In short
The position of a particle in a plane is described by a position vector \vec{r} = x\,\hat{i} + y\,\hat{j}, where x and y are measured from a chosen origin. Displacement is the change in position vector: \Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = (x_2 - x_1)\,\hat{i} + (y_2 - y_1)\,\hat{j}. The magnitude of displacement is the straight-line distance between the two positions. The actual path length (distance) is always greater than or equal to the magnitude of displacement. When position is given as x(t) and y(t), eliminating t gives the trajectory — the curve the particle traces in the plane.
Open a navigation app on your phone and search for a route from your home to a friend's house across town. The app shows two things: the blue dot marking where you are right now, and a winding line showing the route — through lanes, around corners, past the chai stall at the crossing. That blue dot is your position. The straight-line arrow from your home to your friend's house — the crow-fly direction and distance — is your displacement. And the twisting route the app draws? That is your path, the actual trail your autorickshaw follows through the city grid.
In one dimension — say, a straight road — position needed just one number: how far east or west of the bus stop. But the moment you step off that road and into a city with streets running both north-south and east-west, one number is not enough. You need two. You need a plane.
This article extends everything you learned about position, distance, and displacement on a line into two dimensions. The ideas are the same — position is where you are, displacement is the change — but the language shifts from signed numbers to vectors.
Position vector in 2D — where you are on the map
Pick an origin — the centre of a city intersection, say. Lay down two perpendicular axes: the x-axis pointing east, the y-axis pointing north. Now every point on the map has two coordinates: an x (how far east or west of the origin) and a y (how far north or south).
If you are standing 3 km east and 4 km north of the intersection, your position is the point (3, 4). But in physics, you do not just record coordinates — you record them as a vector that stretches from the origin to your location:
Why a vector and not just two numbers: because position in a plane has both magnitude (how far you are from the origin) and direction (which way). The vector captures both. The unit vector \hat{i} points along the positive x-axis, \hat{j} along the positive y-axis.
For the point (3, 4):
The magnitude of this position vector — the straight-line distance from the origin to your location — follows from the Pythagorean theorem:
Why the Pythagorean theorem: the x- and y-components form the two legs of a right triangle. The position vector is the hypotenuse. The magnitude is the length of that hypotenuse.
The direction is described by the angle \theta measured from the positive x-axis:
So you are 5 km from the intersection, in a direction 53.1° north of east.
Notice that the position vector depends on your choice of origin, just like in one dimension. If you shift the origin two kilometres east, every position vector in the city changes — but the physics of how objects move does not. That is why displacement (the change in position) turns out to be more physically meaningful than position itself.
Displacement in 2D — the straight-line shift
An autorickshaw starts at a point A with position \vec{r}_1 = x_1\,\hat{i} + y_1\,\hat{j} and drives through a maze of lanes until it reaches point B with position \vec{r}_2 = x_2\,\hat{i} + y_2\,\hat{j}. The displacement is the vector difference:
Why subtraction: displacement is the change in position. "Change" always means final minus initial. The result is a vector that points from A to B in a straight line, regardless of how winding the actual route was.
Write the components as \Delta x = x_2 - x_1 and \Delta y = y_2 - y_1. Then:
The magnitude of displacement is:
This is the straight-line distance from A to B — the shortest possible path between the two points.
The direction of displacement is:
Why this gives the direction: \Delta x and \Delta y are the horizontal and vertical components of the displacement arrow. The arctangent of their ratio gives the angle this arrow makes with the x-axis — exactly as it does for the position vector.
A key property: displacement does not depend on the choice of origin. If you shift the origin by some vector \vec{d}, both position vectors change — \vec{r}_1 becomes \vec{r}_1 + \vec{d} and \vec{r}_2 becomes \vec{r}_2 + \vec{d} — but their difference stays the same: (\vec{r}_2 + \vec{d}) - (\vec{r}_1 + \vec{d}) = \vec{r}_2 - \vec{r}_1. Displacement is origin-independent. That is one reason why physics prefers to work with displacement rather than position.
Path versus displacement — why they are different in 2D
In one dimension, path and displacement can differ only when you reverse direction. In two dimensions, they can differ even when you never turn back, because any curved path between two points is longer than the straight line connecting them.
Imagine flying a kite on Makar Sankranti. You watch the kite from the ground. It starts at a point A, drifts east in the wind, climbs higher, swings north, dips slightly, and finally ends up at point B. The path is the full curving trail the kite traced in the sky. The displacement is the straight arrow from A to B.
Mathematically:
The equality holds only when the particle moves in a straight line from A to B without doubling back. For every other kind of motion — every curve, every detour, every winding lane the autorickshaw takes — the path is strictly longer than the displacement.
This is exactly the 2D version of what you saw on the number line. The autorickshaw's odometer records path length (distance). The GPS displacement arrow records displacement. The odometer always reads higher.
The trajectory — the curve a particle traces
When a particle moves through a plane, it sweeps out a curve. That curve — the set of all points the particle passes through — is called the trajectory.
Think of a sparkler on Diwali night. You hold it and wave your hand in a circle. The glowing tip traces a bright loop in the air. That loop is the trajectory. It tells you where the particle went, but not when it was at each point or how fast it was moving. The trajectory is purely geometric — it is the shape of the motion, stripped of timing.
If you know the position at every instant — x(t) and y(t) as functions of time — you can find the trajectory by eliminating t. The result is a relation between x and y alone, describing the curve.
Example: a ball kicked at an angle
A cricket ball is hit with a horizontal speed of 20 m/s and a vertical speed of 15 m/s. Neglect air resistance. The position at time t is:
Why these equations: horizontal motion is uniform (no horizontal force), so x grows linearly with time. Vertical motion has gravity pulling downward at g = 9.8 m/s², so y(t) = v_{y0}\,t - \frac{1}{2}g\,t^2. Here v_{y0} = 15 m/s.
To find the trajectory, eliminate t. From the first equation: t = x/20. Substitute into the second:
Why eliminate t: the trajectory is the shape of the path in the x-y plane. Time tells you when the ball is at each point, but the shape itself is a relationship between x and y only. Eliminating t gives you that relationship — here, a downward-opening parabola.
This is a parabola — every projectile near Earth's surface traces a parabolic trajectory (as long as air resistance is negligible). The trajectory equation y = 0.75x - 0.01225x^2 tells you the shape, but not the speed at each point. For that, you need the parametric form x(t), y(t).
Parametric description — position as a function of time
When you describe motion in a plane, you typically give two functions of time: x(t) for the horizontal coordinate and y(t) for the vertical coordinate. Together, these are called the parametric equations of the motion, with time t as the parameter.
This is the complete description of two-dimensional motion. At any instant t, you can read off the position. As t advances, the point (x(t), y(t)) traces out the trajectory.
Why parametric? Because many different motions can produce the same trajectory. A ball thrown gently and a ball thrown hard at the same angle both trace parabolas — but the ball thrown hard covers the parabola faster. The trajectory (shape) is the same; the parametric equations (timing) are different. The parametric description captures both shape and timing.
Here is an example that is not a projectile. An ISRO satellite in a circular orbit around Earth has position:
where R is the orbital radius and \omega is the angular speed. Eliminate t: square both equations and add.
Why squaring and adding works: the identity \cos^2\theta + \sin^2\theta = 1 eliminates the time parameter completely, leaving a purely geometric equation — a circle of radius R.
The trajectory is a circle, as expected. But the parametric form tells you more: the satellite is at angle \omega t at time t, so it sweeps out the circle at a constant angular rate. The trajectory alone cannot tell you that.
How to read a parametric plot
A parametric plot takes the two functions x(t) and y(t) and draws the path (x, y) in the plane as t varies. Each point on the curve corresponds to a specific instant in time. You can think of it as the trail left by a glowing particle — each spot on the trail is labelled with the time the particle was there.
Worked examples
Example 1: Displacement from coordinate change
A delivery drone lifts off from a warehouse at position (2, 3) km and flies to a customer's house at (5, 7) km. Find the displacement vector, its magnitude, and the direction of displacement.
Step 1. Write the position vectors.
Why: each position is given as (x, y), which translates directly into x\,\hat{i} + y\,\hat{j}.
Step 2. Compute the displacement vector.
Why: displacement is final position minus initial position, component by component. The x-component of displacement is \Delta x = 5 - 2 = 3 km. The y-component is \Delta y = 7 - 3 = 4 km.
Step 3. Find the magnitude.
Why: the Pythagorean theorem turns the two perpendicular components into the length of the hypotenuse — the straight-line distance.
Step 4. Find the direction.
Why: the angle is measured from the positive x-axis. Since both \Delta x and \Delta y are positive, the displacement points into the first quadrant — northeast.
Result: \Delta\vec{r} = 3\,\hat{i} + 4\,\hat{j} km, with magnitude 5 km and direction 53.1° from the positive x-axis.
What this shows: Even though the two points have four coordinates between them, the displacement reduces everything to one vector — one magnitude and one direction. This is the power of the vector formulation: it compresses the "from where, to where" question into a single object.
Example 2: Trajectory from parametric equations
A ball is rolled across a flat field. Its position at time t (in seconds) is:
Find: (a) the trajectory equation, (b) the displacement between t = 0 and t = 3 s, and (c) what happens physically.
(a) Trajectory equation.
From x = 4t, solve for t:
Substitute into y = 3t - t^2:
Why: eliminating the time parameter gives a direct relationship between x and y. This is a downward-opening parabola — the shape of the path. The coefficient -0.0625 tells you the curvature; the coefficient 0.75 tells you the initial slope (the ratio of initial vertical to horizontal speed).
The ball reaches maximum height when dy/dx = 0, i.e., 0.75 - 0.125x = 0, giving x = 6 m. At that point, y = 0.75(6) - 0.0625(36) = 4.5 - 2.25 = 2.25 m.
The ball returns to ground level (y = 0) when 0.75x - 0.0625x^2 = 0, i.e., x(0.75 - 0.0625x) = 0. This gives x = 0 (start) or x = 12 m (landing point).
(b) Displacement between t = 0 and t = 3 s.
At t = 0: \vec{r}_0 = 0\,\hat{i} + 0\,\hat{j} = \vec{0}
At t = 3: \vec{r}_3 = 4(3)\,\hat{i} + (3(3) - 3^2)\,\hat{j} = 12\,\hat{i} + 0\,\hat{j}
Why the y-component is zero: the ball started at y = 0 and returned to y = 0 by t = 3 s. The vertical displacement is zero even though the ball went up to 2.25 m and came back down. Only the horizontal displacement survives.
The magnitude is |\Delta\vec{r}| = 12 m, directed purely along the x-axis.
(c) Physical picture. The ball moves at a constant 4 m/s horizontally. Vertically, it starts with an upward component of 3 m/s that decreases due to a deceleration of 2 m/s² (from the -t^2 term, which gives a downward acceleration of 2 m/s²). It rises, peaks at t = 1.5 s, and comes back down. The trajectory is a parabola. The displacement after one complete arc is 12 m horizontally — the ball ends up 12 m from where it started, at the same height.
What this shows: A parametric description x(t), y(t) is richer than the trajectory equation y(x) alone. The trajectory tells you the shape is a parabola. The parametric form tells you when the ball is at each point, how fast it is moving, and when it hits the ground. Both descriptions are useful, but the parametric form is more complete.
Common confusions
-
"Displacement is the shortest path." Not quite. Displacement is not a path at all — it is a vector connecting two positions. The shortest path between two points is a straight line, and the length of that straight line equals the magnitude of displacement. But displacement itself has both magnitude and direction. Calling it just "the shortest path" strips away the direction.
-
"If displacement is zero, the object did not move." Wrong. Zero displacement means the object returned to its starting point — it does not mean the object stayed still. A kite that flies in a loop and returns to the same spot has zero displacement but a very real (and possibly very long) path.
-
"The trajectory tells you everything about the motion." It does not. The trajectory tells you the shape of the path — a parabola, a circle, a figure-eight. But it does not tell you when the particle was at each point or how fast it was moving. Two balls thrown at different speeds along the same parabolic arc have the same trajectory but different parametric descriptions. Speed requires the parametric form.
-
"Path length and displacement magnitude are equal for straight-line motion." Almost right, but with a qualification: they are equal for straight-line motion in one direction. If an object moves in a straight line, reverses, and comes back partway, the path length exceeds the displacement magnitude even though the motion is along a line. In 2D, any curved path guarantees path > displacement, even without reversal.
-
"Position and displacement are the same kind of thing." They are both vectors, but they describe different quantities. Position depends on the choice of origin; displacement does not. If you change the origin, every position vector changes, but every displacement vector stays the same. Displacement is physically more meaningful because it is origin-independent.
If you came here to understand position vectors, displacement, and trajectories in two dimensions, you have what you need. What follows is for readers who want to see successive displacements, the connection to calculus, and a more formal view of trajectories.
Successive displacements and the addition rule
Suppose a particle moves from A to B and then from B to C. The displacement for each leg is:
The total displacement from A to C is:
Why this works: the intermediate position \vec{r}_B cancels when you add the two displacements. This is vector addition — place the tail of the second displacement at the head of the first, and the resulting vector goes directly from A to C.
This generalises to any number of legs. An autorickshaw that takes five turns through a city has five individual displacement vectors. Their vector sum — added head-to-tail — gives the net displacement from pickup to drop-off. The total path length, meanwhile, is the sum of the magnitudes of each leg (which is always larger than or equal to the magnitude of the net displacement).
Instantaneous displacement and the road to velocity
As the time interval shrinks, displacement approaches something fundamental. Consider the position at time t and at time t + \Delta t:
Divide by the time interval:
As \Delta t \to 0, this ratio becomes the velocity vector:
Why this matters: displacement per unit time is velocity. In one dimension, you learned that velocity is the rate of change of position. In two dimensions, it is the same — but now it is a vector, with components v_x = dx/dt and v_y = dy/dt. The velocity vector is always tangent to the trajectory at the particle's current position.
This is the bridge to the next topic — velocity and acceleration in two dimensions. The entire kinematics of 2D motion flows from the position vector \vec{r}(t) by taking successive time derivatives.
Parametric curves beyond parabolas
Not all trajectories are parabolas. The trajectory depends on the forces acting and the initial conditions. Here are a few parametric motions and their trajectory shapes:
| x(t) | y(t) | Trajectory |
|---|---|---|
| v_0 t | v_0 t - \frac{1}{2}g t^2 | Parabola (projectile) |
| R\cos\omega t | R\sin\omega t | Circle (uniform circular motion) |
| a\cos\omega t | b\sin\omega t | Ellipse (orbital motion, with a \neq b) |
| A\cos\omega_1 t | A\cos\omega_2 t | Lissajous figure (two perpendicular oscillations) |
| v_0 t | A\sin(kx) = A\sin(kv_0 t) | Sine wave (transverse wave pattern) |
Each of these will appear in later articles. The key insight is always the same: two independent functions of time, one for each axis, define both the shape and the timing of the motion. The shape (trajectory) is what you get by eliminating t. The timing (parametric form) is what you need for velocity, acceleration, and forces.
Where this leads next
- Velocity and Acceleration in Two Dimensions — the time derivatives of the position vector give you velocity and acceleration as vectors in the plane.
- Projectile Motion Fundamentals — the most important application of 2D kinematics: a body moving under gravity alone, with a parabolic trajectory.
- Vectors: Introduction and Notation — if you need a refresher on unit vectors, components, magnitude, and direction.
- Position, Distance, and Displacement — the one-dimensional version of everything in this article, where displacement is a signed number rather than a vector.