Motion in a plane

A ball thrown at an angle does not care that you learned kinematics on a single straight road. It rises, curves, and falls along a path that lives in two directions at once, and no single number called "velocity" can describe it any more — you need an arrow. A car rounding a bend, a badminton shuttle drifting sideways in the wind, a boat nosing across a flowing river: all of it is motion in a plane, and all of it needs the same upgrade to the toolkit you built for straight-line motion. The upgrade is not new physics. It is the same position, velocity, and acceleration you already know, written as vectors with an $x$ -component and a $y$ -component that are tracked side by side. The payoff is a trick so useful it becomes the backbone of the rest of mechanics: motion along $x$ and motion along $y$ do not talk to each other. You can solve two easy one-dimensional problems instead of one hard two-dimensional one, then stitch the answers back together with the Pythagoras theorem. This article builds that framework from the ground up. We track a particle's position vector $\vec r(t)$ as it slides through the plane, define average and instantaneous velocity and acceleration as vector derivatives, and prove the single most important fact in two-dimensional kinematics: a particle can accelerate even while its speed never changes, simply because its direction is changing. That one idea is the seed from which projectile motion and uniform circular motion — the next two chapters in this story — will grow.

Key takeaways

Position becomes a vector. In a plane, $\vec r(t) = x(t)\hat i + y(t)\hat j$ tracks a particle with two coordinates instead of one, both functions of time.
Average velocity is displacement over time. $\vec v_{avg} = \Delta \vec r/\Delta t$ points along the straight line from start point to end point, not along the actual path.
Instantaneous velocity is the time derivative of position, $\vec v = d\vec r/dt$ , and it always points tangent to the trajectory at that instant.
Acceleration is the time derivative of velocity, $\vec a = d\vec v/dt$ — and it can exist even when speed is constant, if direction is changing.
Components are independent. The $x$ -motion and $y$ -motion can be solved separately using ordinary straight-line kinematics, then combined at the end.
The trajectory (path shape) and the motion (speed and direction along it) are different questions — a $y$ -versus- $x$ graph shows the shape but hides time; you need $\vec v(t)$ to know how fast and which way.
Key idea: writing position, velocity and acceleration as vectors $\vec r=x\hat i+y\hat j$ , $\vec v=d\vec r/dt$ , $\vec a=d\vec v/dt$ lets two-dimensional motion be solved as two independent one-dimensional problems along x and y, with acceleration possible even at constant speed whenever direction changes.

The core insight

In a plane, position, velocity and acceleration are vectors whose $x$ - and $y$ -components obey the ordinary rules of straight-line kinematics independently of each other, so $\vec v = d\vec r/dt$ is always tangent to the path and a particle can accelerate purely by changing direction even at constant speed.

Where it fits in the syllabus

This is the framework-building section of CBSE Class 11 Physics, chapter Motion in a Plane, sitting right after vector addition by the analytical (component) method and right before motion with constant acceleration in a plane. Its job is narrow but essential: define position, displacement, average and instantaneous velocity, and average and instantaneous acceleration as vectors in two dimensions, and establish that $x$ - and $y$ -components can be treated independently. Projectile motion and uniform circular motion, which use this exact machinery, are the next two topics in the chapter and are not the focus of this section — here the goal is only to get comfortable differentiating a position vector.

The physical picture

Picture a particle wandering across the $xy$ -plane. At any instant, its whereabouts is captured by a single arrow drawn from the origin to the particle — the position vector:

\vec r(t) = x(t)\,\hat i + y(t)\,\hat j

where: $x(t)$ and $y(t)$ are the particle's coordinates at time $t$ , and $\hat i$ , $\hat j$ are the fixed unit vectors along the $x$ - and $y$ -axes. As $t$ ticks forward, the tip of $\vec r(t)$ traces a curve in the plane — the particle's trajectory. The trajectory only shows the shape of the path; it says nothing about how fast the particle moved along it or when it was where, because time has been eliminated by plotting $y$ against $x$ . Between two nearby instants $t_1$ and $t_2$ , the particle's displacement is the vector difference $\Delta \vec r = \vec r_2 - \vec r_1$ , and dividing by the elapsed time gives the average velocity:

\vec v_{avg} = \frac{\Delta \vec r}{\Delta t} = \frac{\vec r_2 - \vec r_1}{t_2 - t_1}

where: $\vec v_{avg}$ is a vector pointing along the straight chord joining the start and end points — not along the curved path itself. This is the two-dimensional analogue of "average velocity equals total displacement over total time" from straight-line motion, and it makes the same distinction between displacement (a vector, start to end) and distance travelled (a scalar, along the actual curve). Shrinking the time interval to an instant turns this chord into a tangent, and gives the instantaneous velocity:

\vec v = \lim_{\Delta t \to 0}\frac{\Delta \vec r}{\Delta t} = \frac{d\vec r}{dt} = \frac{dx}{dt}\hat i + \frac{dy}{dt}\hat j = v_x\hat i + v_y\hat j

where: $v_x = dx/dt$ and $v_y = dy/dt$ are the velocity components, obtained by differentiating each coordinate on its own — the $x$ -story and the $y$ -story never mix. Geometrically, as $\Delta t \to 0$ the second point on the trajectory slides toward the first, and the chord direction settles onto the tangent to the curve at that point. This is why instantaneous velocity always points along the direction of motion at that instant, whatever the shape of the path. The same two-step logic gives acceleration. Average acceleration over an interval is the change in velocity divided by the time taken, $\vec a_{avg} = \Delta \vec v/\Delta t$ , and shrinking the interval gives the instantaneous acceleration:

\vec a = \frac{d\vec v}{dt} = \frac{dv_x}{dt}\hat i + \frac{dv_y}{dt}\hat j = a_x\hat i + a_y\hat j

where: $a_x$ and $a_y$ are the rates of change of the two velocity components, again computed independently. Because velocity is a vector with both a size (speed) and a direction, it can change — and so produce a nonzero acceleration — in three distinct ways: the speed can change while direction stays fixed (ordinary speeding up or slowing down along a straight line), the direction can change while the speed stays fixed (as in uniform circular motion, where the arrow keeps turning without growing or shrinking), or both can change together, which is the general case a particle in a plane usually experiences.

Figure

A curved trajectory in the xy plane with the position vector drawn from the origin to a point on the curve and the velocity vector drawn as a short arrow tangent to the curve at that point, together with a nearby second point joined by a chord representing the average velocity over that short interval

A useful special picture is a particle moving at constant speed in a circle. Its velocity vector keeps constant length but constantly turns — so $\vec a = d\vec v/dt \neq 0$ even though $|\vec v|$ never changes. That single fact resolves a very common misconception before it forms: acceleration measures the rate of change of the velocity vector, and a vector changes if either its magnitude or its direction changes — or both.

Derivation AInstantaneous velocity as the derivative of the position vector

The goal of this derivation is to build the instantaneous velocity vector $\vec v = d\vec r/dt$ starting only from the definition of average velocity as displacement over time, showing precisely how the limiting process turns a chord (average velocity, joining two separated points on the trajectory) into a tangent (instantaneous velocity, at a single point), and why this forces the $x$ - and $y$ -components to be differentiated completely independently of one another — the fact that licenses solving planar problems as two separate one-dimensional problems.

Step 1. Set up two nearby instants. Let the particle be at $\vec r_1 = x_1\hat i + y_1\hat j$ at time $t_1$ , and at $\vec r_2 = x_2\hat i + y_2\hat j$ at a slightly later time $t_2 = t_1 + \Delta t$ . The displacement over this interval is the vector subtraction $\Delta \vec r = \vec r_2 - \vec r_1 = (x_2-x_1)\hat i + (y_2-y_1)\hat j = \Delta x\,\hat i + \Delta y\,\hat j$ . Because $\hat i$ and $\hat j$ are fixed, perpendicular, unchanging directions, this subtraction splits cleanly into an $x$ -part and a $y$ -part with no cross-terms between them.

Step 2. Write the average velocity over this interval. Dividing the vector $\Delta \vec r$ by the scalar $\Delta t$ scales each component separately: $\vec v_{avg} = \dfrac{\Delta \vec r}{\Delta t} = \dfrac{\Delta x}{\Delta t}\hat i + \dfrac{\Delta y}{\Delta t}\hat j$ . Geometrically this vector points along the straight chord from the first point to the second point on the trajectory — it is the direction you would walk in if you ignored the actual curved path and simply cut straight from start to end.

Step 3. Shrink the interval. Now let $\Delta t \to 0$ , so that $t_2$ slides back toward $t_1$ and the second point on the trajectory slides back toward the first. As this happens, each ratio $\Delta x/\Delta t$ and $\Delta y/\Delta t$ approaches the ordinary calculus derivative of a single-variable function of time — exactly as in one-dimensional kinematics — because the limiting process for each component only ever involves that component's own coordinate, never the other one. This is the precise reason $x$ - and $y$ -motion decouple: the limit defining $v_x$ never sees $y$ , and the limit defining $v_y$ never sees $x$ .

Step 4. Read off the result. Taking the limit component-by-component gives $\vec v = \displaystyle\lim_{\Delta t\to 0}\vec v_{avg} = \frac{dx}{dt}\hat i + \frac{dy}{dt}\hat j = v_x\hat i + v_y\hat j$ , which is exactly $\vec v = d\vec r/dt$ . Geometrically, as the second point merges into the first, the chord of Step 2 rotates until it lies exactly along the tangent to the trajectory at that point — so the instantaneous velocity vector is always tangent to the path, pointing in the direction the particle is actually moving at that instant. Answer: $\vec v = \dfrac{d\vec r}{dt} = v_x\hat i + v_y\hat j$ , tangent to the trajectory, with $v_x$ and $v_y$ obtainable by differentiating $x(t)$ and $y(t)$ completely independently. The identical two-step limiting argument, applied to $\vec v(t)$ instead of $\vec r(t)$ , gives $\vec a = d\vec v/dt = a_x\hat i + a_y\hat j$ for the instantaneous acceleration.

Worked examples

Board level - velocity and speed from a position vector

A particle's position vector is given by $\vec r(t) = (3t^2)\hat i + (4t)\hat j$ metres, with $t$ in seconds. Find the velocity vector, the speed, and the direction of motion at $t = 2$ s.

Step 1. Differentiate each component of the position vector separately, since $x$ - and $y$ -motion are independent: $v_x = \dfrac{d}{dt}(3t^2) = 6t$ and $v_y = \dfrac{d}{dt}(4t) = 4$ . This gives the general velocity vector $\vec v(t) = 6t\,\hat i + 4\hat j$ m/s, valid at any time $t$ .

Step 2. Substitute $t = 2$ s into the general expression: $\vec v(2) = 6(2)\hat i + 4\hat j = 12\hat i + 4\hat j$ m/s. Notice that the $y$ -component of velocity never changed with time — this particle moves at a constant rate in $y$ while continually speeding up in $x$ .

Step 3. Find the speed, which is the magnitude of the velocity vector, using Pythagoras on the two components: $v = \sqrt{12^2+4^2} = \sqrt{144+16} = \sqrt{160} \approx 12.65$ m/s.

Step 4. Find the direction, the angle $\theta$ the velocity vector makes with the $x$ -axis, using $\tan\theta = v_y/v_x = 4/12 = 1/3$ , so $\theta = \tan^{-1}(1/3) \approx 18.4^\circ$ above the $x$ -axis. Answer: $\vec v(2) = 12\hat i + 4\hat j$ m/s, speed $\approx 12.65$ m/s, directed at about $18.4^\circ$ to the $x$ -axis.

Competitive level - acceleration, average acceleration, and consistency check

A particle moves with velocity components $v_x = 4$ m/s (constant) and $v_y = (3+2t)$ m/s, with $t$ in seconds. Find (a) the instantaneous acceleration vector, (b) the average acceleration between $t=0$ and $t=3$ s, and (c) the speed and direction of motion at $t=3$ s.

Step 1. Differentiate each velocity component to get the instantaneous acceleration: $a_x = \dfrac{dv_x}{dt} = 0$ (since $v_x$ is constant) and $a_y = \dfrac{dv_y}{dt} = 2$ . So $\vec a = 0\hat i + 2\hat j$ m/s², a constant vector pointing purely along $+y$ with magnitude $2$ m/s².

Step 2. Compute the velocity vector at both ends of the interval to find the average acceleration: at $t=0$ , $\vec v(0) = 4\hat i + 3\hat j$ m/s; at $t=3$ s, $\vec v(3) = 4\hat i + (3+6)\hat j = 4\hat i + 9\hat j$ m/s. The change is $\Delta \vec v = \vec v(3)-\vec v(0) = 0\hat i + 6\hat j$ m/s.

Step 3. Divide by the elapsed time to get $\vec a_{avg} = \dfrac{\Delta \vec v}{\Delta t} = \dfrac{0\hat i+6\hat j}{3} = 0\hat i + 2\hat j$ m/s². This matches the instantaneous acceleration from Step 1 exactly, which must happen whenever acceleration itself is constant — a useful self-check to run whenever both quantities are asked for in the same problem.

Step 4. Using $\vec v(3) = 4\hat i+9\hat j$ m/s from Step 2, the speed is $v = \sqrt{4^2+9^2} = \sqrt{16+81} = \sqrt{97} \approx 9.85$ m/s, at an angle $\theta = \tan^{-1}(9/4) \approx 66.0^\circ$ above the $x$ -axis. Answer: $\vec a = 0\hat i+2\hat j$ m/s² (both instantaneous and average, consistently), and at $t=3$ s the speed is $\approx 9.85$ m/s directed at about $66.0^\circ$ to the $x$ -axis. Timing note: In an exam or entrance-test setting, budget no more than about 3-4 minutes for a problem shaped like this — differentiate quickly for part (a), plug the two endpoint velocities into $\Delta \vec v/\Delta t$ for part (b) without re-deriving anything, and only bother finding the magnitude/direction in part (c) if it is explicitly asked, since it costs an extra square root and inverse-tangent that many students needlessly compute for parts that don't require it.

Common traps

Believing constant speed means zero acceleration — in the plane, a change in direction alone, with speed fixed, still counts as acceleration, exactly as in uniform circular motion.
Trying to solve $x$ - and $y$ -motion together instead of separately — the single biggest simplification in two-dimensional kinematics is treating each axis as its own independent one-dimensional problem and only combining the results at the very end (for speed and direction).
Confusing average velocity (net displacement over total time, a straight chord from start to end) with instantaneous velocity (tangent to the actual curved path at one instant) — for a curved trajectory these are generally different vectors, and only coincide for straight-line motion.
Forgetting that instantaneous velocity is always tangent to the trajectory, and instead drawing it pointing toward the origin, toward the next grid point, or along some other unrelated reference direction.
Dropping the vector nature of $\vec a_{avg} = \Delta\vec v/\Delta t$ and subtracting speeds (scalars) instead of subtracting velocity vectors component-by-component, which silently loses the direction information.
Reading a $y$ -versus- $x$ trajectory graph as though it shows speed or timing — it only shows the shape of the path; recovering speed at any point always requires differentiating $x(t)$ and $y(t)$ with respect to time.

You should be able to…

Write the position vector $\vec r(t) = x(t)\hat i + y(t)\hat j$ for a particle moving in a plane, and identify what $x(t)$ and $y(t)$ represent.
Differentiate a position vector's components with respect to time to obtain the velocity vector, and differentiate velocity's components to obtain the acceleration vector.
Distinguish average velocity/acceleration (a chord, over a finite interval) from instantaneous velocity/acceleration (a tangent, at a single instant).
Explain, with the uniform-circular-motion example, why a particle can accelerate even while its speed is exactly constant.
Combine $x$ - and $y$ -components back into a single speed (via Pythagoras) and direction (via $\tan^{-1}$ ) when a problem asks for the full vector rather than just its components.
Justify why treating $x$ -motion and $y$ -motion as two independent one-dimensional problems is valid, using the component-by-component nature of vector addition, subtraction, and differentiation.

Ask the tutor: Want to practice differentiating a position vector to get velocity and acceleration, or work through why a particle can accelerate at constant speed before moving on to projectile motion?

Why do we even need vectors here — can't I just use x and y separately from the start without calling them a vector r?

You actually can solve most problems that way, and in practice many students do split $x$ and $y$ immediately. The vector notation $\vec r = x\hat i+y\hat j$ is there to keep the bookkeeping honest and to let you write one compact equation, like $\vec v=d\vec r/dt$ , that silently means "differentiate both components." The moment you need a single number for speed or a single angle for direction, you're forced to recombine $x$ and $y$ anyway — the vector is just the language that makes that recombination automatic and mistake-resistant.

If velocity is tangent to the path, does that mean acceleration is always perpendicular to velocity?

Not always — only in the special case of constant-speed motion (like uniform circular motion), where acceleration is purely perpendicular to velocity because it can only change direction, not magnitude. In general planar motion, acceleration can have a component along velocity (which changes speed) and a component perpendicular to velocity (which changes direction) at the same time, and both usually happen together. You will meet this split explicitly when you study circular motion later in this chapter.

In the worked examples, why does differentiating v_x = 4 (a constant) give zero instead of just leaving it as 4?

Differentiation with respect to time answers "how fast is this quantity changing," not "what is its value." A constant, by definition, is not changing at all, so its rate of change is zero — this is true whether you're differentiating position to get velocity or velocity to get acceleration. It's the same rule you already used in one-dimensional motion: a body moving at constant velocity has zero acceleration, because velocity, not position, is what's held fixed here.

Is this topic actually tested directly, or is it just setup for projectile motion and circular motion?

It is tested directly and often as a short-answer or numerical problem — giving $\vec r(t)$ or velocity components and asking you to differentiate for velocity or acceleration, exactly like the two worked examples above, is a completely standard board-level question on its own. But you're right that its bigger role is as scaffolding: every projectile-motion and circular-motion formula you'll meet next is built by applying exactly this differentiation machinery to a specific choice of $x(t)$ and $y(t)$ , so getting comfortable here pays off twice.