at the tip of v_i; the final velocity v_f runs from the first tail to the last tip.

This equation can be interpreted as “acceleration changes the velocity vector” (click it to see the addition drawn out). If the acceleration and velocity vectors are aligned, the velocity gets larger in time: the object speeds up. If acceleration points opposite the velocity, the velocity gets smaller: the object slows down. What happens when acceleration is perpendicular to velocity?

Look back at the hedgehog at either tip of its figure eight. At these two points, acceleration is perpendicular to velocity. The hedgehog is turning, and for that instant, its speed isn’t changing at all. Turn on decompose a to see the acceleration vector as: one component parallel to velocity that changes the speed, and another component perpendicular to velocity that turns the hedgehog.

2D Acceleration Sandbox

Apply a constant acceleration to an object in two dimensions.

Problem Solving

Soccer Ball Components

Problem

A soccer ball is kicked at $20\,\text{m/s}$ at an angle of $30^\circ$ above the horizontal. Find its initial velocity components.

Given

$v_0 = 20\,\text{m/s}$
$\theta = 30^\circ$

Use

v_{0x} = v_0\cos\theta, \qquad v_{0y} = v_0\sin\theta.

Solve

v_{0x} = (20)\cos 30^\circ \approx 17.3\,\text{m/s},

v_{0y} = (20)\sin 30^\circ = 10.0\,\text{m/s}.

Check

The launch angle is closer to horizontal than vertical, so the horizontal component should be larger than the vertical component. It is.

Time of Flight and Range

Problem

Using the same soccer ball, find the time to maximum height, total flight time, and horizontal range. Assume it lands at the same height it was kicked from and ignore air resistance.

Given

$v_{0x} \approx 17.3\,\text{m/s}$
$v_{0y} = 10.0\,\text{m/s}$
$g = 9.8\,\text{m/s}^2$

Solve

At maximum height, $v_y = 0$ :

0 = v_{0y} - gt \qquad \Rightarrow \qquad t = \frac{v_{0y}}{g} = \frac{10.0}{9.8} \approx 1.02\,\text{s}.

For a level landing, the trip down mirrors the trip up, so

T \approx 2(1.02) = 2.04\,\text{s}.

The horizontal range is

R = v_{0x}T \approx (17.3)(2.04) \approx 35.3\,\text{m}.

Check

A two-second flight at a little over $17\,\text{m/s}$ horizontally should cover a bit more than $34\,\text{m}$ , so $35.3\,\text{m}$ is reasonable.

Where Is the Cat?

Problem

We are looking down on a flat field of grass. Each square represents a region $1\,\text{m}$ wide and $1\,\text{m}$ long. The black dot represents a cat running along a path across the field. We’ve captured the cat right at the starting time $t = 0.0\,\text{s}$ .

Your task is to draw in dots representing the cat at 1-second intervals, according to these rules:

The cat has an initial velocity in the eastward direction of $4\,\text{m/s}$ , and it keeps this constant horizontal velocity throughout.
The cat has no initial velocity in the northward direction. However, its northward velocity increases steadily by $+1\,\text{m/s}$ each second until it reaches (and stays at) $+4\,\text{m/s}$ .

Place six dots, for $t = 1$ through $6\,\text{s}$ , then check your work.

Tap a grid point to place a dot for the cat; tap it again to remove it.

Show the solution steps

Solve

Treat the two directions separately. East has no acceleration:

x(t) = (4\,\text{m/s})\,t.

North starts from rest with $a_y = 1\,\text{m/s}^2$ . The velocity $v_y = t$ reaches $4\,\text{m/s}$ at $t = 4\,\text{s}$ , so for the first four seconds

y(t) = \tfrac{1}{2}(1\,\text{m/s}^2)\,t^2,

and after that the cat moves north at a steady $4\,\text{m/s}$ :

y(t) = 8\,\text{m} + (4\,\text{m/s})(t - 4\,\text{s}).

$t$ (s)	$v_y$ (m/s)	$x$ (m)	$y$ (m)
1	1	4	0.5
2	2	8	2
3	3	12	4.5
4	4	16	8
5	4	20	12
6	4	24	16

Check

The dots are evenly spaced eastward but spread further apart northward for the first four seconds, so the path curves upward. Once $v_y$ stops changing, each second moves the cat $4\,\text{m}$ east and $4\,\text{m}$ north, and the last dots fall on a straight line.

2D Kinematics Checkpoint1 of 3

In ideal projectile motion with no air resistance, what happens to the horizontal velocity?