NCERT Solutions for Class 11 Physics Chapter 3: Motion in a Plane (NCERT 2026–27)

These Class 11 Physics Chapter 3 solutions cover Motion in a Plane with complete, step-by-step answers to all 22 NCERT exercise questions. Every numerical is solved with full working, correct units and a final answer cross-checked against the official NCERT key, so you can confidently revise vectors, projectile motion and uniform circular motion for the session 2026–27.

Class: 11 Subject: Physics Chapter: 3 Chapter Name: Motion in a Plane Exercises: 3.1 – 3.22 Session: 2026–27
On this page

Class 11 Physics Chapter 3 Solutions – Overview

Chapter 3, Motion in a Plane, extends one-dimensional kinematics into two dimensions. Because two perpendicular directions are now possible, plus/minus signs are no longer enough, so the chapter first builds the language of vectors — addition by the triangle and parallelogram laws, multiplication by a real number, resolution into components, and the analytical (component) method using unit vectors î and ĵ. It then applies these tools to motion in a plane with constant acceleration, and treats two important special cases in detail: projectile motion (a parabolic path that is the superposition of uniform horizontal motion and free fall) and uniform circular motion (constant speed but a centripetal acceleration v2/R directed towards the centre). Mastering this chapter is essential because every later mechanics topic — laws of motion, work-energy, circular dynamics and gravitation — relies on resolving vectors confidently.

Key Concepts & Definitions

Scalar: a quantity with magnitude only (e.g. mass, speed, distance, temperature). It follows ordinary algebra.

Vector: a quantity with both magnitude and direction that obeys the triangle/parallelogram law of addition (e.g. displacement, velocity, acceleration, force).

Equal vectors: two vectors are equal only if they have the same magnitude and the same direction.

Unit vector (n̂): a vector of magnitude 1 that specifies direction only; it has no dimension or unit. Along the axes we use î, ĵ, k̂.

Resolution of a vector: expressing a vector as the sum of components, usually along the x- and y-axes: Ax = A cos θ, Ay = A sin θ.

Projectile: any object in flight after being thrown, moving under gravity alone (ax = 0, ay = −g); its path is a parabola.

Uniform circular motion: motion in a circle at constant speed; the velocity is tangential while the acceleration (centripetal, v2/R) points towards the centre.

Important Formulas

Magnitude & direction: A = √(Ax2 + Ay2),   tan θ = Ay/Ax

Parallelogram law: R = √(A2 + B2 + 2AB cos θ)

Motion with constant acceleration: v = v0 + a·t,   r = r0 + v0t + ½a·t2

Projectile: Time of flight Tf = 2v0sinθ0/g;   Max height hm = (v0sinθ0)2/2g;   Range R = v02sin 2θ0/g (R is maximum at θ0 = 45°, Rmax = v02/g)

Uniform circular motion: ac = v2/R = ω2R;   v = ωR;   ω = 2π/T = 2πν;   ac = 4π2ν2R

NCERT Solutions – Motion in a Plane (Exercises 3.1–3.22)

All questions are reproduced verbatim from the NCERT textbook; the answers below are original, expert-checked and verified against the official NCERT answer key.

3.1 State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

ANSWER Scalars: volume, mass, speed, density, number of moles, angular frequency (these have magnitude only). Vectors: acceleration, velocity, displacement, angular velocity (these have both magnitude and a definite direction).

3.2 Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.

ANSWER The two scalars are work and current. Work is the dot product of two vectors (force and displacement), which yields a scalar; electric current has magnitude and a sense of flow but does not obey the vector law of addition, so it is treated as a scalar.

3.3 Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

ANSWER The only vector is impulse. Impulse = force × time interval; since force is a vector, impulse has both magnitude and direction. All the other listed quantities are scalars.

3.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful: (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.

ANSWER (a) Not always meaningful — two scalars can be added only if they have the same dimensions (e.g. you cannot add a mass to a time). (b) Not meaningful — a scalar and a vector cannot be added even if dimensions match, because a direction cannot be attached to a pure number. (c) Meaningful — multiplying a vector by a scalar gives a vector (e.g. force = mass × acceleration). (d) Meaningful — any two scalars can be multiplied (their units multiply too, e.g. power × time = energy). (e) Not always meaningful — two vectors can be added only if they represent the same physical quantity (same dimensions). (f) Meaningful — a component of a vector is itself a vector with the same dimensions, so it can be added to the vector. Thus only (c) and (d) are unconditionally meaningful.

3.5 Read each statement below carefully and state with reasons, if it is true or false: (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle, (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

ANSWER (a) True — the magnitude of a vector is a pure number with a unit, so it is a scalar. (b) False — a component such as Axî still has a direction, so it is a vector, not a scalar. (c) False — path length equals the displacement magnitude only for straight-line motion without reversal; in general path length ≥ displacement. (d) True — since path length ≥ displacement, average speed ≥ magnitude of average velocity; they are equal only for a straight path. (e) True — three non-coplanar vectors always have a resultant component out of any chosen plane, so they can never cancel to zero.

3.6 Establish the following vector inequalities geometrically or otherwise: (a) |a+b| ≤ |a| + |b|, (b) |a+b| ≥ ||a| − |b||, (c) |a−b| ≤ |a| + |b|, (d) |a−b| ≥ ||a| − |b||. When does the equality sign above apply?

ANSWER Place the vectors head to tail so they form a triangle. By the geometry of any triangle, the sum of any two sides is never less than the third side, and the difference of any two sides is never greater than the third side. (a) The side representing (a+b) cannot exceed the sum of the other two sides ⇒ |a+b| ≤ |a|+|b|. (b) The side representing (a+b) cannot be less than the difference of the other two sides ⇒ |a+b| ≥ ||a|−|b||. (c) and (d) follow the same way by treating subtraction as addition of (−b): |a−b| ≤ |a|+|b| and |a−b| ≥ ||a|−|b||. When equality holds: in (a) and (d) the equality holds when a and b are parallel (same direction); in (b) and (c) it holds when a and b are anti-parallel (opposite directions). In short, equality applies for collinear vectors.

3.7 Given a + b + c + d = 0, which of the following statements are correct: (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of (b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear?

ANSWER (a) Incorrect — the four vectors only need to form a closed polygon; none of them must individually be zero. (b) Correct — since a+b+c+d = 0, we have (a+c) = −(b+d), so their magnitudes are equal. (c) Correct — a = −(b+c+d), and the magnitude of a sum can never exceed the sum of the magnitudes, so |a| ≤ |b|+|c|+|d|. (d) Correct — (b+c) = −(a+d); the vector (a+d) lies in the plane of a and d, hence so does (b+c); if a and d are collinear, (a+d) lies along their line, so (b+c) does too. Thus all statements except (a) are correct.

3.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

ANSWER P and Q are the ends of a diameter, so for every girl the displacement is the straight line PQ regardless of the path taken. Magnitude of displacement = diameter = 2 × radius = 2 × 200 = 400 m for each girl. The displacement equals the actual path length only for the girl who skates straight along the diameter PQ — that is girl B, whose path is the straight line PQ of length 400 m.

3.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

ANSWER (a) Net displacement: the cyclist starts and ends at the centre O, so initial and final positions coincide ⇒ net displacement = 0. (b) Average velocity = net displacement / time = 0 / (10 min) = 0. (c) Average speed: total path length = OP + arc PQ + QO. OP = QO = radius = 1 km. The arc PQ subtends a quarter circle (a quadrant), so arc = (1/4)(2πR) = (πR)/2 = (π × 1)/2 = 1.571 km. Total path = 1 + 1.571 + 1 = 3.571 km. Time = 10 min = 10/60 h = (1/6) h. Average speed = 3.571 / (1/6) = 3.571 × 6 = 21.4 km h−1.

3.10 On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

ANSWER Turning left by 60° each time traces a regular hexagon of side 500 m (exterior turn 60° means the path closes after six 500-m legs). Third turn: after three legs the motorist is at the vertex diametrically opposite the start. Displacement = 2 × (apothem-to-vertex) = diameter of the hexagon = 2 × 500 = 1000 m (1 km), directed at 60° with the initial direction. Path length = 3 × 500 = 1500 m (1.5 km). So magnitude of displacement (1 km) < path length (1.5 km). Sixth turn: after six legs the path is complete and the motorist returns to the starting point, so displacement = null vector (0). Path length = 6 × 500 = 3000 m (3 km). Eighth turn: after eight legs the motorist is two legs beyond the start. By geometry the displacement = 866 m at 30° with the initial direction (since 2 × 500 × cos 30° = 1000 × 0.866 = 866 m). Path length = 8 × 500 = 4000 m (4 km).

3.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

ANSWER Time = 28 min = 28/60 h = 0.4667 h. (a) Average speed = total path / time = 23 / 0.4667 = 49.3 km h−1. (b) Magnitude of average velocity = displacement / time = 10 / 0.4667 = 21.4 km h−1 (the displacement is the straight-line distance 10 km). No, the two are not equal. They are equal only when the motion is along a straight path; here the path is circuitous, so average speed exceeds the magnitude of average velocity.

3.12 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s−1 can go without hitting the ceiling of the hall?

ANSWER To get maximum horizontal range without touching the ceiling, the ball must just reach the maximum height hm = 25 m. Using hm = (v0sinθ)2/2g: 25 = (40)2 sin2θ / (2 × 9.8) ⇒ sin2θ = (25 × 2 × 9.8) / 1600 = 490/1600 = 0.30625. sinθ = 0.5534 ⇒ θ = 33.6°; then cosθ = 0.8329. Horizontal range R = v02 sin 2θ / g = v02 (2 sinθ cosθ) / g = (1600 × 2 × 0.5534 × 0.8329) / 9.8. R = (1600 × 0.9218) / 9.8 = 1474.9 / 9.8 = 150.5 m.

3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

ANSWER Maximum horizontal range occurs at θ = 45°: Rmax = v02/g = 100 m ⇒ v02 = 100 × g = 100g. To throw highest, the ball is thrown straight up (θ = 90°), where all the speed becomes vertical. Maximum height H = v02/(2g) = 100g/(2g) = 100/2 = 50 m. (The answer is independent of the exact value of g.)

3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

ANSWER R = 80 cm = 0.80 m. Frequency ν = 14/25 rev s−1; time period T = 25/14 s. Angular speed ω = 2π/T = 2π × 14/25 = 88/25 = 3.52 rad s−1. Centripetal acceleration ac = ω2R = (3.52)2 × 0.80 = 12.39 × 0.80 = 9.9 m s−2. Direction: along the radius at every point, directed towards the centre of the circle (centripetal).

3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

ANSWER Convert speed: v = 900 km/h = 900 × (1000/3600) = 250 m s−1. Radius R = 1.00 km = 1000 m. ac = v2/R = (250)2 / 1000 = 62500 / 1000 = 62.5 m s−2. Ratio ac/g = 62.5 / 9.8 = 6.4. So the centripetal acceleration is about 6.4 times g.

3.16 Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre, (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point, (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

ANSWER (a) False — this is true only for uniform circular motion. If the speed changes, there is also a tangential acceleration, so the net acceleration is not purely radial. (b) True — the velocity at any point is always directed along the tangent to the path in the direction of motion. (c) True — over one complete cycle the centripetal acceleration points in every direction equally, so the vector average over the full cycle is zero (a null vector).

3.17 The position of a particle is given by r = 3.0t î − 2.0t2 ĵ + 4.0 k̂ m, where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

ANSWER (a) v(t) = dr/dt = 3.0 î − 4.0t ĵ m s−1 (the constant k̂ term differentiates to zero). a(t) = dv/dt = −4.0 ĵ m s−2 (constant acceleration along the −y direction). (b) At t = 2.0 s: v = 3.0 î − 4.0(2.0) ĵ = 3.0 î − 8.0 ĵ m s−1. Magnitude |v| = √(3.02 + 8.02) = √(9 + 64) = √73 = 8.54 m s−1. Direction: tanθ = vy/vx = −8.0/3.0 = −2.667 ⇒ θ = −69.4°, i.e. about 70° below the x-axis (in the fourth quadrant).

3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m/s and moves in the x-y plane with a constant acceleration of (8.0 î + 2.0 ĵ) m s−2. (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time?

ANSWER Initial velocity: v0x = 0, v0y = 10.0 m s−1. Acceleration: ax = 8.0, ay = 2.0 m s−2. Start at origin (x0 = y0 = 0). (a) x = v0xt + ½axt2 = 0 + ½(8.0)t2 = 4.0t2. Set x = 16: 4.0t2 = 16 ⇒ t2 = 4 ⇒ t = 2 s. y = v0yt + ½ayt2 = 10.0(2) + ½(2.0)(2)2 = 20 + 4 = 24 m. (b) vx = axt = 8.0(2) = 16 m s−1; vy = v0y + ayt = 10.0 + 2.0(2) = 14 m s−1. Speed = √(162 + 142) = √(256 + 196) = √452 = 21.26 m s−1.

3.19 î and ĵ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors î + ĵ, and î − ĵ? What are the components of a vector A = 2 î + 3 ĵ along the directions of î + ĵ and î − ĵ? [You may use graphical method]

ANSWER (î + ĵ): magnitude = √(12 + 12) = √2; direction tanθ = 1/1 = 1 ⇒ θ = 45° with the x-axis. (î − ĵ): magnitude = √(12 + (−1)2) = √2; direction tanθ = −1/1 = −1 ⇒ θ = −45° with the x-axis. Component of A along (î + ĵ): unit vector n̂ = (î + ĵ)/√2. A·n̂ = (2 + 3)/√2 = 5/√2 ⇒ component = 5/√2. Component of A along (î − ĵ): unit vector = (î − ĵ)/√2. A·n̂ = (2 − 3)/√2 = −1/√2 ⇒ component = −1/√2.

3.20 For any arbitrary motion in space, which of the following relations are true: (a) vaverage = (1/2) (v(t1) + v(t2)), (b) vaverage = [r(t2) − r(t1)]/(t2 − t1), (c) v(t) = v(0) + a t, (d) r(t) = r(0) + v(0) t + (1/2) a t2, (e) aaverage = [v(t2) − v(t1)]/(t2 − t1). (The ‘average’ stands for average of the quantity over the time interval t1 to t2)

ANSWER (a) False — the average of the two end velocities equals the average velocity only when acceleration is uniform, not for arbitrary motion. (b) True — this is the definition of average velocity (displacement divided by time interval) and holds for any motion. (c) False — v = v(0) + a t holds only for constant acceleration, not for arbitrary motion. (d) False — this kinematic equation also assumes constant acceleration, so it is not generally true. (e) True — this is the definition of average acceleration (change in velocity divided by time interval) and holds for any motion. Thus only (b) and (e) are true for arbitrary motion.

3.21 Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that (a) is conserved in a process, (b) can never take negative values, (c) must be dimensionless, (d) does not vary from one point to another in space, (e) has the same value for observers with different orientations of axes.

ANSWER (a) False — energy is a scalar and is conserved, but kinetic energy alone is not always conserved (e.g. in inelastic collisions). Conservation is not what defines a scalar. (b) False — many scalars can be negative, for example temperature in °C or electric potential. (c) False — most scalars have dimensions (mass, speed, energy). Being dimensionless is not required. (d) False — scalars such as temperature or gravitational potential do vary from point to point in space. (e) True — the defining property of a scalar is that its value is unchanged (invariant) when the coordinate axes are rotated. Only (e) is true.

3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?

ANSWER The aircraft flies horizontally at height h = 3400 m. The observation point is directly below the mid-path, so the 30° angle is split symmetrically into 15° on each side of the vertical. Horizontal distance covered in 10 s = 2 × h × tan 15° = 2 × 3400 × 0.2679 = 1821.7 m. Speed = distance / time = 1821.7 / 10.0 = 182.2 ≈ 182 m s−1.

Extra Practice Questions

Short Answer Type Questions

Q1. Why can a quantity like electric current be treated as a scalar even though it has a direction of flow?

ANSWERAlthough current has a sense of flow, it does not obey the vector law of addition. When wires meet at a junction, currents add algebraically (Kirchhoff’s rule), not by the parallelogram law, so current is treated as a scalar.

Q2. A body is projected horizontally and another is dropped from the same height at the same instant. Which reaches the ground first?

ANSWERBoth reach the ground at the same time. The horizontal motion has no effect on the vertical fall; the vertical component of motion is identical free fall (a = g) for both, so the time of descent is the same.

Q3. State two differences between distance and displacement.

ANSWER(i) Distance is a scalar (magnitude only) while displacement is a vector (magnitude and direction). (ii) Distance depends on the actual path and is always ≥ the magnitude of displacement, which depends only on the start and end points.

Q4. A projectile is fired at 30° and another at 60° with the same speed. How do their horizontal ranges compare?

ANSWERTheir ranges are equal. Range R = v02sin 2θ/g, and sin 60° = sin 120°, so complementary angles (30° and 60°) give the same range, though the 60° throw goes higher and stays in the air longer.

Q5. Why is uniform circular motion called an accelerated motion even though the speed is constant?

ANSWERVelocity is a vector; in circular motion its direction changes continuously even though the magnitude (speed) stays constant. A changing velocity means the body is accelerating — the acceleration is centripetal, directed towards the centre.

Long Answer Type Questions

Q1. Derive the expression for the maximum height and the time of flight of a projectile launched with speed v0 at angle θ0.

ANSWERResolve the initial velocity: v0x = v0cosθ0 (constant) and v0y = v0sinθ0. Vertically the body decelerates at g. At maximum height the vertical velocity is zero: vy = v0sinθ0 − g tm = 0, giving the time to reach the top tm = v0sinθ0/g. Substituting this in y = v0sinθ0·t − ½g t2 gives the maximum height hm = (v0sinθ0)2/2g. By the symmetry of the parabola the total time of flight is twice the time to the top: Tf = 2v0sinθ0/g. These two results show that higher launch angles give greater height and longer flight time.

Q2. Explain the analytical (component) method of adding two vectors and obtain the magnitude and direction of the resultant.

ANSWERResolve each vector into rectangular components. For A = Axî + Ayĵ and B = Bxî + Byĵ, the resultant R = A + B has components Rx = Ax + Bx and Ry = Ay + By; that is, each component of the resultant is the sum of the corresponding components. The magnitude is R = √(Rx2 + Ry2) and the direction with the x-axis is given by tanθ = Ry/Rx. This method is more accurate than the graphical method and extends easily to three dimensions and to any number of vectors.

Q3. Show that the path of a projectile is a parabola.

ANSWERTaking the launch point as origin, the horizontal motion is uniform: x = (v0cosθ0)t, so t = x/(v0cosθ0). The vertical motion is uniformly accelerated: y = (v0sinθ0)t − ½g t2. Substituting for t gives y = (tanθ0)x − [g/(2v02cos2θ0)]x2. Since v0, θ0 and g are constants, this has the form y = a x − b x2, which is the equation of a parabola. Hence the trajectory of a projectile is parabolic.

MCQs & Assertion–Reason

1. Which of the following is a vector quantity?

(a) work    (b) current    (c) impulse    (d) energy

2. For a projectile, the horizontal range is maximum when the angle of projection is:

(a) 30°    (b) 45°    (c) 60°    (d) 90°

3. In uniform circular motion, the centripetal acceleration is directed:

(a) along the tangent    (b) away from the centre    (c) towards the centre    (d) along the velocity

4. The magnitude of the resultant of two vectors A and B with angle θ between them is:

(a) A + B    (b) √(A2 + B2)    (c) √(A2 + B2 + 2AB cosθ)    (d) A B cosθ

5. A unit vector has:

(a) magnitude 1 and no dimension    (b) magnitude 0    (c) the unit of length    (d) variable magnitude

6. The horizontal component of a projectile’s velocity during flight (ignoring air resistance):

(a) increases    (b) decreases    (c) remains constant    (d) becomes zero at the top

7. Two vectors of equal magnitude are added; the resultant is zero when the angle between them is:

(a) 0°    (b) 90°    (c) 180°    (d) 60°

8. If A = Axî + Ayĵ makes angle θ with the x-axis, then Ay equals:

(a) A cosθ    (b) A sinθ    (c) A tanθ    (d) A

9. The acceleration of a projectile at the highest point of its path is:

(a) zero    (b) g, horizontal    (c) g, vertically downward    (d) maximum and upward

10. The relation between linear speed v and angular speed ω in circular motion is:

(a) v = ω/R    (b) v = ωR    (c) v = ω2R    (d) v = R/ω

Answer key: 1-(c), 2-(b), 3-(c), 4-(c), 5-(a), 6-(c), 7-(c), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: In uniform circular motion the body is accelerating.

Reason: The direction of velocity changes continuously even though the speed is constant.

A-R 2. Assertion: The horizontal and vertical motions of a projectile are independent of each other.

Reason: Gravity acts only vertically and has no horizontal component.

A-R 3. Assertion: The magnitude of displacement can be greater than the path length.

Reason: Displacement is the shortest distance between the initial and final positions.

A-R 4. Assertion: A scalar quantity has the same value for observers with differently oriented axes.

Reason: A scalar is invariant under rotation of the coordinate axes.

A-R 5. Assertion: Two vectors can be added only if they represent the same physical quantity.

Reason: Vector addition requires the quantities to have the same dimensions.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A). (A-R 3: the Assertion is false — displacement is never greater than path length — while the Reason is a true statement.)

Common Mistakes to Avoid

Watch out for these

  • Adding vector magnitudes directly (A + B) instead of using R = √(A2 + B2 + 2AB cosθ).
  • Forgetting that the horizontal velocity of a projectile stays constant — only the vertical component changes.
  • Writing the acceleration at the top of a projectile as zero; it is always g, vertically downward.
  • Confusing distance (scalar, path length) with displacement (vector, straight line between ends).
  • Treating uniform circular motion as unaccelerated because the speed is constant — the direction (and hence velocity) changes.
  • Mixing units — always convert km/h to m s−1 (divide by 3.6) and minutes to hours before substituting.
  • Calling a vector component a scalar; Axî is still a vector.

How to score full marks in this chapter

Always resolve vectors into perpendicular components before calculating, and clearly label what is along x and what is along y. For projectile numericals, write the standard formulas (Tf, hm, R) first, then substitute with units. Remember the complementary-angle rule (30° and 60° give equal ranges) for quick checks. In true/false and reasoning questions, justify with one crisp physics reason. Convert all quantities to SI units at the start, and quote the final answer with its correct unit and direction — markers award method marks for clear, step-by-step working.

Frequently Asked Questions

What is Class 11 Physics Chapter 3 Motion in a Plane about?

It teaches the vector tools needed for two-dimensional motion — vector addition, resolution and the analytical method — and applies them to motion in a plane with constant acceleration, projectile motion (a parabolic path) and uniform circular motion (centripetal acceleration v²/R towards the centre).

How many exercises are there in Class 11 Physics Chapter 3?

The NCERT textbook lists 22 exercise questions, numbered 3.1 to 3.22, including conceptual true/false questions and several numericals on projectile and circular motion. All 22 are solved step by step on this page.

What is the formula for the range of a projectile?

The horizontal range is R = v02 sin 2θ0/g. It is maximum when θ0 = 45°, giving Rmax = v02/g. Complementary launch angles (such as 30° and 60°) give the same range.

Are these Class 11 Physics Chapter 3 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27, with every numerical verified against the NCERT answer key.

← Chapter 2: Motion in a Straight Line All Class 11 Physics Chapter 4: Laws of Motion →
Scroll to Top