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Physics · Motion in a Plane

Motion in a Plane with Constant Acceleration

One dimension is forgiving — a plus or minus sign carries all the direction you need. Two dimensions are not. This deep-dive builds the framework NCERT §3.7 and §3.8 use — position vector $\mathbf{r}(t)$, instantaneous velocity $\mathbf{v} = d\mathbf{r}/dt$, and the vector equations $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}\,t$ and $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \tfrac{1}{2}\mathbf{a}\,t^2$ — and pins down the single insight that makes every projectile, dropped-ball-from-a-car and accelerated-charge problem tractable: when $\mathbf{a}$ is constant, the $x$- and $y$-motions decouple into two independent 1-D problems. NEET PYQs from 2021, 2017 and 2016 close the article.

Position vector and displacement in a plane

Pick an origin $O$. The instantaneous position of a particle $P$ moving in the $xy$-plane is described by the position vector $\mathbf{r}$, drawn from $O$ to $P$. NCERT §3.7.1 writes it in component form as

$$\mathbf{r} = x\,\hat{\imath} + y\,\hat{\jmath}$$

where $x$ and $y$ are the Cartesian coordinates of $P$, now reinterpreted as scalar projections of $\mathbf{r}$ along $\hat{\imath}$ and $\hat{\jmath}$. If the particle moves from $P$ to $P'$ between $t$ and $t'$, the displacement vector is

$$\Delta\mathbf{r} = \mathbf{r}' - \mathbf{r} = (x' - x)\,\hat{\imath} + (y' - y)\,\hat{\jmath} = \Delta x\,\hat{\imath} + \Delta y\,\hat{\jmath}$$

The straight-line segment from $P$ to $P'$ is the displacement, regardless of the curving path actually followed. So $|\Delta\mathbf{r}|$ is less than or equal to the path length, with equality only for straight-line motion — a distinction that becomes essential in 2-D, because a curving path will always have $|\Delta\mathbf{r}|$ strictly less than the arc length.

Velocity in two dimensions

Average velocity is displacement over time — a vector in the direction of $\Delta\mathbf{r}$:

$$\bar{\mathbf{v}} = \frac{\Delta\mathbf{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\,\hat{\imath} + \frac{\Delta y}{\Delta t}\,\hat{\jmath}$$

Taking $\Delta t \to 0$ gives the instantaneous velocity (NCERT Eq. 3.28):

$$\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\,\hat{\imath} + \frac{dy}{dt}\,\hat{\jmath} = v_x\,\hat{\imath} + v_y\,\hat{\jmath}$$

Two facts follow. First, the direction of $\mathbf{v}$ at any point is tangential to the path there — NCERT Fig. 3.13. Second, speed and direction recover the standard way:

$$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\,,\qquad \theta = \tan^{-1}\!\left(\frac{v_y}{v_x}\right)$$

So if $x(t)$ and $y(t)$ are explicit functions of time, $v_x$ and $v_y$ follow by ordinary differentiation.

Acceleration in two dimensions

Average acceleration is $\Delta\mathbf{v}/\Delta t$ and instantaneous acceleration is its limit (NCERT Eqs. 3.31–3.32):

$$\mathbf{a} = \frac{d\mathbf{v}}{dt} = a_x\,\hat{\imath} + a_y\,\hat{\jmath},\qquad a_x = \frac{d^2 x}{dt^2},\ a_y = \frac{d^2 y}{dt^2}$$

Unlike 1-D, where $\mathbf{v}$ and $\mathbf{a}$ always lie along the same line, in 2-D they can make any angle between $0°$ and $180°$. NCERT highlights this explicitly. When $\mathbf{a}$ is perpendicular to $\mathbf{v}$, the speed stays constant and only the direction turns — exactly the case of uniform circular motion.

The vector kinematic equations

Now restrict to constant acceleration — the case NCERT §3.8 treats. "Constant" means both magnitude and direction of $\mathbf{a}$ are fixed; the average acceleration over any interval equals the instantaneous value. Let $\mathbf{v}_0$ be the velocity at $t = 0$ and $\mathbf{v}$ the velocity at time $t$. From the definition of acceleration,

$$\mathbf{a} = \frac{\mathbf{v} - \mathbf{v}_0}{t}\ \Longrightarrow\ \boxed{\mathbf{v} = \mathbf{v}_0 + \mathbf{a}\,t}\qquad (\text{NCERT Eq. 3.33a})$$

Because $\mathbf{a}$ is constant, the velocity increases linearly with $t$ in the direction of $\mathbf{a}$. For the position, the average velocity over $[0, t]$ is $(\mathbf{v}_0 + \mathbf{v})/2$, and the displacement is the average velocity times the elapsed time:

$$\mathbf{r} - \mathbf{r}_0 = \left(\frac{\mathbf{v}_0 + \mathbf{v}}{2}\right) t = \frac{\mathbf{v}_0 + \mathbf{v}_0 + \mathbf{a}\,t}{2}\,t$$ $$\boxed{\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0\,t + \tfrac{1}{2}\mathbf{a}\,t^2}\qquad (\text{NCERT Eq. 3.34a})$$

You can verify that differentiating this position once with respect to time returns $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}\,t$, and that at $t = 0$ it reduces to $\mathbf{r} = \mathbf{r}_0$. So the two boxed equations form a self-consistent pair. Both are vector equations — every symbol except $t$ is a vector. They look identical to the 1-D constant-acceleration formulae, and that is the point: vectors let you write 2-D and 3-D kinematics in the same algebraic shape as 1-D kinematics.

"Motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions."

NCERT Class 11 Physics, §3.8

Decoupling — the central NCERT result

Take NCERT Eq. 3.34a and resolve it along $\hat{\imath}$ and $\hat{\jmath}$. The vector equation

$$\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0\,t + \tfrac{1}{2}\mathbf{a}\,t^2$$

becomes two scalar equations — one for the $x$-component, one for the $y$-component — because vector equality means component-wise equality:

$$x = x_0 + v_{0x}\,t + \tfrac{1}{2}a_x\,t^2$$ $$y = y_0 + v_{0y}\,t + \tfrac{1}{2}a_y\,t^2$$

Similarly, $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}\,t$ splits into

$$v_x = v_{0x} + a_x\,t,\qquad v_y = v_{0y} + a_y\,t$$

Each pair $\{x, v_x, v_{0x}, a_x\}$ involves only the $x$-direction; each pair $\{y, v_y, v_{0y}, a_y\}$ involves only the $y$-direction. No equation mixes $x$- and $y$-quantities. Both motions are governed by the familiar 1-D constant-acceleration equations from the previous chapter. Galileo first articulated this for projectile motion in 1632; the modern statement extends to any constant $\mathbf{a}$.

Quantity $x$-direction $y$-direction
Initial position $x_0$ $y_0$
Initial velocity $v_{0x}$ $v_{0y}$
Acceleration (constant) $a_x$ $a_y$
Velocity at time $t$ $v_x = v_{0x} + a_x t$ $v_y = v_{0y} + a_y t$
Position at time $t$ $x = x_0 + v_{0x} t + \tfrac{1}{2}a_x t^2$ $y = y_0 + v_{0y} t + \tfrac{1}{2}a_y t^2$
v–s (third equation) $v_x^2 = v_{0x}^2 + 2 a_x (x - x_0)$ $v_y^2 = v_{0y}^2 + 2 a_y (y - y_0)$

The procedure for any NEET 2-D constant-acceleration problem reduces to a four-step recipe. (1) Set up axes and read off $x_0, y_0, v_{0x}, v_{0y}, a_x, a_y$. (2) Write the four scalar equations above. (3) Use the single time $t$ to couple the two directions — it is the only quantity that appears in both. (4) Recombine the components at the end using $|\mathbf{r}| = \sqrt{x^2 + y^2}$ or $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$.

Worked examples

Worked example · NCERT Exercise 3.18

A particle starts from the origin at $t = 0$ with a velocity $\mathbf{v}_0 = 10\,\hat{\jmath}$ m s$^{-1}$ and moves in the $xy$-plane with constant acceleration $\mathbf{a} = (8\,\hat{\imath} + 2\,\hat{\jmath})$ m s$^{-2}$. (a) At what time is the $x$-coordinate of the particle 16 m, and what is the $y$-coordinate at that instant? (b) What is the speed of the particle at that time?

Set up. Origin at $t = 0$: $x_0 = y_0 = 0$. Initial velocity components: $v_{0x} = 0$, $v_{0y} = 10$ m s$^{-1}$. Acceleration components: $a_x = 8$ m s$^{-2}$, $a_y = 2$ m s$^{-2}$.

(a) Find $t$ when $x = 16$ m. The $x$-equation alone (no $y$ involved):

$$x = \tfrac{1}{2}(8)\,t^2 = 4t^2 = 16 \Rightarrow t = 2\text{ s}$$

Plug this $t$ into the $y$-equation:

$$y = (10)(2) + \tfrac{1}{2}(2)(2)^2 = 24\text{ m}$$

(b) Speed at $t = 2$ s.

$$v_x = 8(2) = 16,\quad v_y = 10 + 2(2) = 14\text{ m s}^{-1}$$

$$|\mathbf{v}| = \sqrt{16^2 + 14^2} = \sqrt{452} \approx 21.3\text{ m s}^{-1}$$

The $x$-equation gave $t$ without mentioning $y$; the two motions share only the clock.

Worked example · NCERT Exercise 3.17 paraphrase

The position of a particle is $\mathbf{r}(t) = (3.0\,t)\,\hat{\imath} + (2.0\,t^2)\,\hat{\jmath}$ metres, where $t$ is in seconds. (a) Find $\mathbf{v}(t)$ and $\mathbf{a}(t)$. (b) Find the magnitude and direction of the velocity at $t = 2.0$ s.

(a) Differentiate component-wise:

$$\mathbf{v} = 3.0\,\hat{\imath} + 4.0\,t\,\hat{\jmath}\ \text{m s}^{-1},\qquad \mathbf{a} = 4.0\,\hat{\jmath}\ \text{m s}^{-2}$$

$x$-velocity is constant; $y$-velocity grows linearly; $\mathbf{a}$ is constant along $+y$ with $a_x = 0$.

(b) At $t = 2.0$ s. $\mathbf{v} = 3.0\,\hat{\imath} + 8.0\,\hat{\jmath}$, $|\mathbf{v}| = \sqrt{73} \approx 8.54$ m s$^{-1}$, $\theta = \tan^{-1}(8/3) \approx 69.4°$.

Worked example · NEET-style numeric

A particle starts from rest at the origin and moves in the $xy$-plane under constant acceleration $\mathbf{a} = (4\,\hat{\imath} - 3\,\hat{\jmath})$ m s$^{-2}$. Find (a) the position vector at $t = 5$ s, (b) the velocity vector at $t = 5$ s, and (c) the speed.

Set up. $\mathbf{r}_0 = \mathbf{0}$, $\mathbf{v}_0 = \mathbf{0}$, $a_x = 4$, $a_y = -3$ m s$^{-2}$.

(a) Position. With $\mathbf{v}_0 = \mathbf{0}$, $\mathbf{r} = \tfrac{1}{2}\mathbf{a}\,t^2$ gives $x = 50$ m, $y = -37.5$ m. So $\mathbf{r}(5) = 50\,\hat{\imath} - 37.5\,\hat{\jmath}$ m.

(b) Velocity. $\mathbf{v} = \mathbf{a}\,t = 20\,\hat{\imath} - 15\,\hat{\jmath}$ m s$^{-1}$.

(c) Speed. $|\mathbf{v}| = \sqrt{20^2 + 15^2} = 25$ m s$^{-1}$.

Because $\mathbf{v}_0 = \mathbf{0}$, $\mathbf{v}$ stays parallel to $\mathbf{a}$ for all $t$ — the path is a straight line. A 2-D constant-acceleration problem curves only when $\mathbf{v}_0$ is not parallel to $\mathbf{a}$.

The dot-product third equation

The 1-D kinematic toolkit has three equations: $v = u + at$, $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$. The first two have direct vector analogues, written above. The third generalises to

$$v^2 = v_0^2 + 2\,\mathbf{a}\cdot(\mathbf{r} - \mathbf{r}_0)$$

where $v^2 = \mathbf{v}\cdot\mathbf{v} = v_x^2 + v_y^2$ and $\mathbf{a}\cdot(\mathbf{r} - \mathbf{r}_0) = a_x(x - x_0) + a_y(y - y_0)$. NCERT does not box this equation, but it follows immediately by squaring $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}\,t$ and using $\mathbf{r} - \mathbf{r}_0 = \mathbf{v}_0 t + \tfrac{1}{2}\mathbf{a}\,t^2$ to eliminate $t$. In practice, NEET problems on this subtopic rarely require the third equation in vector form — solving the two scalar third equations (one for each axis, as in the table above) is more direct.

Bridge to projectile motion

Everything in this article is the general framework. Projectile motion is the specific case where

$$\mathbf{a} = -g\,\hat{\jmath}\ \Longleftrightarrow\ a_x = 0,\ a_y = -g$$

Substituting this into the boxed equations gives the projectile-motion formulae NCERT derives in §3.9. With $a_x = 0$, the $x$-equation degenerates to constant velocity: $x = v_{0x}\,t$. With $a_y = -g$, the $y$-equation becomes $y = v_{0y}\,t - \tfrac{1}{2}g\,t^2$. The path is a parabola, the trajectory equation $y(x)$ is quadratic, and the range-time-of-flight-maximum-height results are direct consequences. Master the framework here and the projectile chapter reduces to substitution.

Quick recap

What this subtopic locked in

  • Position vector. $\mathbf{r} = x\hat{\imath} + y\hat{\jmath}$; displacement $\Delta\mathbf{r} = \Delta x\,\hat{\imath} + \Delta y\,\hat{\jmath}$.
  • Velocity is the time-derivative of position. $\mathbf{v} = d\mathbf{r}/dt = v_x\hat{\imath} + v_y\hat{\jmath}$; magnitude $\sqrt{v_x^2 + v_y^2}$; direction $\tan^{-1}(v_y/v_x)$. Tangential to the path.
  • Acceleration is the time-derivative of velocity. Unlike 1-D, $\mathbf{v}$ and $\mathbf{a}$ can make any angle.
  • Vector equations for constant $\mathbf{a}$. $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}\,t$ and $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \tfrac{1}{2}\mathbf{a}\,t^2$ — NCERT Eqs. 3.33a and 3.34a.
  • Decoupling. Constant $\mathbf{a}$ in 2-D = two independent 1-D constant-acceleration problems sharing only the clock.
  • Recipe. Resolve into components, solve each axis, recombine at the end.
  • Projectile = special case. Set $a_x = 0$, $a_y = -g$.

NEET PYQ Snapshot — 2-D Constant Acceleration

Three PYQs that test the vector kinematics of this subtopic directly (projectile-specific PYQs sit in the sibling article).

NEET 2021

A car starts from rest and accelerates at $5$ m s$^{-2}$. At $t = 4$ s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t = 6$ s? (Take $g = 10$ m s$^{-2}$.)

  1. $20\sqrt{2}$ m s$^{-1}$, $10$ m s$^{-2}$
  2. $20$ m s$^{-1}$, $5$ m s$^{-2}$
  3. $20$ m s$^{-1}$, $0$
  4. $20\sqrt{2}$ m s$^{-1}$, $0$
Answer: (1) $20\sqrt{2}$ m s$^{-1}$, $10$ m s$^{-2}$

Why: At release ($t = 4$ s) the car's speed is $5 \times 4 = 20$ m s$^{-1}$. The ball then carries this horizontal velocity into free fall: $v_{0x} = 20$, $v_{0y} = 0$, $a_x = 0$, $a_y = -g$. Two seconds later: $v_x = 20$, $v_y = 10 \times 2 = 20$. Speed $= \sqrt{20^2 + 20^2} = 20\sqrt{2}$ m s$^{-1}$; acceleration is $g = 10$ m s$^{-2}$ downward. Distractor (2) keeps the car's $5$ m s$^{-2}$ for the ball; (3) ignores the vertical velocity; (4) drops gravity. 2-D decoupling — $a_x = 0$, $a_y = -g$ — is the spine.

NEET 2017

The $x$ and $y$ coordinates of the particle at any time are $x = 5t - 2t^2$ and $y = 10t$ respectively, where $x$ and $y$ are in metres and $t$ in seconds. The acceleration of the particle at $t = 2$ s is:

  1. $-8$ m s$^{-2}$
  2. $0$
  3. $5$ m s$^{-2}$
  4. $-4$ m s$^{-2}$
Answer: (4) $-4$ m s$^{-2}$

Why: Differentiate component-wise. $v_x = dx/dt = 5 - 4t$; $v_y = dy/dt = 10$. Then $a_x = dv_x/dt = -4$ m s$^{-2}$ and $a_y = dv_y/dt = 0$. The acceleration vector is $\mathbf{a} = -4\,\hat{\imath}$, with magnitude $4$ m s$^{-2}$ along $-x$. The signed value $-4$ is option (4). Distractor (1) doubles the answer (perhaps confusing $a_x$ with $2a_x$); (2) picks up $a_y$ alone; (3) uses the initial $x$-velocity $5$ m s$^{-1}$ as an "acceleration".

NEET 2016

A particle moves so that its position vector is given by $\mathbf{r} = \cos\omega t\,\hat{x} + \sin\omega t\,\hat{y}$, where $\omega$ is a constant. Which of the following is true?

  1. Velocity and acceleration both are parallel to $\mathbf{r}$
  2. Velocity is perpendicular to $\mathbf{r}$ and acceleration is directed towards the origin
  3. Velocity is perpendicular to $\mathbf{r}$ and acceleration is directed away from the origin
  4. Velocity and acceleration both are perpendicular to $\mathbf{r}$
Answer: (2)

Why: $\mathbf{v} = -\omega\sin\omega t\,\hat{x} + \omega\cos\omega t\,\hat{y}$. Dot product $\mathbf{r}\cdot\mathbf{v} = 0$, so $\mathbf{v}\perp\mathbf{r}$. Differentiating again, $\mathbf{a} = -\omega^2\mathbf{r}$ — anti-parallel to $\mathbf{r}$, directed toward the origin. (Uniform circular motion in disguise: radius 1, angular speed $\omega$.) The acceleration here is not constant in direction, but the differentiation technique — $d\mathbf{r}/dt$ and $d\mathbf{v}/dt$ component-wise — is the core skill of this subtopic.

FAQs — Motion in a Plane with Constant Acceleration

Short answers to the questions NEET aspirants ask most about 2-D constant-acceleration kinematics.

What does it mean to say x- and y-motions decouple under constant acceleration?
When a is constant, r = r0 + v0 t + ½ a t² splits into two independent scalar equations — one for x(t) using a_x, v0x, x0, and one for y(t) using a_y, v0y, y0. Each is a 1-D constant-acceleration problem. NCERT §3.8 states this explicitly.
What are the vector kinematic equations for motion in a plane with constant acceleration?
Two: v = v0 + a t (NCERT Eq. 3.33a) and r = r0 + v0 t + ½ a t² (NCERT Eq. 3.34a). Resolving gives four scalar equations: v_x = v0x + a_x t, v_y = v0y + a_y t, x = x0 + v0x t + ½ a_x t², y = y0 + v0y t + ½ a_y t². The third relation v² = v0² + 2 a · (r − r0) uses the dot product.
Can the 1-D kinematic equations with signs be used directly for 2-D motion?
Only if motion is restricted to a single axis. In 2-D you must resolve all vectors into perpendicular components and apply the scalar equations independently to each component. Adding magnitudes of vectors along different directions is a classic NEET trap.
What is the difference between average velocity and the average of initial and final velocities?
Average velocity is v_avg = Δr/Δt. The expression v_avg = (v_initial + v_final)/2 holds only when a is constant. Even then, |v_avg| = (|v_i| + |v_f|)/2 is wrong — the average is a vector average, not a magnitude average.
How is projectile motion related to motion in a plane with constant acceleration?
Projectile motion is the special case a = (0, −g). The vector equations, decoupling and trajectory equation all specialise directly when a_x = 0 and a_y = −g. NCERT §3.9 builds projectile motion on §3.8 in exactly this way.
If r(t) = 3t î + 2t² ĵ, what are v and a?
Differentiate component-wise. v = 3 î + 4t ĵ; a = 4 ĵ. The x-velocity is constant (3 m/s), the y-velocity grows linearly, and a is a constant 4 m/s² along +y. This is NCERT Exercise 3.17.
Related Topics
Motion in a Plane — Parent chapter Full chapter overview: vectors, projectile, circular motion, relative velocity, PYQs. Scalars and Vectors The language vectors are written in — magnitude, direction, equality, null vectors. Resolution of Vectors Breaking any vector into rectangular components using $\hat{\imath}, \hat{\jmath}, \hat{k}$. Vector Addition — Graphical & Analytical Triangle, parallelogram and polygon laws; the resultant formula. Projectile Motion The constant-acceleration framework with $a_x = 0$, $a_y = -g$. Uniform Circular Motion Acceleration of constant magnitude but changing direction — not constant $\mathbf{a}$. Relative Velocity in Two Dimensions $\mathbf{v}_{AB} = \mathbf{v}_A - \mathbf{v}_B$ and the river/rain–umbrella problems.