Class 11 Physics Chapter 3: Motion in a Plane — Vectors, Projectile & Circular Motion

01 / Chapter Theory

Motion in a Plane

Official NCERT syllabus: Scalars and vectors, position and displacement vectors, equality of vectors, multiplication by real numbers, graphical and analytical addition, resolution into rectangular components, motion in 2D with constant acceleration, projectile trajectory & range, and uniform circular motion.

10 SECTIONS

Chapter Navigation

3.1Introduction 3.2Scalars and Vectors 3.3Multiplication of Vectors 3.4Vector Addition (Graphical) 3.5Resolution of Vectors 3.6Vector Addition (Analytical) 3.7Motion in a Plane 3.8Motion with Constant Acceleration 3.9Projectile Motion 3.10Uniform Circular Motion

3.1 - 3.2

NCERT Sections

Scalars and Vectors

Foundations

3.2.1 Scalars vs. Vectors

Scalar Quantity: A physical quantity specified completely by its magnitude (a single real number along with proper unit) and obeying ordinary algebraic rules (e.g. mass, length, time, temperature, work, density, electric current).
Vector Quantity: A physical quantity possessing both magnitude and direction that strictly obeys the triangle law or parallelogram law of addition (e.g. displacement, velocity, acceleration, force, momentum).
Notation: Vectors are represented by boldface letters (A, v) or with an overhead arrow (A⃗). Magnitude is denoted as |A| = A.

Figure 3.1: (a) Position vectors r, r' and displacement vector Δr = r' − r. (b) Displacement is path-independent (|Δr| ≤ Path Length).

3.3 - 3.4

NCERT Sections

Vector Operations & Graphical Addition

Vector Algebra

3.4.1 Graphical Addition & Subtraction

Multiplication by a Real Number λ: |λA| = λ|A|. If λ > 0, direction is unchanged; if λ < 0, direction is reversed.
Triangle Law (Head-to-Tail): Place the tail of B at the head of A. The vector joining the tail of A to the head of B is the resultant R = A + B.
Commutative & Associative: A + B = B + A | (A + B) + C = A + (B + C).
Null / Zero Vector (0): Vector with zero magnitude and unspecified direction. A − A = 0.
Vector Subtraction: A − B = A + (−B).

Figure 3.4 & 3.6: Vector addition methods: (a) Triangle method (head-to-tail), (b) Parallelogram method (co-initial tails).

Example 3.1 • Umbrella Angle in Wind and Rain

NCERT Solved

Rain is falling vertically with a speed of 35 m s⁻¹. Wind starts blowing after some time with a speed of 12 m s⁻¹ in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?

Answer: Hold umbrella at ≈ 19° with the vertical towards the East (Resultant speed R = 37 m s⁻¹).

Step 1: Identify vector directions.
Rain velocity v_r = 35 m s⁻¹ (vertically downward).
Wind velocity v_w = 12 m s⁻¹ (East to West, horizontal).

Step 2: Magnitude of resultant velocity.
R = √(v_r² + v_w²) = √(35² + 12²) = √(1225 + 144) = √1369 = 37 m s⁻¹.

Step 3: Direction with the vertical.
tan θ = v_w / v_r = 12 / 35 = 0.343 → θ = tan⁻¹(0.343) ≈ 19° towards East.

Interactive 3D Relative Velocity & Rain-Man Umbrella Angle Simulator

Relative Motion Lab

Drag to change perspective

Relative Speed & Umbrella Orientation Angle

Relative Speed: |v_rm| = 26.9 m/s • Umbrella Tilt: θ = tan⁻¹(vₘ/vᵣ) = 21.8° with vertical

Man Speed ($v_\text{m}$):

Rain Fall Speed ($v_\text{r}$):

3.5 - 3.6

NCERT Sections

Resolution of Vectors & Analytical Addition

Component Analysis

3.5.1 Rectangular Components & Analytical Formulas

A vector A in the x-y plane making an angle θ with the x-axis resolves into mutually perpendicular components:

A = A_x î + A_y ĵ
where A_x = A cos θ | A_y = A sin θ
Magnitude: A = √(A_x² + A_y²) | Direction: tan θ = A_y / A_x

Figure 3.10: Analytical addition of two vectors at angle θ: R = √(A² + B² + 2AB cos θ).

Law of Cosines (Magnitude of Resultant):
R = √(A² + B² + 2AB cos θ)

Law of Sines (Direction α with vector A):
tan α = (B sin θ) / (A + B cos θ) | R / sin θ = A / sin β = B / sin α

Interactive 3D Vector Addition & Parallelogram Law

Vector Lab

Drag to rotate in 3D

Resultant Vector Magnitude & Direction

Resultant |R| = √(A² + B² + 2AB cosθ) = 10.44 units • Angle with A: α = 24.5°

Vector A:

Vector B:

Angle (θ):

Example 3.2 • Derivation of Law of Cosines & Sines

NCERT Solved

Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle θ between them.

Step 1: Express geometry with perpendicular dropped.
In right ΔOSN: OS² = ON² + SN² = (OP + PN)² + SN².
Since PN = B cos θ and SN = B sin θ:
R² = (A + B cos θ)² + (B sin θ)² = A² + 2AB cos θ + B²(cos²θ + sin²θ).
R = √(A² + B² + 2AB cos θ).

Step 2: Direction angle α.
tan α = SN / ON = (B sin θ) / (A + B cos θ).

Example 3.3 • Motorboat and River Current

NCERT Solved

A motorboat is racing towards north at 25 km/h and the water current is 10 km/h in the direction of 60° east of south. Find the resultant velocity of the boat.

Answer: Resultant speed R ≈ 21.8 km/h at φ ≈ 23.4° East of North.

Step 1: Angle between boat and current.
Boat is North (0°), Current is 60° East of South (180° − 60° = 120° from North). So θ = 120°.

Step 2: Magnitude of resultant.
R = √[25² + 10² + 2(25)(10) cos 120°] = √[625 + 100 + 500(−0.5)] = √[725 − 250] = √475 ≈ 21.8 km/h ≈ 22 km/h.

Step 3: Direction using Law of Sines.
sin φ = (v_c sin 120°) / R = [10 × (√3/2)] / 21.8 ≈ 0.397 → φ ≈ 23.4° East of North.

3.7 - 3.8

NCERT Sections

2D Kinematics & Constant Acceleration

2D Superposition Principle

3.8.1 Independence of Perpendicular Coordinates

Motion in a plane with constant acceleration a can be treated as two independent simultaneous 1D motions along x and y axes:

Position Vector: r(t) = r₀ + v₀t + ½ at²
x(t) = x₀ + v_0xt + ½ a_xt² | y(t) = y₀ + v_0yt + ½ a_yt²

Velocity Vector: v(t) = v₀ + at
v_x = v_0x + a_xt | v_y = v_0y + a_yt

Example 3.4 • Position, Velocity, and Acceleration Vectors

NCERT Solved

The position of a particle is given by r = 3.0t î + 2.0t² ĵ + 5.0 k̂ (m). (a) Find v(t) and a(t). (b) Find the magnitude and direction of v(t) at t = 1.0 s.

Answers: (a) v(t) = 3.0 î + 4.0t ĵ, a(t) = 4.0 ĵ m s⁻² | (b) |v(1.0 s)| = 5.0 m s⁻¹ at θ ≈ 53° with x-axis

(a) Derivatives:
v(t) = dr/dt = 3.0 î + 4.0t ĵ.
a(t) = dv/dt = 4.0 ĵ m s⁻² (constant along +y direction).

(b) At t = 1.0 s:
v(1.0) = 3.0 î + 4.0 ĵ.
Magnitude: v = √(3.0² + 4.0²) = √25 = 5.0 m s⁻¹.
Direction: tan θ = 4.0 / 3.0 = 1.333 → θ = tan⁻¹(4/3) ≈ 53.1° ≈ 53° with x-axis.

Example 3.5 • 2D Kinematic Motion under Constant Force

NCERT Solved

A particle starts from origin at t = 0 with velocity 5.0 î m/s and moves in the x-y plane with constant acceleration (3.0 î + 2.0 ĵ) m/s². (a) What is the y-coordinate when its x-coordinate is 84 m? (b) What is its speed at this time?

Answers: (a) y = 36.0 m (at t = 6 s) | (b) Speed v ≈ 26.0 m s⁻¹

Step 1: Set up component position equations.
x(t) = v_0xt + ½ a_xt² = 5.0t + 1.5t².
y(t) = v_0yt + ½ a_yt² = 0 + ½(2.0)t² = 1.0t².

Step 2: Find time t when x = 84 m.
1.5t² + 5.0t − 84 = 0 → 3t² + 10t − 168 = 0 → (t − 6)(3t + 28) = 0 → t = 6.0 s.

Step 3: Calculate y-coordinate and velocity at t = 6.0 s.
y(6.0) = 1.0(6)² = 36.0 m.
v_x = 5.0 + 3.0(6) = 23.0 m s⁻¹ | v_y = 0 + 2.0(6) = 12.0 m s⁻¹.
Speed: v = √(23² + 12²) = √(529 + 144) = √673 ≈ 25.94 m s⁻¹ ≈ 26 m s⁻¹.

3.9

NCERT Section

Projectile Motion

Parabolic Trajectory

3.9.1 Projectile Trajectory, Height & Range

When an object is launched with initial velocity v₀ at an elevation angle θ₀ above the horizontal, gravity acts strictly downward (a_x = 0, a_y = −g):

Figure 3.17: Parabolic trajectory of a projectile with maximum height h_m, time of flight T_f, and horizontal range R.

Projectile Kinematic Master Equations $$y = (\tan\theta_0) x - \left[\frac{g}{2 v_0^2 \cos^2\theta_0}\right] x^2 \quad (\text{Trajectory Parabola})$$ $$t_\text{m} = \frac{v_0 \sin\theta_0}{g}, \qquad T_\text{f} = 2t_\text{m} = \frac{2 v_0 \sin\theta_0}{g}$$ $$h_\text{m} = \frac{v_0^2 \sin^2\theta_0}{2g}, \qquad R = \frac{v_0^2 \sin 2\theta_0}{g} \quad \left(R_\text{max} = \frac{v_0^2}{g} \text{ at } \theta_0 = 45^\circ\right)$$

Galileo's Range Symmetry: Projectile ranges are identical for complementary launch angles $\theta_0 = 45^\circ + \alpha$ and $\theta_0 = 45^\circ - \alpha$ because: $$\sin 2(45^\circ + \alpha) = \sin(90^\circ + 2\alpha) = \cos 2\alpha = \sin(90^\circ - 2\alpha) = \sin 2(45^\circ - \alpha)$$

Interactive 3D Projectile Trajectory, Max Height & Range Simulator

Ballistics Lab

Drag to change perspective

Trajectory Kinematic Parameters

Range: R = 26.1m | Max Height: hₘ = 6.5m | Flight Time: T = 2.31s

Elevation ($\theta_0$):

Initial Speed ($v_0$):

Example 3.6 • Equal Ranges for Complementary Angles

NCERT Solved

Galileo stated that “for elevations which exceed or fall short of $45^\circ$ by equal amounts, the ranges are equal”. Prove this statement.

Proof: For launch angles $\theta_1 = 45^\circ + \alpha$ and $\theta_2 = 45^\circ - \alpha$: $$\sin 2\theta_1 = \sin(2(45^\circ + \alpha)) = \sin(90^\circ + 2\alpha) = \cos 2\alpha$$ $$\sin 2\theta_2 = \sin(2(45^\circ - \alpha)) = \sin(90^\circ - 2\alpha) = \cos 2\alpha$$ Since $\sin 2\theta_1 = \sin 2\theta_2 = \cos 2\alpha$, the horizontal range: $$R = \frac{v_0^2 \sin 2\theta_0}{g}$$ is strictly identical for both elevations.

Example 3.7 • Horizontal Projection from a Cliff

NCERT Solved

A hiker stands on the edge of a cliff 490 m high and throws a stone horizontally with an initial speed of 15 m s⁻¹. Neglecting air resistance, find the time taken to reach the ground and the speed with which it hits the ground (g = 9.8 m s⁻²).

Answers: Time of fall t = 10.0 s | Impact speed v ≈ 99.1 m s⁻¹

Step 1: Vertical fall time.
Vertical initial velocity v_0y = 0. Using y = ½gt² → 490 = ½(9.8)t² = 4.9t² → t² = 100 → t = 10.0 s.

Step 2: Velocity components upon impact.
v_x = v_0x = 15 m s⁻¹ (constant).
v_y = 0 − gt = −(9.8)(10) = −98 m s⁻¹.
Impact speed: v = √(v_x² + v_y²) = √(15² + 98²) = √(225 + 9604) = √9829 ≈ 99.14 m s⁻¹ ≈ 99 m s⁻¹.

Example 3.8 • Full Projectile Calculation

NCERT Solved

A cricket ball is thrown at a speed of 28 m s⁻¹ at 30° above the horizontal. Calculate: (a) maximum height, (b) time of flight, (c) horizontal range (g = 9.8 m s⁻²).

Answers: (a) h_m = 10.0 m | (b) T_f = 2.86 s ≈ 2.9 s | (c) R = 69.3 m ≈ 69 m

(a) Maximum height:
h_m = (v₀ sin θ₀)² / (2g) = (28 sin 30°)² / (2 × 9.8) = (14)² / 19.6 = 196 / 19.6 = 10.0 m.

(b) Time of flight:
T_f = (2 v₀ sin θ₀) / g = (2 × 28 × 0.5) / 9.8 = 28 / 9.8 = 2.86 s ≈ 2.9 s.

(c) Horizontal range:
R = (v₀² sin 60°) / g = (28² × 0.866) / 9.8 = (784 × 0.866) / 9.8 = 69.28 m ≈ 69 m.

3.10

NCERT Section

Uniform Circular Motion

Centripetal Dynamics

3.10.1 Centripetal Acceleration & Angular Speed

When a particle moves in a circle of radius R with constant speed v, its direction changes continuously, generating a centripetal acceleration directed towards the centre of the circle:

Figure 3.18: Uniform circular motion: Tangential velocity v ⊥ Radius R • Centripetal acceleration a_c directed towards centre C.

Interactive 3D Uniform Circular Motion & Centripetal Acceleration

Centripetal Lab

Rotate orbital plane

Kinematic Parameters ($v = \omega R$, $a_\text{c} = \omega^2 R$)

Angular Speed: 1.5 rad/s • Linear Speed: v = 12.0 m/s • Centripetal Acc: a_c = 18.0 m/s²

Angular Velocity ($\omega$): Observe tangential $\mathbf{v}$ (blue) vs centripetal $\mathbf{a}_\text{c}$ (rose)

Example 3.9 • Insect in a Circular Groove

NCERT Solved

An insect trapped in a circular groove of radius 12 cm moves steadily and completes 7 revolutions in 100 s. (a) What is the angular speed and linear speed? (b) Is the acceleration vector constant? What is its magnitude?

Answers: (a) ω = 0.44 rad/s, v = 5.3 cm/s | (b) Acceleration vector is NOT constant (direction rotates continuously), |a| = 2.3 cm s⁻²

(a) Angular & Linear Speed:
Frequency ν = 7 / 100 s⁻¹ = 0.07 Hz.
Angular speed ω = 2πν = 2(22/7)(7/100) = 44/100 = 0.44 rad s⁻¹.
Linear speed v = ω R = 0.44 × 12 cm = 5.28 cm s⁻¹ ≈ 5.3 cm s⁻¹.

(b) Centripetal Acceleration:
Magnitude: a_c = ω² R = (0.44)² × 12 = 0.1936 × 12 = 2.32 cm s⁻² ≈ 2.3 cm s⁻².
The direction points continuously towards the centre of the circle, hence the acceleration vector is not constant.

02 / Self-Assessment Lab

Concept Check Questions

Select any option to reveal stepwise explanations for every individual choice, detailing why the correct answer is valid and why the alternatives are incorrect.

20 QUESTIONS

Progress: 0 / 20 Answered

Score: 0

Q01 UNANSWERED

Which of the following physical quantities is a scalar?

Option A is incorrect. Linear momentum p = mv is a vector quantity having direction.

Option B is correct. Work is the dot product of force and displacement (W = F · d), which is a pure scalar quantity.

Option C is incorrect. Angular velocity ω is an axial vector directed along the axis of rotation.

Option D is incorrect. Magnetic moment is a vector pointing from South to North pole.

Core Rule

Scalars have magnitude only (e.g. work, energy, mass, current, temperature).

Q02 UNANSWERED

The resultant of two equal vectors A and B inclined at 120° has a magnitude equal to:

Option A is incorrect. 2A corresponds to θ = 0° (parallel vectors).

Option B is correct. R = √[A² + A² + 2A² cos 120°] = √[2A² + 2A²(−0.5)] = √[A²] = A.

Option C is incorrect. √2 A occurs at θ = 90°.

Option D is incorrect. Zero resultant occurs at θ = 180° (opposite vectors).

Core Rule

Two equal vectors at 120° produce a resultant equal in magnitude to either vector.

Q03 UNANSWERED

At the peak (maximum height) of a standard projectile trajectory, the velocity and acceleration vectors are:

Option A is incorrect. Parallel vectors occur in 1D free fall, not at 2D projectile peak.

Option B is correct. At the peak, vertical velocity v_y = 0, so velocity is purely horizontal (v = v₀ cos θ₀ î), while acceleration is purely downward (a = −g ĵ), making them strictly perpendicular (90°).

Option C is incorrect. Velocity is horizontal, not upward.

Option D is incorrect. Neither v_x nor g is zero at the peak.

Core Rule

At projectile apex: Velocity is horizontal (v_x) ⊥ Acceleration is downward (g).

Q04 UNANSWERED

For a projectile, the maximum horizontal range R_max is related to the maximum height h_m attained at 45° by:

Option A is incorrect. Incorrect multiplier.

Option B is correct. At θ₀ = 45°: R_max = v₀²/g. Maximum height h_m = (v₀² sin²45°)/(2g) = v₀²/(4g). Thus R_max = 4 h_m.

Option C is incorrect. R = h occurs at tan θ₀ = 4 (θ₀ ≈ 76°).

Option D is incorrect. Incorrect factor.

Core Rule

At 45° projection angle, the horizontal range is exactly 4 times the maximum height (R_max = 4h_m).

Q05 UNANSWERED

In uniform circular motion, which of the following quantities remains constant?

Option A is incorrect. Velocity direction continuously changes tangentially.

Option B is incorrect. Acceleration direction continuously rotates towards the centre.

Option C is correct. The scalar magnitude of velocity (speed v) is constant, hence kinetic energy (½mv²) is strictly constant.

Option D is incorrect. Direction of momentum vector changes continuously.

Core Rule

Uniform circular motion has constant speed, constant kinetic energy, and constant magnitude of acceleration, but continuously changing vectors.

Q06 UNANSWERED

The average acceleration vector of a particle in uniform circular motion averaged over one complete cycle is:

Option A is incorrect. v²/R is the instantaneous magnitude, not the cycle average vector.

Option B is correct. Δv over one complete revolution is zero because final velocity equals initial velocity (v(T) = v(0)). Thus a_avg = Δv/T = 0.

Option C is incorrect. Vector average of symmetric radial vectors sums to zero.

Option D is incorrect. Incorrect expression.

Core Rule

Over one full period T, total change in velocity vector Δv = 0, so average acceleration is a null vector.

Q07 UNANSWERED

Can three vectors not lying in the same plane add up to give a null vector?

Option A is incorrect. Non-coplanar vectors have a net component perpendicular to any plane.

Option B is correct. The resultant of any two non-collinear vectors lies in their common plane. To cancel this resultant, the third vector must be equal and opposite, which requires it to lie in the exact same plane.

Option C is incorrect. Coplanarity is mandatory for 3 vectors to sum to zero.

Option D is incorrect. At least 4 non-coplanar vectors are required to form a closed 3D polygon.

Core Rule

Minimum number of non-coplanar vectors required for a null vector sum is 4 (for coplanar it is 3).

Q08 UNANSWERED

If a stone is thrown horizontally from a height with speed u while another identical stone is simply dropped from the same height simultaneously:

Option B is correct. Vertical motion is completely independent of horizontal motion. Both start with vertical initial velocity v_0y = 0 and fall under gravity g, taking time t = √(2h/g).

Option B is incorrect. Horizontal speed does not alter vertical downward acceleration.

Option C is incorrect. Vertical distance and acceleration are identical for both.

Option D is incorrect. Gravitational acceleration g is independent of mass.

Core Rule

Vertical free-fall time t = √(2h/g) is independent of horizontal launch velocity.

Q09 UNANSWERED

The unit vector along the direction of vector A = 3 î + 4 ĵ is:

Option A is incorrect. This vector has magnitude 5, not 1.

Option B is correct. Unit vector n̂ = A / |A| = (3 î + 4 ĵ) / √(3² + 4²) = (3 î + 4 ĵ) / 5 = 0.6 î + 0.8 ĵ.

Option C is incorrect. Points at 45°, not along A.

Option D is incorrect. Perpendicular vector.

Core Rule

Unit vector â = A / |A| has magnitude 1 and indicates direction only.

Q10 UNANSWERED

If position is r = 3.0t î − 2.0t² ĵ + 4.0 k̂ (m), the acceleration vector is:

Option A is incorrect. Velocity contains 3.0 î; its derivative is zero.

Option B is correct. v = dr/dt = 3.0 î − 4.0t ĵ. Then a = dv/dt = −4.0 ĵ m s⁻².

Option C is incorrect. Sign is negative (−4.0 ĵ).

Option D is incorrect. The quadratic term produces non-zero constant acceleration.

Core Rule

Acceleration is the second time-derivative of position (a = d²r/dt²).

Q11 UNANSWERED

A cricketer throws a ball to a maximum horizontal range of 100 m. How high above the ground can he throw the same ball vertically?

Option A is correct. R_max = v₀²/g = 100 m. For vertical throw (θ = 90°): H_max = v₀² / (2g) = R_max / 2 = 100 / 2 = 50 m.

Option B is incorrect. H_max is half of R_max.

Option C is incorrect. 25 m is height attained during 45° throw (R_max/4).

Option D is incorrect. Incorrect calculation.

Core Rule

Maximum vertical throw height H_max = R_max / 2 = v₀² / (2g).

Q12 UNANSWERED

An aircraft executes a horizontal loop of radius 1.0 km at 900 km/h (250 m/s). Its centripetal acceleration compared to g (9.8 m s⁻²) is approximately:

Option A is incorrect. Too small.

Option B is correct. v = 900 × (5/18) = 250 m s⁻¹, R = 1000 m. a_c = v²/R = (250)²/1000 = 62500 / 1000 = 62.5 m s⁻². Ratio = 62.5 / 9.8 ≈ 6.38 g.

Option C is incorrect. Incorrect radius conversion.

Option D is incorrect. Missed dividing by radius.

Core Rule

Centripetal acceleration a_c = v²/R (convert km/h to m/s and km to m).

Q13 UNANSWERED

The vector inequality |a + b| ≤ |a| + |b| becomes an exact equality when:

Option A is correct. When θ = 0°, cos θ = 1, so |a + b| = √(a² + b² + 2ab) = a + b.

Option B is incorrect. At 90°, |a + b| = √(a² + b²) < a + b.

Option C is incorrect. At 180°, |a + b| = |a − b|.

Option D is incorrect. Triangle inequality is strict for non-collinear vectors.

Core Rule

Vector triangle inequality |a + b| = |a| + |b| holds if and only if vectors are parallel.

Q14 UNANSWERED

The horizontal distance covered by a projectile launched at 15° is 50 m. What would be its range if launched at 45° with the same speed?

Option A is incorrect. 15° gives sin 30° = 0.5, which is not maximum.

Option B is correct. R(15°) = (v₀² sin 30°)/g = 0.5 (v₀²/g) = 50 m → R_max = v₀²/g = 100 m (at 45° where sin 90° = 1).

Option C is incorrect. Range doubles from sin 30° (0.5) to sin 90° (1.0).

Option D is incorrect. Incorrect multiplier.

Core Rule

R ∝ sin 2θ₀. Since sin 90° / sin 30° = 1 / 0.5 = 2, range doubles.

Q15 UNANSWERED

The angle made by vector A = î + ĵ with the positive x-axis is:

Option A is incorrect. tan 30° = 1/√3.

Option B is correct. tan θ = A_y / A_x = 1 / 1 = 1 → θ = 45°. Magnitude is √(1² + 1²) = √2.

Option C is incorrect. tan 60° = √3.

Option D is incorrect. 90° is along the y-axis (ĵ only).

Core Rule

Vector (î + ĵ) bisects the first quadrant at 45°; (î − ĵ) is at −45° (315°).

Q16 UNANSWERED

If a stone tied to an 80 cm string completes 14 revolutions in 25 s, its centripetal acceleration is:

Option A is correct. ν = 14/25 s⁻¹ = 0.56 Hz. ω = 2πν = 2 × (22/7) × (14/25) = 88/25 = 3.52 rad/s. a_c = ω²R = (3.52)² × 0.80 m = 12.39 × 0.80 = 9.91 m s⁻².

Option B is incorrect. Incorrect calculation.

Option C is incorrect. Double the actual value.

Option D is incorrect. Circular motion always requires non-zero centripetal acceleration.

Core Rule

a_c = 4π² ν² R (with R in metres).

Q17 UNANSWERED

The trajectory equation of a projectile y = ax − bx² represents a:

Option A is incorrect. Quadratic dependency creates curvature.

Option B is correct. A quadratic equation of the form y = ax + bx² with b < 0 defines an inverted parabola.

Option C is incorrect. Hyperbola requires reciprocal relation xy = constant.

Option D is incorrect. Ellipse requires quadratic in both x and y.

Core Rule

Eliminating time t gives y = (tan θ₀)x − [g/(2v₀²cos²θ₀)]x², which is parabolic.

Q18 UNANSWERED

An aircraft flying at height 3400 m subtends an angle of 30° at ground over 10.0 s. Its speed is:

Option A is correct. Half angle θ/2 = 15°. Distance d = 2 × 3400 × tan 15° = 6800 × 0.2679 = 1822 m. Speed = 1822 m / 10.0 s = 182.2 m s⁻¹ ≈ 182 m s⁻¹.

Option B is incorrect. Incorrect trigonometry.

Option C is incorrect. Missed factor of 2 for total flight distance.

Option D is incorrect. Incorrect calculation.

Core Rule

Total distance d = 2h tan(θ/2), then speed = d/Δt.

Q19 UNANSWERED

Given a + b + c + d = 0, which of the following is definitely true?

Option A is incorrect. Non-zero vectors can form a closed polygon.

Option B is correct. Rearranging gives (a + c) = −(b + d). Taking magnitude: |a + c| = |−(b + d)| = |b + d|.

Option C is incorrect. |a| ≤ |b| + |c| + |d| (polygon inequality).

Option D is incorrect. Closed polygons exist in 2D and 3D.

Core Rule

If A + B = 0, then A = −B, meaning their magnitudes are strictly equal.

Q20 UNANSWERED

Which of the following operations is physical and mathematically meaningful?

Option A is meaningless. Scalars and vectors cannot be added together.

Option B is meaningful. Multiplying a vector (e.g. velocity v) by a scalar (e.g. mass m or time t) yields a valid new vector (momentum p = mv, displacement d = vt).

Option C is meaningless. Quantities with different dimensions cannot be added (e.g. mass + temperature).

Option D is meaningless. Scalar component cannot be added directly to the parent vector.

Core Rule

Vector scaling: Multiplying a vector by a scalar is always well-defined and scales magnitude/dimensions.

03 / Textbook Solutions

NCERT Exercises 3.1 – 3.22

Complete stepwise solutions for every textbook exercise question in NCERT Class 11 Physics Chapter 3 Reprint 2026-27.

22 QUESTIONS

Ex 3.1 Scalar vs. Vector Classification

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.

Physical Quantity	Classification	Reason
Volume	Scalar	Magnitude only
Mass	Scalar	Magnitude only
Speed	Scalar	Magnitude of velocity
Acceleration	Vector	Magnitude and direction
Density	Scalar	Mass per unit volume
Number of moles	Scalar	Amount of substance
Velocity	Vector	Rate of change of displacement
Angular frequency	Scalar	Magnitude 2πν (rad/s)
Displacement	Vector	Directed position change
Angular velocity	Vector	Axial vector along rotation axis

Ex 3.2 Pick Scalars

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.

The two scalar quantities are: Work and Current.

Work: Scalar product of force and displacement (W = F · d).

Current: Even though current has a direction along a wire, it does not obey vector algebra (triangle law); it adds algebraically, hence it is a scalar quantity.

Ex 3.3 Pick Single Vector

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.

The only vector quantity is: Impulse.

Impulse: Defined as change in linear momentum (J = F Δt = Δp), which has both magnitude and direction. (Note: Pressure is a scalar because normal force per unit area has no independent directional orientation).

Ex 3.4 Meaningful Operations

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

(a) Meaningful only if dimensions/units match: Adding mass to temperature is meaningless. Adding mass to mass is meaningful.

(b) Not meaningful: A scalar cannot be added to a vector.

(c) Meaningful: Multiplying a vector by a scalar yields a valid new vector (e.g. ma = F).

(d) Meaningful: Product of two scalars yields a valid scalar (e.g. mass × specific heat × ΔT = heat energy).

(e) Meaningful only if they represent same physical quantity: You can add two velocity vectors, but not a velocity vector to a force vector.

(f) Not meaningful: A scalar component cannot be directly added to the vector itself (a component vector A_xî can be added, but not scalar A_x).

Ex 3.5 True / False Statements

3.5 State with reasons if each statement 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 is either greater or equal to the magnitude of average velocity over the same interval.

(e)

Three vectors not lying in a plane can never add up to give a null vector.

(a) True: Magnitude is a pure number with unit, which is a scalar.

(b) True: The components A_x, A_y are real numbers (scalars), whereas A_xî is a component vector.

(c) False: Path length ≥ |Displacement| (only equal for unidirectional straight line motion).

(d) True: Average speed = Path Length / Δt ≥ |Displacement| / Δt = |Average Velocity|.

(e) True: The resultant of two vectors lies in their common plane; the third vector must be in that same plane to cancel it.

Ex 3.6 Vector Triangle Inequalities

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 apply?

Geometric Proof: In any triangle with sides |a|, |b|, and |a+b|:
• The sum of any two sides is strictly greater than the third side: |a+b| < |a| + |b|.
• The difference of any two sides is strictly less than the third side: |a+b| > ||a| − |b||.

Equality Conditions:
• For (a) and (d): Equality holds when a and b are in the same direction (θ = 0°).
• For (b) and (c): Equality holds when a and b are in opposite directions (θ = 180°).

Ex 3.7 Four Vector Sum Analysis

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 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.

Correct statements: (b), (c), and (d) are correct. (a) is incorrect.

(a) Incorrect: Four non-zero vectors can form a closed polygon whose vector sum is zero.

(b) Correct: (a + c) = −(b + d) → |a + c| = |b + d|.

(c) Correct: a = −(b + c + d) → |a| = |b + c + d| ≤ |b| + |c| + |d|.

(d) Correct: (b + c) = −(a + d). The vector (a + d) lies in the plane of a and d, hence −(a + d) must also lie in the same plane.

Ex 3.8 Skating Paths on Circular Ice

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

Figure 3.19: Paths A, B, C connecting diametrically opposite points P and Q on circular ground (R = 200 m).

Displacement for each girl = 400 m | Path length equals displacement for Girl B.

Displacement: Diameter = 2R = 2 × 200 m = 400 m from P to Q (same for all three girls).

Path Equality: Girl B skates straight along the diameter PQ, so for Girl B, path length = displacement = 400 m.

Ex 3.9 Circular Park Round Trip

3.9 A cyclist starts from centre O of a circular park of radius 1 km, reaches edge P, cycles along circumference to Q, and returns to centre along QO in 10 min. Find: (a) net displacement, (b) average velocity, (c) average speed.

(a) Net Displacement = 0 • (b) Average Velocity = 0 • (c) Average Speed ≈ 21.4 km/h (5.95 m/s)

(a) Net Displacement: Cyclist returns to starting point O → Δr = 0.

(b) Average Velocity: Δr / Δt = 0 km/h.

(c) Average Speed:
Path length = OP + Arc PQ (quarter circle) + QO = 1 km + (¼ × 2π × 1 km) + 1 km = 2 + 1.571 = 3.571 km.
Time = 10 min = 10/60 h = 1/6 h.
Average speed = 3.571 / (1/6) = 21.43 km/h (or 5.95 m s⁻¹).

Ex 3.10 Hexagonal Track Motorist

3.10 A motorist follows a track that turns left by 60° after every 500 m (regular hexagon). Specify displacement and compare with total path length at: (i) 3rd turn, (ii) 6th turn, (iii) 8th turn.

(i) 3rd turn: Disp = 1000 m (Path = 1500 m) • (ii) 6th turn: Disp = 0 m (Path = 3000 m) • (iii) 8th turn: Disp = 866 m (Path = 4000 m)

(i) At 3rd turn (diametrically opposite vertex):
Displacement = 2 × side = 2(500) = 1000 m (1 km) at 60° left of initial path. Path length = 3 × 500 = 1500 m.

(ii) At 6th turn (completed hexagon back to start):
Displacement = 0 m. Path length = 6 × 500 = 3000 m (3 km).

(iii) At 8th turn (vertex 2 of second round):
Displacement = √[500² + 500² + 2(500)(500) cos 60°] = 500 √3 = 866 m (0.866 km) at 30° to initial path. Path length = 8 × 500 = 4000 m (4 km).

Ex 3.11 Dishonest Cabman

3.11 A passenger wishes to go from station to hotel 10 km away along a straight road. A dishonest cabman takes a circuitous path 23 km long in 28 min. What is: (a) average speed, (b) magnitude of average velocity? Are they equal?

(a) Average Speed = 49.3 km/h • (b) |Average Velocity| = 21.4 km/h • No, they are not equal.

(a) Average Speed: Path length / Time = 23 km / (28/60 h) = (23 × 60) / 28 = 49.29 km/h ≈ 49.3 km/h.

(b) |Average Velocity|: |Displacement| / Time = 10 km / (28/60 h) = 600 / 28 = 21.43 km/h ≈ 21.4 km/h.

Comparison: They are not equal because the path is circuitous (Path Length > |Displacement|).

Ex 3.12 Hall Ceiling Projectile Limit

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? (g = 9.8 m s⁻²).

Answer: Maximum Range = 150.5 m ≈ 151 m

Step 1: Determine maximum allowable angle from ceiling height.
h_m = (v₀² sin²θ) / (2g) ≤ 25 m.
sin²θ = (25 × 2 × 9.8) / (40)² = 490 / 1600 = 0.30625 → sin θ ≈ 0.5534 (θ ≈ 33.6°).
cos²θ = 1 − 0.30625 = 0.69375 → cos θ ≈ 0.8329.

Step 2: Calculate horizontal range at this angle.
R = (v₀² × 2 sin θ cos θ) / g = (1600 × 2 × 0.5534 × 0.8329) / 9.8 = 1475.2 / 9.8 = 150.53 m ≈ 151 m.

Ex 3.13 Max Range to Max Height

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 vertical height = 50 m

Step 1: Relate R_max to launch speed.
R_max = v₀²/g = 100 m.

Step 2: Vertical throw (θ = 90°).
H_max = v₀² / (2g) = R_max / 2 = 100 / 2 = 50 m.

Ex 3.14 Whirled Stone Acceleration

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

Answer: Magnitude = 9.9 m s⁻², directed towards the centre of the circle along the string.

Step 1: Calculate angular frequency.
R = 0.80 m, ν = 14 / 25 s⁻¹ = 0.56 Hz.
ω = 2πν = 2 × (22/7) × (14/25) = 88/25 = 3.52 rad s⁻¹.

Step 2: Centripetal acceleration.
a_c = ω² R = (3.52)² × 0.80 = 12.39 × 0.80 = 9.91 m s⁻² ≈ 9.9 m s⁻².

Ex 3.15 Aircraft Loop-the-Loop

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 (g = 9.8 m s⁻²).

Answer: Centripetal acceleration = 62.5 m s⁻² = 6.38 g ≈ 6.4 g

Step 1: Convert to SI units.
v = 900 × (5/18) = 250 m s⁻¹ | R = 1000 m.

Step 2: Calculate a_c and ratio.
a_c = v² / R = (250)² / 1000 = 62500 / 1000 = 62.5 m s⁻².
a_c / g = 62.5 / 9.8 = 6.38 (i.e. 6.38 times gravity).

Ex 3.16 Circular Motion True/False

3.16 State, with reasons, if each statement is True or False:
(a) The net acceleration of a particle in circular motion is always along the radius towards the centre.
(b) The velocity vector of a particle at a point is always along the tangent to the path.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

(a) False: Only in UNIFORM circular motion is net acceleration purely radial. In non-uniform circular motion, tangential acceleration a_t exists, so net acceleration is inclined to the radius.

(b) True: The velocity vector at any point on any curved path is always tangential to the trajectory.

(c) True: Over one complete period, for every radial acceleration vector, there is an equal and opposite vector diametrically opposite. Thus the vector average over one cycle is a null vector.

Ex 3.17 Position, Velocity & Acceleration Vectors

3.17 The position of a particle is given by r = 3.0t î − 2.0t² ĵ + 4.0 k̂ (m). (a) Find v and a of the particle. (b) What is the magnitude and direction of velocity at t = 2.0 s?

(a) v(t) = 3.0 î − 4.0t ĵ, a = −4.0 ĵ m s⁻² • (b) |v(2.0 s)| = 8.54 m s⁻¹ at θ ≈ −69.4° with x-axis

(a) Velocity & Acceleration:
v(t) = dr/dt = 3.0 î − 4.0t ĵ.
a(t) = dv/dt = −4.0 ĵ m s⁻².

(b) At t = 2.0 s:
v(2.0) = 3.0 î − 8.0 ĵ.
Magnitude: v = √(3.0² + (−8.0)²) = √(9 + 64) = √73 = 8.54 m s⁻¹.
Direction: tan θ = −8.0 / 3.0 = −2.667 → θ = −69.4° (below positive x-axis).

Ex 3.18 2D Constant Acceleration Motion

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

(a) t = 2.0 s, y = 24.0 m • (b) Speed v = 21.26 m s⁻¹ ≈ 21.3 m s⁻¹

(a) Time and y-coordinate:
x(t) = ½ a_x t² = ½(8.0)t² = 4.0 t².
For x = 16 m → 4.0 t² = 16 → t² = 4 → t = 2.0 s.
y(2.0) = v_0yt + ½ a_yt² = (10.0)(2) + ½(2.0)(2)² = 20 + 4 = 24.0 m.

(b) Speed at t = 2.0 s:
v_x = a_xt = 8.0(2) = 16 m s⁻¹.
v_y = v_0y + a_yt = 10.0 + 2.0(2) = 14 m s⁻¹.
Speed: v = √(16² + 14²) = √(256 + 196) = √452 = 21.26 m s⁻¹ ≈ 21.3 m s⁻¹.

Ex 3.19 Unit Vectors & Projections

3.19 î and ĵ are unit vectors along x- and y- axes. What is the magnitude and direction of vectors (î + ĵ) and (î − ĵ)? What are the components of A = 2 î + 3 ĵ along the directions of (î + ĵ) and (î − ĵ)?

|î+ĵ| = √2 at 45° • |î−ĵ| = √2 at −45° • Component along (î+ĵ) = 5/√2 • Component along (î−ĵ) = −1/√2

Magnitudes and Directions:
• î + ĵ: Magnitude = √(1² + 1²) = √2. Angle θ = tan⁻¹(1/1) = 45° with x-axis.
• î − ĵ: Magnitude = √(1² + (−1)²) = √2. Angle θ = tan⁻¹(−1/1) = −45° (315°).

Components of A = 2 î + 3 ĵ:
Unit vector along (î+ĵ) is û₁ = (î+ĵ)/√2.
Component = A · û₁ = (2×1 + 3×1)/√2 = 5 / √2 = 2.5 √2 ≈ 3.54.
Unit vector along (î−ĵ) is û₂ = (î−ĵ)/√2.
Component = A · û₂ = (2×1 + 3×(−1))/√2 = −1 / √2 ≈ −0.71.

Ex 3.20 Arbitrary Motion Validity

3.20 For any arbitrary motion in space, which of the following relations are true:
(a) v_avg = ½ (v(t₁) + v(t₂))
(b) v_avg = [r(t₂) − r(t₁)] / (t₂ − t₁)
(c) v(t) = v(0) + at
(d) r(t) = r(0) + v(0)t + ½ at²
(e) a_avg = [v(t₂) − v(t₁)] / (t₂ − t₁)

Only (b) and (e) are true for arbitrary motion. Relations (a), (c), and (d) hold only for uniform acceleration.

(b) True: Definition of average velocity (displacement divided by time interval).

(e) True: Definition of average acceleration (velocity change divided by time interval).

(a), (c), (d) False in general: These are valid strictly when acceleration a is constant.

Ex 3.21 Scalar Invariance Properties

3.21 State with reasons if each statement 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.

(a) False: Many scalars are not conserved (e.g. kinetic energy in inelastic collisions).

(b) False: Temperature (−10 °C), electric charge, and potential energy can be negative.

(c) False: Mass, density, and energy have physical dimensions.

(d) False: Gravitational potential, temperature, and density vary from point to point in space.

(e) True: A scalar is invariant under coordinate axis rotations; it has the exact same value for all observers.

Ex 3.22 Aircraft Subtended Angle Speed

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: Speed of aircraft = 182.2 m s⁻¹ ≈ 182 m s⁻¹ (656 km/h)

Step 1: Geometric setup.
Let O be the observer on the ground, A and B the aircraft positions at times t = 0 and t = 10 s. Height OC ⊥ AB = 3400 m. Angle ∠AOB = 30° → ∠AOC = 15°.

Step 2: Distance travelled AB.
In right ΔACO: AC = OC tan 15° = 3400 × 0.26795 = 911.0 m.
Total distance AB = 2 × AC = 2 × 911.0 = 1822.0 m.

Step 3: Speed calculation.
Speed v = AB / Δt = 1822.0 m / 10.0 s = 182.2 m s⁻¹ (approx 656 km/h).

04 / Rapid Reference

Chapter Summary & Formulas

Complete 2D vectors formulas, projectile motion cheat sheets, uniform/non-uniform circular motion relationships, conditions, and exam tricks.

100% SYLLABUS

3.1 - 3.4 • Vector Mechanics

Vector Operations, Resultant & Resolution

Scalars have only magnitude, while vectors have both magnitude and direction and obey vector laws of addition.

The Scalar Traps: Electric current has direction but is a scalar because it adds algebraically (Kirchhoff's current law) rather than via vector addition. Pressure is also a scalar because it acts uniformly in all directions at a point.

Vector Addition & Resolution Formulas

Vector Concept	Mathematical Formula	Tricks, Conditions & Limits
Resultant Magnitude	$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$	$\theta = 0^\circ \implies R_{\text{max}} = A+B$ $\theta = 180^\circ \implies R_{\text{min}} = \|A-B\|$
Resultant Angle ($\alpha$)	$\tan \alpha = \frac{B\sin\theta}{A + B\cos\theta}$	$\alpha$ is the angle of resultant $\vec{R}$ with vector $\vec{A}$.
Resolution in 2D	$\vec{A} = A_x\hat{i} + A_y\hat{j}$ $A_x = A\cos\theta$, $A_y = A\sin\theta$	Mnemonic: The component touching the angle $\theta$ is always $\cos \theta$. The opposite component is $\sin \theta$.
Dot Product	$\vec{A} \cdot \vec{B} = AB\cos\theta = A_x B_x + A_y B_y$	Result is a scalar. Zero if vectors are perpendicular ($\theta = 90^\circ$).
Cross Product	$\vec{A} \times \vec{B} = (AB\sin\theta)\hat{n}$	Result is a vector perpendicular to both $\vec{A}$ and $\vec{B}$. Direction given by right-hand thumb rule.

3.5 • Projectile Dynamics

Projectile Motion (Ideal Trajectory under Gravity)

A projectile has a constant horizontal velocity and a uniform vertical acceleration ($a_x = 0$, $a_y = -g$).

Trajectory Equation: $y = x\tan\theta_0 - \frac{gx^2}{2v_0^2\cos^2\theta_0}$ (Parabolic path)

Core Formulas & Variables

Kinematic Value	Formula	Exponents & Mnemonic Shortcuts
Time of Flight ($T$)	$$T = \frac{2v_0\sin\theta_0}{g}$$	Mnemonic (2-1-2 Rule): Factor of $2$ is in front.
Maximum Height ($H$)	$$H = \frac{v_0^2\sin^2\theta_0}{2g}$$	Mnemonic (2-1-2 Rule): Factor of $2$ is in denominator ($2g$) and as a square.
Horizontal Range ($R$)	$$R = \frac{v_0^2\sin 2\theta_0}{g}$$	Mnemonic (2-1-2 Rule): Factor of $2$ is inside the angle ($2\theta$).

Important Range Tricks for Exams

Complementary Launch Angles: The range $R$ is identical for complementary angles $\theta_0$ and $90^\circ - \theta_0$ (e.g. $30^\circ$ and $60^\circ$).
Maximum Range Angle: Range is maximum at $\theta_0 = 45^\circ$, where $R_{\text{max}} = \frac{v_0^2}{g}$.
Range-Height Link at $45^\circ$: When thrown for maximum range ($\theta_0 = 45^\circ$), we have: $$R_{\text{max}} = 4 H_{\text{max}}$$
Velocity at Highest Point: Vertical velocity is zero ($v_y = 0$). Speed is minimum but non-zero: $v_{\text{min}} = v_0\cos\theta_0$. Here, velocity is perpendicular to acceleration ($\vec{v} \perp \vec{g}$).

3.6 • Circular Mechanics

Uniform & Non-Uniform Circular Motion

When an object moves in a circle of radius $R$ at constant speed $v$:

Centripetal Acceleration ($a_c$): Directed radially inwards, changing velocity direction: $$a_c = \frac{v^2}{R} = \omega^2 R = 4\pi^2 \nu^2 R$$ (where $\omega$ is angular velocity and $\nu$ is rotation frequency in rev/s).

Non-Uniform Circular Motion Condition: If speed also changes, the object has both centripetal acceleration ($a_c$) and tangential acceleration ($a_t = \frac{dv}{dt}$). The net acceleration vector is: $$a_{\text{net}} = \sqrt{a_c^2 + a_t^2}$$

Trick: Standard equations of motion ($v = u+at$, etc.) do not apply to circular motion because the direction of the acceleration vector changes continuously (it is not constant in direction).

NCERT Official

Points to Ponder & Exam Tips

Vector equations are independent of coordinate system choice. Components change when coordinate axes rotate, but the vector itself remains invariant.
In projectile motion, acceleration is constant ($g$ downwards) in both magnitude and direction, but velocity changes at every point.
In uniform circular motion, acceleration is constant in magnitude ($v^2/R$) but not in direction (always points to the moving center).
Average velocity can equal instantaneous velocity only if the motion is in a straight line with constant speed.
To find relative velocity in 2D, use vector subtraction: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. Solve using components or parallelogram law.

05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to calculate score and review answers.

24 QUESTIONS

1. Which of the following is a vector quantity?

2. The maximum horizontal range of a projectile is achieved at an angle of:

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

4. The magnitude of a unit vector is:

5. The path of a projectile in air (neglecting air resistance) is a:

6. The relation between linear velocity v and angular velocity ω is:

7. The horizontal component of velocity in ideal projectile motion is:

8. Minimum number of non-coplanar vectors required to give a null vector is:

1. If two vectors of magnitude 3 and 4 are perpendicular, their resultant magnitude is:

2. A ball thrown at 28 m/s at 30° has a maximum height of (g = 9.8 m/s²):

3. An insect in a groove of R = 12 cm with ω = 0.44 rad/s has speed:

4. The magnitude of A = î + ĵ is:

5. If R_max = 100 m, the maximum vertical height achievable with same speed is:

6. Ranges for launch angles 35° and 55° with the same initial speed are:

7. Rain 35 m/s vertically and wind 12 m/s horizontally gives resultant speed:

8. For r = 3.0t î + 2.0t² ĵ, velocity magnitude at t = 1.0 s is:

1. Hall ceiling is 25 m high. Ball thrown at 40 m/s has maximum range of:

2. Stone thrown horizontally at 15 m/s from 490 m cliff hits the ground after:

3. Aircraft looping 1.0 km at 900 km/h has centripetal acceleration of:

4. Component of A = 2 î + 3 ĵ along (î + ĵ) is:

5. If a + b + c + d = 0, then |a + c| equals:

6. Aircraft at 3400 m subtends 30° over 10 s. Its speed is:

7. Particle with v₀ = 10 ĵ and a = 8 î + 2 ĵ has y-coordinate at x = 16 m of:

8. Range at 15° is 50 m. Range at 45° with same speed is:

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