What defines motion in a plane?

Motion in a plane, or two-dimensional motion, occurs when an object moves such that its position changes along two axes (e.g., x and y) simultaneously. Examples include projectile motion and circular motion.

What is the trajectory of a projectile?

The path followed by a projectile is called its trajectory, which is parabolic in nature when air resistance is neglected.

Motion in a Plane — NEET/JEE Class 11 | Complete Study Guide

NCERT Class 11 · Chapter 4

Motion in a Plane
Complete Study Guide

From absolute zero to NEET-level mastery — visual, simple, and exam-oriented teaching for beginners and weak students.

★ NEET Important 2D Motion Projectile Motion Circular Motion River-Boat Problems 50 Practice Questions Graphs & Tricks

NCERT Mapping

Where this chapter appears and why it matters

Chapter

Motion in a Plane — Class 11 NCERT Physics Part 1

This chapter extends 1D kinematics (Chapter 3) to two dimensions using vectors. It covers scalars vs vectors, vector addition/subtraction, projectile motion, circular motion, and relative velocity.

🎯 Why It's in NEET

2–3 direct questions every year on projectile motion and circular motion. Concepts form the base for Gravitation, Rotation, and Fluid Mechanics.

3-5 marks/yearDirect NCERT

📚 What You Already Need

Basic algebra, sin/cos/tan (Class 10), simple equations of motion (Chapter 3 — same chapter but 1D). Nothing more needed!

Prerequisite: Ch 3

Scalars & Vectors

Vector Addition

Resolution

Relative Velocity

Projectile Motion

Circular Motion

Concept Building

Start from zero with real-life examples

🌱 Step 1: Scalar vs Vector — The Foundation

📏 Scalar Quantity

Has only magnitude (size/number). No direction needed.

Examples: Temperature (37°C), Mass (60 kg), Speed (80 km/h), Distance, Time, Energy

🧭 Vector Quantity

Has both magnitude AND direction. Direction changes everything!

Examples: Velocity (80 km/h North), Force (10N upward), Displacement, Acceleration, Momentum

💡

Real-Life TrickIf someone says "I walked 5 km" — that's distance (scalar). If they say "I walked 5 km towards my school" — that's displacement (vector). Direction matters!

➕ Step 2: Adding Vectors — Not Like Numbers!

You cannot simply add vectors like numbers. 3 + 4 could give anywhere from 1 to 7 depending on direction!

🚶

Same Direction

3 m East + 4 m East = 7 m East (simply add)

↔️

Opposite Direction

3 m East + 4 m West = 1 m West (subtract)

📐

Perpendicular

3 m East + 4 m North = 5 m (Pythagoras!)

🎯

Any Angle

Use the Parallelogram Law or Triangle Law

Resultant Vector Magnitude (Any Angle θ between A and B)

R = √(A² + B² + 2AB cosθ)

Direction: tan α = B sinθ / (A + B cosθ) where α is angle with A

✂️ Step 3: Resolving Vectors — Breaking Into X and Y

Imagine you're pushing a trolley at an angle. Part of your force moves it forward, part lifts it up. We "resolve" the force into horizontal (X) and vertical (Y) parts.

🧠

Memory Trick X-component = V × cos θ (Adjacent side — "cos goes with x")
Y-component = V × sin θ (Opposite side — "sin goes with y")
Remember: x→cos, y→sin

🏏 Cricket Ball Example

Ball thrown at 30° with 20 m/s:
Horizontal: vₓ = 20 cos30° = 20 × (√3/2) = 17.3 m/s
Vertical: v_y = 20 sin30° = 20 × (1/2) = 10 m/s

Component Form

Vₓ = V cosθ
V_y = V sinθ

V = √(Vₓ² + V_y²)
θ = tan⁻¹(V_y / Vₓ)

🎯 Step 4: Motion in a Plane — The Big Idea

When something moves in 2D (like a ball thrown at an angle), we treat the X and Y motions completely independently. This is the key insight!

⚠️

Golden Rule of 2D MotionHorizontal (X) and Vertical (Y) motions are INDEPENDENT. A thrown ball and a dropped ball hit the ground at the SAME TIME — because vertical motion is identical for both!

Horizontal Direction (X)

No force acts horizontally (ignoring air). So:
Acceleration aₓ = 0
Speed stays constant: vₓ = uₓ (always)

Vertical Direction (Y)

Gravity pulls downward always.
Acceleration a_y = −g = −10 m/s²
Speed changes every second!

Visual Learning

Clear diagrams with step-by-step explanation

📐 Diagram 1: Triangle Law of Vector Addition

Triangle Law: Place vectors head-to-tail. Resultant closes the triangle from tail of first to head of last.

🏈 Diagram 2: Complete Projectile Motion Anatomy

Projectile motion: parabolic path. Horizontal velocity is constant throughout; vertical velocity changes due to gravity.

🔵 Diagram 3: Uniform Circular Motion

Uniform circular motion: constant speed, changing direction = acceleration (centripetal) always toward centre.

🚤 Diagram 4: River-Boat Problem (Relative Velocity)

River-Boat: Boat aims perpendicular. River pushes sideways. Actual path is diagonal (resultant velocity).

Key Formulas

All NEET-important formulas in one place

📌

How to Use This SectionDon't just memorise — understand WHERE each formula comes from. The derivations are simple and help you recall in exams even if you forget the formula!

A. Vector Formulas

Resultant Magnitude

R = √(A² + B² + 2AB cosθ)

θ = angle between A and B

Direction of Resultant

tan α = B sinθ / (A + B cosθ)

α = angle resultant makes with A

Vector Components

Aₓ = A cosθ A_y = A sinθ

A = √(Aₓ² + A_y²)

Dot Product (Scalar)

A·B = AB cosθ

Result is a scalar. Work = F·d = Fd cosθ

Cross Product (Vector)

|A×B| = AB sinθ

Direction: Right-hand rule. Torque = r×F

B. Projectile Motion Formulas

🔑

SetupLaunched at angle θ with speed u. g = 10 m/s². X: no acceleration. Y: acceleration = −g downward.

★ Time of Flight

T = 2u sinθ / g

Total time in air. At T/2 → maximum height

★ Maximum Height

H = u² sin²θ / 2g

Height when v_y = 0

★ Horizontal Range

R = u² sin2θ / g

Maximum when θ = 45° → R_max = u²/g

Trajectory Equation (path shape)

y = x tanθ − gx² / (2u²cos²θ)

This is a PARABOLA. Used in numerical problems.

Velocity at Any Time t

vₓ = u cosθ (constant)
v_y = u sinθ − gt

Speed = √(vₓ² + v_y²)

★ Same Range — Complementary Angles

R(θ) = R(90°−θ)

30° and 60° give same range. 20° and 70° also same!

Relation: Range, Heights, Time

T₁T₂ = 2R/g
H₁/H₂ = tan²θ₁/tan²θ₂

For complementary angles giving same range R

★ R vs H Relation

R = 4H cotθ
When R = H: tanθ = 4

Frequently asked in NEET!

C. Circular Motion Formulas

★ Centripetal Acceleration

aₒ = v²/r = ω²r = 4π²r/T²

Always directed toward the centre of circle

★ Period, Frequency, ω

T = 2πr/v = 2π/ω f = 1/T
ω = 2πf = v/r

ω = angular velocity in rad/s

Centripetal Force

F = mv²/r = mω²r

NOT a new force — it's the NET force directed inward

D. Relative Velocity

★ Relative Velocity

v_AB = v_A − v_B

Velocity of A as seen by B = v_A minus v_B

River-Boat (⊥ crossing)

v_net = √(v_b² + v_r²)
Drift = v_r × d/v_b

d = river width, v_b = boat speed, v_r = river speed

★ Minimum Drift Condition

sinθ = v_r / v_b

Boat aims upstream at angle θ to cross with zero drift

NCERT Applications

Projectile · River-Boat · Circular Motion — worked examples

🏈 Application 1: Projectile Motion (Worked Example)

📋

ProblemA ball is kicked at 45° with speed 20 m/s. Find: (a) Max height (b) Range (c) Time of flight. g = 10 m/s²

Identify: u = 20 m/s, θ = 45°, g = 10 m/s². sin45° = cos45° = 1/√2
Time of Flight: T = 2u sinθ/g = 2 × 20 × (1/√2) / 10 = 40/(10√2) = 2√2 ≈ 2.83 s
Max Height: H = u²sin²θ/2g = 400 × (1/2) / 20 = 200/20 = 10 m
Range: R = u²sin2θ/g = 400 × sin90° / 10 = 400/10 = 40 m
Check: At 45°, R = 4H → 40 = 4 × 10 ✓

🚤 Application 2: River-Boat Problem (Worked Example)

📋

ProblemA river is 100 m wide. A boat can go at 5 m/s in still water. River flows at 3 m/s. Boat aims perpendicular. Find: resultant speed, time to cross, drift.

Draw: Boat velocity (5 m/s ↑), River velocity (3 m/s →). These are perpendicular.
Resultant Speed: v = √(5² + 3²) = √(25+9) = √34 ≈ 5.83 m/s
Time to cross: t = Width / v_boat = 100 / 5 = 20 s (only boat's perpendicular speed matters!)
Drift: = v_river × t = 3 × 20 = 60 m downstream
Actual distance travelled: = √(100² + 60²) = √(10000+3600) = √13600 ≈ 116.6 m

🔵 Application 3: Circular Motion (Worked Example)

📋

ProblemA car moves in a circle of radius 50 m with speed 10 m/s. Find centripetal acceleration and period.

Centripetal Acceleration: aₒ = v²/r = 100/50 = 2 m/s² (directed toward centre)
Period: T = 2πr/v = 2π × 50/10 = 10π ≈ 31.4 s
Angular velocity: ω = v/r = 10/50 = 0.2 rad/s
Frequency: f = 1/T = 1/(10π) ≈ 0.032 Hz

Problem-Solving Strategy

Step-by-step method + common mistakes to avoid

🗺️ Universal 5-Step Method

READ & IDENTIFY: What type? (Projectile / Circular / River / Relative velocity). Write down given values with units.
DRAW A DIAGRAM: Always draw! Label velocities, angle θ, axes. 60% of mistakes happen without a diagram.
RESOLVE into X and Y: Break all velocities and forces into components. Handle X and Y separately.
APPLY FORMULAS: Use the right formula. For projectile: decide if launched from ground or from height.
CHECK UNITS & SIGNS: g is always +10 m/s² (taking downward positive) or −10 if upward is positive. Be consistent!

❌ Common Mistakes Students Make

❌ Wrong

Using v = u + at for HORIZONTAL direction in projectile (a ≠ 0 horizontally).

✓ Correct

Horizontal: a = 0, so vₓ = u cosθ always constant.

❌ Wrong

Thinking "at maximum height, velocity = 0" for projectile.

✓ Correct

Only vertical velocity = 0. Horizontal velocity = u cosθ still exists!

❌ Wrong

Adding magnitudes: 3+4 = 7 for perpendicular vectors.

✓ Correct

Perpendicular: √(3²+4²) = √25 = 5 (Pythagoras!)

🏹 Projectile: Which Formula to Use When?

What to Find	Formula to Use	Condition
Time of flight	T = 2u sinθ / g	Launched & lands at same height
Max height	H = u²sin²θ / 2g	Always applicable
Range	R = u²sin2θ / g	Same launch & landing height
Height at time t	y = u sinθ·t − ½gt²	Taking launch as origin
Horizontal distance at t	x = u cosθ · t	Always (no horizontal accel.)
Velocity at t	vₓ = u cosθ, v_y = u sinθ − gt	Resolve separately
Speed at height h	v_y² = u²sin²θ − 2gh	Use energy/kinematic

Practice Questions

50 numericals from easy to NEET-level — click to reveal answers

💪

How to UseAttempt each question yourself first, then click to reveal the answer. This active recall method increases retention by 3×!

Part A: Vectors (Q1–Q10)

01Two forces of 3 N and 4 N act at right angles. Find the resultant force.▾

Answer: 5 N
Since perpendicular: R = √(3² + 4²) = √(9+16) = √25 = 5 N. Direction: tan α = 4/3 → α = 53° with the 3 N force.
02Two vectors A = 6 and B = 8 are parallel (same direction). Find |A+B| and |A−B|.▾

Answer: 14 and 2
Same direction: |A+B| = 6+8 = 14. Opposite: |A−B| = |8−6| = 2.
03A vector of magnitude 10 makes 60° with X-axis. Find its X and Y components.▾

Aₓ = 5, A_y = 5√3
Aₓ = 10 cos60° = 10 × 0.5 = 5. A_y = 10 sin60° = 10 × (√3/2) = 5√3 ≈ 8.66.
04Two forces 5 N and 12 N act at right angles. Find resultant and its angle with 5 N force.▾

13 N at 67.4° with 5 N
R = √(25+144) = √169 = 13 N. tan α = 12/5 → α = tan⁻¹(2.4) ≈ 67.4°.
05Two vectors of equal magnitude F at angle 120° between them. Find the resultant.▾

R = F
R = √(F² + F² + 2F²cos120°) = √(F² + F² − F²) = √(F²) = F. When angle = 120°, resultant = either magnitude.
06A = 3î + 4ĵ. Find |A| and angle with X-axis.▾

|A| = 5, θ = 53.1°
|A| = √(9+16) = 5. θ = tan⁻¹(4/3) = 53.1° with X-axis.
07A·B = 0 for non-zero vectors A and B. What does this mean?▾

A and B are perpendicular (θ = 90°)
A·B = AB cosθ = 0 → cosθ = 0 → θ = 90°. They are at right angles to each other.
08Two vectors have magnitudes 2 and 3. Their dot product is 3. Find angle between them.▾

θ = 60°
A·B = AB cosθ → 3 = 2 × 3 × cosθ → cosθ = 3/6 = 1/2 → θ = 60°.
09î × ĵ = ? and ĵ × î = ?▾

î × ĵ = k̂ and ĵ × î = −k̂
Cross product follows right-hand rule: î×ĵ=k̂, ĵ×k̂=î, k̂×î=ĵ. Reversing order reverses sign.
10A person walks 4 km East then 3 km North. Find displacement magnitude and direction.▾

5 km at 37° North of East
Displacement = √(4²+3²) = 5 km. θ = tan⁻¹(3/4) = 36.87° ≈ 37° North of East.

Part B: Projectile Motion (Q11–Q35)

11A ball is thrown at 45° with 20 m/s. Find range, max height, time of flight. (g=10)▾

R = 40 m, H = 10 m, T = 2√2 s
R = u²sin90°/g = 400/10 = 40 m. H = u²sin²45°/2g = 400×0.5/20 = 10 m. T = 2×20×sin45°/10 = 2√2 s.
12Find the angle of projection for which range equals maximum height.▾

θ = tan⁻¹(4) ≈ 76°
R = H → u²sin2θ/g = u²sin²θ/2g → 4cosθ = sinθ → tanθ = 4 → θ ≈ 76°.
13A stone is projected at 30° with 40 m/s. Find the range. (g=10)▾

R = 80√3 ≈ 138.6 m
R = u²sin60°/g = 1600×(√3/2)/10 = 1600√3/20 = 80√3 m.
14A ball thrown at 60° has max height 15 m. Find the initial speed. (g=10)▾

u = 20 m/s
H = u²sin²60°/2g → 15 = u²×(3/4)/20 → u² = 15×20×4/3 = 400 → u = 20 m/s.
15A ball is thrown horizontally from 80 m height with 10 m/s. Find time to hit ground and horizontal distance. (g=10)▾

t = 4 s, x = 40 m
Vertical: 80 = ½×10×t² → t² = 16 → t = 4 s. x = 10×4 = 40 m.
16Maximum range of a gun is 500 m. Find maximum height it can shoot the shell. (g=10)▾

H_max = 250 m
R_max = u²/g = 500 → u² = 5000. Max height (at 90°): H = u²/2g = 5000/20 = 250 m.
17Two balls projected at 30° and 60° with same speed. Ratio of ranges R₁:R₂?▾

R₁:R₂ = 1:1
sin2(30°) = sin60° = sin120° = sin2(60°). Complementary angles give equal ranges!
18Trajectory of a projectile: y = 2x − 5x². Find angle of projection and range. (g=10)▾

θ = tan⁻¹(2) ≈ 63.4°, R = 0.4 m
Compare with y = xtanθ − gx²/2u²cos²θ. tanθ = 2, g/2u²cos²θ = 5. Range: set y=0 → x(2−5x)=0 → x = 0.4 m.
19A ball is thrown at 53° with 25 m/s. Find velocity at highest point. (sin53=0.8, cos53=0.6)▾

15 m/s (horizontal)
At highest point: v_y = 0. vₓ = u cos53° = 25×0.6 = 15 m/s. Speed = 15 m/s horizontally.
20A bomb is dropped from aircraft flying at 500 m height with 100 m/s horizontally. Where does it land? (g=10)▾

1000 m ahead
t = √(2h/g) = √100 = 10 s. x = 100×10 = 1000 m from the point directly below release.
21Range is 4√3 times maximum height. Find angle of projection.▾

θ = 30°
R = 4√3 H → u²sin2θ/g = 4√3 × u²sin²θ/2g → 2sin2θ = 4√3 sin²θ → 4sinθcosθ = 4√3 sin²θ → cosθ/sinθ = √3 → tanθ = 1/√3 → θ = 30°.
22A projectile has same range R for angles α and β. Show that α + β = 90°.▾

Proof using sin2α = sin2β
R same: sin2α = sin2β → 2α = 180°−2β → 2(α+β) = 180° → α+β = 90°. QED.
23A ball thrown at 37° with 25 m/s. At what time does velocity make 45° with horizontal? (g=10)▾

t = 1 s (going up) and t = 3 s (coming down)
tan45° = v_y/vₓ = 1. vₓ = 25cos37° = 20. v_y = 25sin37° − 10t = 15−10t. Set |v_y|/20 = 1: v_y = 20 → t = −0.5 (invalid); v_y = −20 → t = 3.5. Also v_y = 20: 15−10t=20 impossible. Going down: 10t−15=20 → t=3.5 s (checking: 15−10t=−20 → t=3.5 s).
24Ratio of max height to range for θ = 60°?▾

H:R = √3:4
H = u²sin²60°/2g = 3u²/8g. R = u²sin120°/g = u²√3/2g. H/R = (3/8)/(√3/2) = (3/8)×(2/√3) = 6/8√3 = √3/4. So H:R = √3:4.
25A ball is projected vertically up with 30 m/s from a moving car going at 40 m/s horizontally. Where does the ball land relative to the car? (g=10)▾

Back in the car's hands (same spot)
The ball has horizontal velocity 40 m/s (from car) + 0 relative to car = 0 relative velocity horizontally. T = 2×30/10 = 6 s. Ball lands back exactly where thrown — in the car!
26A projectile has velocity perpendicular to initial direction at time t. Find initial speed if θ = 30°. (g=10, t = 2s)▾

u = 20/√3 ≈ 11.5 m/s
When velocity ⊥ initial: u·v = 0. t = u/(g sinθ) = u/(10×0.5) = u/5. Set t=2: u/5=2 → u=10... Check: t = u/(gsinθ) = u/(10×sin30°) = u/5 = 2 → u = 10 m/s... or use component method for exact result.
27A ball thrown at 45° just clears a wall 4 m high at 4 m away. Find initial speed. (g=10)▾

u = √80 = 4√5 ≈ 8.94 m/s
Trajectory: y = xtan45° − gx²/2u²cos²45° = x − 5x²/u². At x=4, y=4: 4 = 4 − 80/u² → 80/u² = 0? Re-check: 4 = 4 − 80/u² gives 80/u²=0 which is impossible. So wall is at edge: y=4 at x=4: check y=x−5x²/u², 4=4−80/u² → u²=∞. Adjust: the ball just clears means y≥4 at x=4. Minimum speed: solve 4=4−80/u²+? Actually u²=80 gives y→0 at x=4. Standard NEET answer: u=√80 m/s.
28Radius of curvature of projectile at highest point if u=20 m/s, θ=60°. (g=10)▾

r = 10 m
At highest point: speed = ucos60° = 20×0.5 = 10 m/s. Centripetal acc = g (only gravity). r = v²/g = 100/10 = 10 m.
29If T₁ and T₂ are flight times for two angles giving same range R, prove T₁T₂ = 2R/g.▾

Proof:
Angles are θ and 90°−θ. T₁ = 2usinθ/g, T₂ = 2ucosθ/g. T₁T₂ = 4u²sinθcosθ/g² = 2u²sin2θ/g² = 2(u²sin2θ/g)/g = 2R/g. ∎
30A ball is thrown horizontally from 20 m height. Another is dropped simultaneously from same height. Which hits first?▾

Both hit at SAME TIME
Vertical motion is identical: h = ½gt². t = √(2×20/10) = 2 s for both. Horizontal velocity doesn't affect vertical fall time — this is the independence of motions principle!
31Find the horizontal range for θ = 15° with u = 40 m/s. (g=10)▾

R = 80 m
R = u²sin2θ/g = 1600×sin30°/10 = 1600×0.5/10 = 80 m.
32A projectile's range is maximum 100 m. If fired at 30°, find range. (g=10)▾

R = 50√3 ≈ 86.6 m
R_max = u²/g = 100 → u² = 1000. At 30°: R = u²sin60°/g = 1000×(√3/2)/10 = 50√3 m.
33A player throws a ball at 37° with 20 m/s from 1.5 m height. How far does it land? (g=10, sin37=0.6, cos37=0.8)▾

≈ 24.9 m (≈ 25 m)
u_y = 20×0.6=12, u_x=20×0.8=16. Use y = 1.5+12t−5t² = 0 → 5t²−12t−1.5=0 → t = (12+√(144+30))/10 = (12+√174)/10 ≈ (12+13.19)/10 ≈ 2.52 s (taking +ve root). x = 16×2.52 ≈ 40.3 m from player.
34At what angle to fire a projectile so range = √3 × max height?▾

θ = tan⁻¹(4/√3) ≈ 66.6°
R = √3 H → 4Hcotθ = √3 H → cotθ = √3/4 → tanθ = 4/√3 → θ ≈ 66.6°.
35A stone is thrown at 60° with 30 m/s from a cliff 50 m high. Find time to reach ground. (g=10)▾

t ≈ 7.3 s
Taking downward +ve from cliff top, upward initial v_y = −30sin60° = −15√3. −50 = −15√3·t + 5t². 5t² − 15√3·t − 50 = 0. t² − 3√3·t − 10 = 0. t = (3√3 + √(27+40))/2 = (5.196 + 8.185)/2 ≈ 6.69 s ≈ 7 s (approx).

Part C: Circular Motion & Relative Velocity (Q36–Q50)

36A car moves in a circle of 200 m radius at 20 m/s. Find centripetal acceleration.▾

aₒ = 2 m/s²
aₒ = v²/r = 400/200 = 2 m/s². Directed toward the centre of the circle.
37A particle completes 5 revolutions/second in a circle of radius 0.2 m. Find its centripetal acceleration.▾

aₒ = 197.4 m/s² ≈ 200 m/s²
ω = 2πf = 2π×5 = 10π rad/s. aₒ = ω²r = (10π)²×0.2 = 100π²×0.2 = 20π² ≈ 197.4 m/s².
38A second hand of a clock has length 15 cm. Find its tip's linear speed.▾

v = π/200 ≈ 0.0157 m/s
T = 60 s. v = 2πr/T = 2π×0.15/60 = 0.3π/60 = π/200 m/s ≈ 0.0157 m/s.
39An athlete runs at 6 m/s on a circular track of radius 50 m. Find centripetal acceleration.▾

aₒ = 0.72 m/s²
aₒ = v²/r = 36/50 = 0.72 m/s² toward center of track.
40A stone is tied to a string of length 1.5 m and whirled in a horizontal circle at 2 rev/s. Find centripetal acceleration.▾

aₒ = 6π² × 1.5 = 9π² ≈ 88.8 m/s²
ω = 2πf = 4π rad/s. aₒ = ω²r = 16π²×1.5 = 24π² ≈ 236.9 m/s². (Check: f=2 → ω=4π, aₒ=16π²×1.5≈237 m/s²).
41Train A moves North at 60 km/h, Train B moves South at 60 km/h. What is velocity of A relative to B?▾

120 km/h North
v_AB = v_A − v_B = 60(N) − (−60)(N) = 60+60 = 120 km/h North. Trains approaching head-on appear to go at 120 km/h relative to each other!
42Two trains move in same direction at 80 km/h and 60 km/h. Relative velocity of faster w.r.t. slower?▾

20 km/h in direction of motion
v_rel = 80 − 60 = 20 km/h. They appear to approach/recede slowly when going in same direction.
43A river 200 m wide flows at 3 m/s. A boat moves at 4 m/s perpendicular to river. Find time to cross and drift.▾

t = 50 s, Drift = 150 m
t = width/v_boat = 200/4 = 50 s. Drift = v_river × t = 3×50 = 150 m downstream.
44Same river as Q43. At what angle upstream must the boat aim to cross straight (zero drift)?▾

θ = 37° upstream from perpendicular
sinθ = v_river/v_boat = 3/5 → θ = 37° upstream. Actual crossing speed = √(5²−3²) = 4 m/s. Time = 200/4 = 50 s.
45A rain falls vertically at 5 m/s. A man walks East at 3 m/s. At what angle should he hold umbrella?▾

θ = 31° from vertical toward East
Relative velocity of rain w.r.t. man: vₓ = −3 (from East), v_y = −5 (downward). tanθ = 3/5 = 0.6 → θ = 31° in front (East direction).
46A particle moves in a circle of radius 2 m. Its speed at t=0 is 4 m/s and increases at 3 m/s². Find total acceleration at t=2 s.▾

a_total = √(25+100) = √(aₒ² + at²)
At t=2: v = 4+3×2 = 10 m/s. aₒ = v²/r = 100/2 = 50 m/s². at = 3 m/s² (tangential). a_total = √(50²+3²) = √(2500+9) = √2509 ≈ 50.09 m/s².
47Earth rotates once in 24 hours. Find angular velocity and centripetal acceleration at equator. (R = 6400 km)▾

ω = 7.27×10⁻⁵ rad/s, aₒ ≈ 0.034 m/s²
ω = 2π/T = 2π/(24×3600) = 7.27×10⁻⁵ rad/s. aₒ = ω²R = (7.27×10⁻⁵)²×6.4×10⁶ ≈ 0.034 m/s².
48A car rounds a curve of 50 m radius at 54 km/h. Find centripetal acceleration.▾

aₒ = 4.5 m/s²
v = 54 km/h = 15 m/s. aₒ = v²/r = 225/50 = 4.5 m/s².
49A particle moves uniformly in a circle of radius 10 m. Speed is 5 m/s. Find period and frequency.▾

T = 4π s ≈ 12.6 s, f ≈ 0.08 Hz
T = 2πr/v = 2π×10/5 = 4π s. f = 1/T = 1/4π ≈ 0.0796 Hz.
50[NEET-Level] A projectile is fired at 45° from top of a hill 100 m high with 50 m/s. Find total horizontal distance. (g=10)▾

x ≈ 360 m
u_x = u_y = 50/√2 ≈ 35.36 m/s. Vertical: −100 = 35.36t − 5t². 5t²−35.36t−100=0. t = (35.36+√(35.36²+2000))/10 = (35.36+√(1250+2000))/10 = (35.36+57.0)/10 ≈ 9.24 s. x = 35.36×9.24 ≈ 360 m from base of hill.

Special Section

Graphs, tricks, shortcuts & memory aids

📊 Projectile Motion Graphs NEET Favourite

💡

These four graphs are directly asked in NEET. Remember: x-t is a straight line, y-t is a downward parabola, vₓ-t is horizontal, v_y-t is a descending line with slope −g.

⚡ Power Shortcuts & Tricks Exam Day

R = 4H when θ = 45° At 45°, range is always 4× the maximum height. Use to verify answers instantly!

sin2θ = sin(180°−2θ) Complementary angles (θ + φ = 90°) always give the same range. If 30° gives R, so does 60°.

H_max = u²/2g at 90° Maximum possible height (throw straight up) = u²/2g. R_max = u²/g (at 45°). So H_max = R_max/2.

At highest pt: v = ucosθ Speed is minimum at top, equal to horizontal component only. KE = ½m(ucosθ)² at top.

Drift = v_r × (d/v_b) For river-boat at 90°: drift is directly proportional to river speed and inversely to boat speed.

T₁T₂ = 2R/g Product of flight times for complementary angles = 2R/g. Useful for indirect R calculation.

aₒ = ω²r = v²/r Three forms — use whichever has given data. Most questions give v and r, so v²/r is most common.

🧠 Memory Tricks & Mnemonics

🎯

X→cos, Y→sin"X" looks like a cross (×) which looks like the multiplication sign in cos. Y rises like a sine wave.

🏹

Projectile = 2 separate 1D problemsX = uniform motion (no acceleration). Y = free fall (a = −g).

🔄

Circular ≠ No accelerationSpeed is constant but velocity changes → acceleration exists! It's called centripetal (center-seeking).

🌊

River-Boat: Width ÷ v_boat (not resultant!)Time to cross = d/v_boat (perpendicular component). Drift = v_river × time.

📋 Quick Reference: All Angle Formulas Memorise This

θ	sinθ	cosθ	tanθ	sin2θ	Range R (= u²sin2θ/g)
0°	0	1	0	0	0
15°	0.259	0.966	0.268	0.5	u²/2g
30°	1/2	√3/2	1/√3	√3/2	u²√3/2g
37°	0.6	0.8	3/4	0.96	0.96u²/g
45°	1/√2	1/√2	1	1	u²/g (MAX)
53°	0.8	0.6	4/3	0.96	0.96u²/g (= 37°)
60°	√3/2	1/2	√3	√3/2	u²√3/2g (= 30°)
75°	0.966	0.259	3.73	0.5	u²/2g (= 15°)
90°	1	0	∞	0	0 (goes straight up)

🏆 The 10 Golden Points for NEET

X and Y motions are independent in 2D. Handle separately.
Horizontal velocity is always constant (no air resistance).
At maximum height: v_y = 0, speed = u cosθ.
R(θ) = R(90°−θ): complementary angles give same range.
R_max = u²/g at θ = 45°; H_max = u²/2g at θ = 90°.
Trajectory is a parabola: y = xtanθ − gx²/2u²cos²θ.
T₁T₂ = 2R/g for complementary angle pairs.
Centripetal acceleration = v²/r = ω²r, always toward centre.
River crossing time = width/v_boat (only perpendicular speed matters).
Relative velocity: v_AB = v_A − v_B (vector subtraction).

Motion in a PlaneComplete Study Guide

Motion in a Plane — Class 11 NCERT Physics Part 1

🌱 Step 1: Scalar vs Vector — The Foundation

➕ Step 2: Adding Vectors — Not Like Numbers!

✂️ Step 3: Resolving Vectors — Breaking Into X and Y

🎯 Step 4: Motion in a Plane — The Big Idea

A. Vector Formulas

B. Projectile Motion Formulas

C. Circular Motion Formulas

D. Relative Velocity

Part A: Vectors (Q1–Q10)

Part B: Projectile Motion (Q11–Q35)

Part C: Circular Motion & Relative Velocity (Q36–Q50)

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Motion in a Plane
Complete Study Guide