This MCQ module is based on: Motion in Plane
TOPIC 11 OF 33
Motion in Plane
🎓 Class 11
Physics
CBSE
Theory
Ch 3 – Motion in a Plane
⏱ ~14 min
🌐 Language: [gtranslate]
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Motion in Plane
3.7 Motion in a Plane
In this section we shall see how to describe motion in two dimensions using vectors.
3.7.1 Position Vector and Displacement
The position vector \(\vec{r}\) of a particle P located in a plane with reference to the origin of an x-y reference frame (Fig. 3.12) is given by:
\[\vec{r} = x\hat{i} + y\hat{j}\]
where x and y are components of \(\vec{r}\) along x- and y-axes or simply they are the coordinates of the object.
Suppose a particle moves along the curve shown by the thick line and is at P at time \(t\) and P' at time \(t'\). Then, the displacement is:
\[\Delta\vec{r} = \vec{r}' - \vec{r}\]
And is directed from P to P'. We can write \(\Delta\vec{r}\) in component form:
\[\Delta\vec{r} = (x' - x)\hat{i} + (y' - y)\hat{j} = \Delta x\,\hat{i} + \Delta y\,\hat{j}\]
3.7.2 Velocity
The average velocity \(\vec{\bar{v}}\) of an object is the ratio of the displacement and the corresponding time interval:
\[\vec{\bar{v}} = \frac{\Delta\vec{r}}{\Delta t} = \frac{\Delta x\,\hat{i} + \Delta y\,\hat{j}}{\Delta t} = \bar{v_x}\,\hat{i} + \bar{v_y}\,\hat{j}\]
The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero:
\[\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta\vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}\]
The meaning of the limiting process: The direction of average velocity is along the chord (e.g., PP' in Fig. 3.13a). As Δt is made smaller and smaller, P' approaches P, and the chord approaches the tangent at P. So the instantaneous velocity at P is along the tangent at that point.
In component form:
\[\vec{v} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\,\hat{i} + v_y\,\hat{j}\]
The magnitude is \(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\) and direction \(\theta = \tan^{-1}(v_y/v_x)\).
3.7.3 Acceleration
The average acceleration \(\vec{\bar{a}}\) is the ratio of the change in velocity \(\Delta\vec{v}\) and the corresponding time interval \(\Delta t\):
\[\vec{\bar{a}} = \frac{\Delta\vec{v}}{\Delta t}\]
Instantaneous acceleration:
\[\vec{a} = \lim_{\Delta t \to 0} \frac{\Delta\vec{v}}{\Delta t} = \frac{d\vec{v}}{dt} = a_x\,\hat{i} + a_y\,\hat{j}\]
Crucial difference from 1D motion: In 1D, velocity and acceleration are along the same line. In 2D, the acceleration vector \(\vec{a}\) can have any direction relative to \(\vec{v}\) — and this is why curved trajectories occur.
3.8 Motion in a Plane with Constant Acceleration
Suppose that an object is moving in x-y plane with velocity \(\vec{v}_0\) at time \(t = 0\) and constant acceleration \(\vec{a}\). At time \(t\), the velocity is given by:
\[\vec{v} = \vec{v}_0 + \vec{a}\,t\]
In component form:
\[v_x = v_{0x} + a_x\,t \qquad v_y = v_{0y} + a_y\,t\]
Position vector at time \(t\):
\[\vec{r} = \vec{r}_0 + \vec{v}_0\,t + \frac{1}{2}\vec{a}\,t^2\]
In component form:
\[x = x_0 + v_{0x}\,t + \frac{1}{2}a_x\,t^2\]
\[y = y_0 + v_{0y}\,t + \frac{1}{2}a_y\,t^2\]
Key Insight: Motion in a plane (2D) with constant acceleration can be treated as TWO independent 1D motions — one along x, one along y. Each follows the standard kinematic equations from Chapter 2. This is the principle behind projectile motion analysis.
Worked Example 1 (NCERT Example 3.4): Velocity and Position
The position of a particle is given by \(\vec{r} = 3t\hat{i} + 2t^2\hat{j} + 5\hat{k}\) where t is in seconds and coefficients have proper units. (a) Find \(\vec{v}(t)\) and \(\vec{a}(t)\). (b) Find the magnitude and direction of \(\vec{v}(t)\) at t = 1.0 s.
(a) Velocity and Acceleration:
\[\vec{v} = \frac{d\vec{r}}{dt} = 3\hat{i} + 4t\,\hat{j}\]
\[\vec{a} = \frac{d\vec{v}}{dt} = 4\,\hat{j} \text{ (constant)}\]
(b) At t = 1 s: \[\vec{v}(1) = 3\hat{i} + 4\hat{j}\] \[|\vec{v}| = \sqrt{3^2 + 4^2} = 5 \text{ m/s}\] \[\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx \boxed{53.13° \text{ from x-axis}}\]
(b) At t = 1 s: \[\vec{v}(1) = 3\hat{i} + 4\hat{j}\] \[|\vec{v}| = \sqrt{3^2 + 4^2} = 5 \text{ m/s}\] \[\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx \boxed{53.13° \text{ from x-axis}}\]
Worked Example 2 (NCERT Example 3.5): Motion with Acceleration
A particle starts from origin at t = 0 with a velocity of \(5.0\hat{i}\) m/s and moves in x-y plane under the action of force which produces constant acceleration of \((3.0\hat{i} + 2.0\hat{j})\) m/s². (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m? (b) What is the speed of the particle at this time?
Given: \(\vec{v}_0 = 5.0\hat{i}\), \(\vec{a} = 3.0\hat{i} + 2.0\hat{j}\), \(\vec{r}_0 = 0\).
Position at time t: \[x = 0 + 5.0t + \frac{1}{2}(3.0)t^2 = 5t + 1.5t^2\] \[y = 0 + 0 + \frac{1}{2}(2.0)t^2 = t^2\]
(a) When x = 84 m: \(1.5t^2 + 5t - 84 = 0\). Using quadratic formula: \[t = \frac{-5 + \sqrt{25 + 4 \cdot 1.5 \cdot 84}}{3} = \frac{-5 + \sqrt{529}}{3} = \frac{-5 + 23}{3} = 6 \text{ s}\] \[y = (6)^2 = \boxed{36 \text{ m}}\]
(b) At t = 6 s: \[v_x = 5 + 3 \times 6 = 23 \text{ m/s}\] \[v_y = 0 + 2 \times 6 = 12 \text{ m/s}\] \[|\vec{v}| = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \approx \boxed{25.9 \text{ m/s}}\]
Position at time t: \[x = 0 + 5.0t + \frac{1}{2}(3.0)t^2 = 5t + 1.5t^2\] \[y = 0 + 0 + \frac{1}{2}(2.0)t^2 = t^2\]
(a) When x = 84 m: \(1.5t^2 + 5t - 84 = 0\). Using quadratic formula: \[t = \frac{-5 + \sqrt{25 + 4 \cdot 1.5 \cdot 84}}{3} = \frac{-5 + \sqrt{529}}{3} = \frac{-5 + 23}{3} = 6 \text{ s}\] \[y = (6)^2 = \boxed{36 \text{ m}}\]
(b) At t = 6 s: \[v_x = 5 + 3 \times 6 = 23 \text{ m/s}\] \[v_y = 0 + 2 \times 6 = 12 \text{ m/s}\] \[|\vec{v}| = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \approx \boxed{25.9 \text{ m/s}}\]
🎯 Interactive Simulation: 2D Motion Trajectory
Set initial velocity components and constant acceleration. Watch the particle's trajectory unfold in time.
Try: Set aᵧ = −9.8 (gravity), aₓ = 0 → projectile! Set both a = 0 → straight-line motion.
3.9 Relative Velocity in Two Dimensions
The concept of relative velocity in two dimensions is similar to that in 1D except that we have to use vector addition more carefully to subtract the vectors. Suppose two objects A and B are moving with velocities \(\vec{v_A}\) and \(\vec{v_B}\) (each with respect to some common ground). Then:
\[\vec{v_{AB}} = \vec{v_A} - \vec{v_B}\]
and similarly:
\[\vec{v_{BA}} = \vec{v_B} - \vec{v_A} = -\vec{v_{AB}}\]
Worked Example 3 (NCERT Example 3.6): Relative Velocity Application
Rain is falling vertically with a speed of 35 m/s. Wind starts blowing after some time with a speed of 12 m/s in east-to-west direction. In which direction should a boy waiting at a bus stop hold his umbrella?
Setup: Take east as +x, vertical up as +y.
\(\vec{v_{rain,\,ground}} = -35\,\hat{j}\) m/s
\(\vec{v_{wind}} = -12\,\hat{i}\) m/s (east-to-west = −x)
The combined velocity of rain in still air relative to wind isn't quite what we want — actually we want velocity of rain relative to the boy (at rest):
\[\vec{v_{rain,\,boy}} = -12\,\hat{i} - 35\,\hat{j}\]
\[|\vec{v_{rain,\,boy}}| = \sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37 \text{ m/s}\]
The angle from vertical: \[\tan\theta = \frac{12}{35} = 0.343 \Rightarrow \theta \approx \boxed{18.9°}\] The boy should tilt the umbrella \(\boxed{\approx 19°}\) towards the west (i.e., into the wind direction).
\(\vec{v_{rain,\,ground}} = -35\,\hat{j}\) m/s
\(\vec{v_{wind}} = -12\,\hat{i}\) m/s (east-to-west = −x)
The combined velocity of rain in still air relative to wind isn't quite what we want — actually we want velocity of rain relative to the boy (at rest):
\[\vec{v_{rain,\,boy}} = -12\,\hat{i} - 35\,\hat{j}\]
\[|\vec{v_{rain,\,boy}}| = \sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37 \text{ m/s}\]
The angle from vertical: \[\tan\theta = \frac{12}{35} = 0.343 \Rightarrow \theta \approx \boxed{18.9°}\] The boy should tilt the umbrella \(\boxed{\approx 19°}\) towards the west (i.e., into the wind direction).
Worked Example 4: Boat in River Crossing
A river flows due east at 5 m/s. A boat that can travel at 12 m/s in still water heads due north across the river. Find: (a) the boat's velocity relative to ground, (b) the angle the boat's actual path makes with north.
Setup: Take north = +y, east = +x.
\(\vec{v_{boat,\,water}} = 12\,\hat{j}\) m/s
\(\vec{v_{water,\,ground}} = 5\,\hat{i}\) m/s
\(\vec{v_{boat,\,ground}} = \vec{v_{boat,\,water}} + \vec{v_{water,\,ground}} = 5\hat{i} + 12\hat{j}\)
(a) Magnitude: \[|\vec{v_{boat,\,ground}}| = \sqrt{5^2 + 12^2} = \sqrt{169} = \boxed{13 \text{ m/s}}\]
(b) Angle east of north: \[\tan\alpha = \frac{5}{12} \Rightarrow \alpha = \tan^{-1}(0.417) \approx \boxed{22.6°}\] The boat drifts 22.6° east of north despite heading north.
\(\vec{v_{boat,\,water}} = 12\,\hat{j}\) m/s
\(\vec{v_{water,\,ground}} = 5\,\hat{i}\) m/s
\(\vec{v_{boat,\,ground}} = \vec{v_{boat,\,water}} + \vec{v_{water,\,ground}} = 5\hat{i} + 12\hat{j}\)
(a) Magnitude: \[|\vec{v_{boat,\,ground}}| = \sqrt{5^2 + 12^2} = \sqrt{169} = \boxed{13 \text{ m/s}}\]
(b) Angle east of north: \[\tan\alpha = \frac{5}{12} \Rightarrow \alpha = \tan^{-1}(0.417) \approx \boxed{22.6°}\] The boat drifts 22.6° east of north despite heading north.
📐 Activity 3.3 — Walking on Moving Train (Relative Velocity)
L4 Analyse
Imagine a thought experiment:
- A train moves due east at 60 km/h.
- A passenger walks within the train towards the back at 4 km/h.
- A bird flies overhead at 8 km/h due south (relative to ground).
Predict: Calculate (a) passenger's velocity relative to ground, (b) bird's velocity relative to passenger.
Setting up axes: East = +x, North = +y.
Train: \(\vec{v_T} = 60\hat{i}\) km/h. Passenger relative to train: \(\vec{v_{P,T}} = -4\hat{i}\) km/h.
(a) Passenger relative to ground: \(\vec{v_P} = \vec{v_T} + \vec{v_{P,T}} = 60\hat{i} - 4\hat{i} = 56\hat{i}\) km/h. So 56 km/h due east.
(b) Bird's velocity (ground): \(\vec{v_B} = -8\hat{j}\) km/h.
Bird relative to passenger: \(\vec{v_{B,P}} = \vec{v_B} - \vec{v_P} = -8\hat{j} - 56\hat{i} = -56\hat{i} - 8\hat{j}\) km/h.
Magnitude: \(\sqrt{56^2 + 8^2} = \sqrt{3136 + 64} = \sqrt{3200} \approx 56.6\) km/h.
Direction: \(\tan\theta = 8/56 \approx 0.143 \Rightarrow \theta \approx 8.1°\) south of west. So the bird seems to fly mostly westward at 56.6 km/h from the passenger's frame.
🎯 Competency-Based Questions
A particle moves in xy-plane with position vector \(\vec{r}(t) = (4t)\hat{i} + (3t^2 - 2)\hat{j}\) where r is in meters and t is in seconds.
Q1. Find the velocity of the particle at t = 2 s.L3 Apply
Answer: \(\vec{v} = d\vec{r}/dt = 4\hat{i} + 6t\hat{j}\). At t = 2 s: \(\vec{v} = 4\hat{i} + 12\hat{j}\) m/s. \(|\vec{v}| = \sqrt{16 + 144} = \sqrt{160} \approx \boxed{12.65 \text{ m/s}}\).
Q2. Determine whether the acceleration is constant. If yes, find its magnitude.L4 Analyse
Answer: \(\vec{a} = d\vec{v}/dt = 0\hat{i} + 6\hat{j}\) m/s² — independent of time, so YES the acceleration is constant. Magnitude = \(\boxed{6 \text{ m/s}^2}\), directed along +y.
Q3. Two cars A and B move with velocities \(\vec{v_A} = 30\hat{i}\) m/s and \(\vec{v_B} = 20\hat{j}\) m/s. Find the velocity of A as observed from B. L3 Apply
Answer: \(\vec{v_{AB}} = \vec{v_A} - \vec{v_B} = 30\hat{i} - 20\hat{j}\). Magnitude: \(\sqrt{900 + 400} = \sqrt{1300} \approx \boxed{36.06 \text{ m/s}}\). Angle below x-axis: \(\tan\theta = 20/30 \Rightarrow \theta \approx 33.7°\).
Q4. Evaluate: A passenger in train A (moving east at 50 km/h) sees train B moving north. Train B's actual ground velocity is 30 km/h NE. Critique the passenger's observation. L5 Evaluate
Answer: Compute relative velocity. \(\vec{v_A} = 50\hat{i}\). \(\vec{v_B} = 30\cos45°\hat{i} + 30\sin45°\hat{j} \approx 21.2\hat{i} + 21.2\hat{j}\). \(\vec{v_{B,A}} = \vec{v_B} - \vec{v_A} = -28.8\hat{i} + 21.2\hat{j}\). Direction: 2nd quadrant, mostly west-north. Magnitude ≈ 35.8 km/h. So passenger sees B moving northwest, NOT due north. The passenger's claim is incorrect — B has both northward AND westward components in A's frame.
Q5. HOT: Design an experimental setup to measure the wind velocity using two boats: one anchored, one moving. Outline the measurements needed. L6 Create
Sample Design:
- Anchor Boat 1 — measure flag direction & angle of windsock relative to compass north (gives wind direction).
- Move Boat 2 with known velocity \(\vec{v_2}\). Measure apparent wind direction and angle on Boat 2 (using its windsock).
- Apparent wind on Boat 2: \(\vec{v_{wind,2}} = \vec{v_{wind,ground}} - \vec{v_2}\).
- From two angle measurements (on stationary and moving boats) and known \(\vec{v_2}\), solve vectorially for \(\vec{v_{wind,ground}}\).
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: The instantaneous velocity vector is always tangent to the trajectory.
R: As Δt → 0, the chord PP' rotates and approaches the tangent at P.
Answer: (A). Both true; R correctly explains A using the limit definition.
A: An object can have zero velocity but non-zero acceleration.
R: Velocity is the rate of change of position, while acceleration is the rate of change of velocity — independently defined.
Answer: (A). Both true; R correctly explains A. Example: a ball thrown straight up — at the highest point velocity = 0, but acceleration = g (downward).
A: Relative velocity of A with respect to B has the same magnitude as that of B with respect to A.
R: \(\vec{v_{AB}} = -\vec{v_{BA}}\), and the magnitudes of opposite vectors are equal.
Answer: (A). Both true; R correctly explains A. Magnitudes equal; only the direction is reversed.
Frequently Asked Questions - Motion in Plane
What is the main concept covered in Motion in Plane?
In NCERT Class 11 Physics Chapter 3 (Motion in a Plane), "Motion in Plane" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Motion in Plane useful in real-life applications?
Real-life applications of Motion in Plane from NCERT Class 11 Physics Chapter 3 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Motion in Plane?
Key formulas in Motion in Plane (NCERT Class 11 Physics Chapter 3 Motion in a Plane) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 3?
NCERT Class 11 Physics Chapter 3 (Motion in a Plane) is structured so each part builds on the previous one. Motion in Plane connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Motion in Plane?
CBSE board questions from Motion in Plane typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Motion in Plane lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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