Motion in a Plane with Constant Acceleration (NCERT 3.7–3.8)
1. What Changes When Motion Gets a Second Axis?
In Chapter 2 the position was a single number x(t); now it is a vector r(t) = x(t)î + y(t)ĵ, and the definitions upgrade automatically: velocity v = dr/dt = vxî + vyĵ and acceleration a = dv/dt. Differentiating a vector differentiates each component — so kinematics in a plane is not new physics; it is Chapter 2's kinematics running twice, once per axis, with the answers recombined by Pythagoras. The average definitions survive too: vavg = Δr/Δt (a vector) and average speed = total path/Δt (a scalar), with average speed ≥ |vavg| exactly as in 1-D (Chapter 2's mean-trap page).
The entire section is governed by one sentence: if a is constant, every Chapter 2 equation applies to each component independently. That sentence is the source of projectile motion's behaviour, and NTA's statement-based questions test precisely this independence idea.
2. Complete Theory: The Vector Equations and Their Components
Vector equations (constant a only). Integrating a = dv/dt gives v = v0 + at; integrating again gives r = r0 + v0t + ½at². These are vector equations — the additions are triangle-law additions — and they are valid only for constant acceleration, the same validity condition as their 1-D parents. A useful third form eliminates t: r − r0 = ½(v0 + v)t.
Component equations. Choosing rectangular axes splits each vector equation into two ordinary 1-D sets: x0, v0x, ax produce the x-motion; y0, v0y, ay produce the y-motion; the two share only the parameter t. Average acceleration in 2-D is aavg = Δv/Δt with Δv = v2 − v1 computed component-wise — subtract arrows, not magnitudes.
Trajectory. Eliminating t between x(t) and y(t) yields the path y(x). When a is constant and nonzero, the path is always a parabola (or a straight line in the degenerate case where v0 is parallel to a — including a = 0). This elimination technique is exactly how the projectile page obtains y = x tan θ − gx²/(2u²cos²θ), so it is worth practising on arbitrary r(t) first, where the algebra is friendlier.
3. Visualising Component Independence
Figure 3.6 — One problem, two independent 1-D problems
Read top to bottom: each axis runs its own Chapter 2 calculation; only at the end do the rows merge into vectors.
uses only x0, v0x, ax — never sees y
uses only y0, v0y, ay — never sees x
Exam read-out: the merging step is where marks are lost — √, not +. For vx = 3 and vy = 4, speed is 5 m/s, never 7. The bullet-and-block demonstration (dropped vs fired bullet landing together) is the classic assertion–reason wrapper for this figure.
4. Solved Examples
Example 1 — Full workup of r = 3tî + (4t − 2t²)ĵ (metres, seconds)
Find v(t), a(t), the maximum y-coordinate and the particle's position when it occurs.
Solution (step by step): Differentiate component-wise: v = 3î + (4 − 4t)ĵ m/s; differentiate again: a = −4ĵ m/s² — constant, downward, so the kinematic equations are legitimate. The y-velocity vanishes when 4 − 4t = 0 → t = 1 s; there y = 4(1) − 2(1)² = 2 m (the y-peak), while x(1) = 3 m keeps advancing — the particle is at (3, 2). At t = 0, speed = √(3² + 4²) = 5 m/s. Note the x-velocity never changes: the horizontal world is force-free.
Example 2 — Constant acceleration from the origin
A particle starts at the origin with v0 = 6î m/s and moves with constant a = 2ĵ m/s². Find its position and speed at t = 2 s.
Solution: x-row: ax = 0 → x = 6 × 2 = 12 m. y-row: starts from rest vertically → y = ½ × 2 × 2² = 4 m. Position (12, 4) m. Velocities: vx = 6 (unchanged), vy = 2 × 2 = 4 → v = 6î + 4ĵ, speed = √(36 + 16) = √52 = 2√13 ≈ 7.21 m/s. Direction: tan α = 4/6 → α ≈ 33.7° above the x-axis. The trajectory: eliminate t = x/6 → y = 2(x/6)²/2... i.e., y = x²/36 — a parabola, as promised.
5. Practice Questions
Q1. For r = 2t²î + 3tĵ, find the average velocity between t = 0 and t = 2 s.
ANSWER: r(0) = 0; r(2) = 8î + 6ĵ. vavg = Δr/Δt = (8î + 6ĵ)/2 = 4î + 3ĵ m/s, magnitude 5 m/s at tan⁻¹(3/4) ≈ 36.9°. Endpoints only — the curved path is irrelevant to vavg.
Q2. A particle moves with constant velocity v = 3î + 4ĵ m/s. Find its displacement in 5 s and its acceleration.
ANSWER: Δr = vΔt = 15î + 20ĵ m, magnitude √(225+400) = 25 m. Acceleration is 0 — constant velocity means constant speed AND constant direction.
Q3 (MCQ). In a plane motion with constant acceleration, which statement is FALSE? (a) trajectory is a parabola (b) x and y equations are independent (c) velocity is always parallel to acceleration (d) v = v0 + at applies
ANSWER: (c). Velocity is parallel to acceleration only in 1-D or when v0 ∥ a; generally v tilts gradually toward a. (a) holds when a ⊥ component of v0 exists; straight-line motion is the degenerate parabola-free case.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| r(t) = x(t)î + y(t)ĵ | Position vector; components are ordinary functions |
| v = dr/dt = vxî + vyĵ | Differentiate component-wise; speed = |v| = √(vx²+vy²) |
| a = dv/dt | Second derivative of r; constant a unlocks the next rows |
| v = v0 + at | Vector form; constant a only; split per axis |
| r = r0 + v0t + ½at² | Vector form; parabolic path unless v0 ∥ a |
| r − r0 = ½(v0 + v)t | t-free form; needs constant a |
| vavg = Δr/Δt | Vector from endpoints; average speed ≥ |vavg| |
| aavg = Δv/Δt | Δv component-wise; subtract arrows, not magnitudes |
7. Frequently Asked Questions
What are the equations of motion in a plane with constant acceleration?
In vector form they look exactly like the one-dimensional equations: v = v0 + at and r = r0 + v0 t + ½ a t², valid only when the acceleration vector a is constant. Written in components, the single vector equation splits into two independent one-dimensional sets: x = x0 + v0x t + ½ ax t² and y = y0 + v0y t + ½ ay t², with vx = v0x + ax t and vy = v0y + ay t. Each axis is solved as a separate Chapter 2 problem, then the results are recombined.
What does it mean that the x and y motions are independent?
The acceleration along x (a_x) influences only the x-equations, and a_y influences only the y-equations — the axes do not talk to each other when chosen along rectangular directions. A bullet fired horizontally and a bullet dropped from the same height reach the ground together, because the vertical motion is identical for both regardless of the horizontal velocity. Independence is why two-dimensional problems reduce to two one-dimensional problems.
How do you find the trajectory of a particle from its r(t)?
Write the position components x(t) and y(t), eliminate the parameter t between them, and the resulting relation y(x) is the trajectory. For example, if x = 3t and y = 4t − 2t², then t = x/3 and y = (4/3)x − (2/9)x², which is a parabola opening downward. A straight-line trajectory results only when the velocity direction never changes, i.e., when v and a are parallel (or a = 0).
Is the speed of a particle equal to the magnitude of its average velocity?
Not in general. Instantaneous speed is the magnitude of instantaneous velocity, |v| = √(vx² + vy²). But average velocity is displacement over time — a vector — while average speed is total path length over time — a scalar. Whenever the path curves or reverses, the path length exceeds the displacement magnitude, so average speed is strictly greater than the magnitude of average velocity, exactly as in the one-dimensional case of Chapter 2.
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