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CBSE Class 11 Physics★ 5-7 marks · CBSE Class 11 Physics Chapter 2; kinematics is foundational for JEE and NEET

Motion in a Straight Line

The kinematics of one-dimensional motion — position, velocity, acceleration, equations of motion, and relative velocity. CBSE Class 11 Physics Chapter 2.

⏱ 15 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Distinguish distance from displacement and speed from velocity
2Define instantaneous velocity and acceleration using calculus
3Apply the three kinematic equations for constant acceleration
4Analyse motion using position-time and velocity-time graphs
5Solve free-fall and relative-velocity problems

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Why this chapter matters

Kinematics describes how things move without asking why. Displacement, velocity, acceleration, the equations of motion, and graphical analysis are the foundation of all mechanics -- mastering one-dimensional motion makes projectile, rotational, and orbital motion far easier.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Units and Measurements

Units and significant figures for kinematics calculations

Motion in a Straight Line

'Motion is the change of position with time.' — Basic definition

1. Chapter Overview

KINEMATICS describes motion WITHOUT asking about its cause. This chapter focuses on RECTILINEAR motion — motion along a straight line. You will learn the KEY VARIABLES (position, displacement, velocity, acceleration), their relationships, the THREE EQUATIONS of motion for constant acceleration, and how to analyse motion through GRAPHS.

2. Position, Path Length, and Displacement

Position (x): Location of an object relative to a chosen origin (can be + or -)
Distance (Path Length): TOTAL length of the ACTUAL path travelled (scalar)
Displacement: CHANGE in position = final position - initial position (vector, can be +, -, or 0)

Key Distinction

Quantity	Scalar/Vector	Depends on Path?
Distance	Scalar	YES
Displacement	Vector	NO

3. Speed and Velocity

Average Speed = Total path length / Total time taken (SCALAR)
Average Velocity = Displacement / Total time taken (VECTOR)
Instantaneous Velocity v = dx/dt (velocity at an INSTANT)
Instantaneous Speed = |v| (magnitude of instantaneous velocity)

Uniform Motion

When an object covers EQUAL distances in EQUAL intervals of time
Velocity is CONSTANT, acceleration is ZERO

4. Acceleration

Average Acceleration ā = (v — u) / t = Δv/Δt
Instantaneous Acceleration a = dv/dt = d²x/dt²
Uniform Acceleration: Velocity changes by EQUAL amounts in EQUAL time intervals
Deceleration: Negative acceleration (velocity decreasing)

Important

Acceleration is a VECTOR; direction matters
SI unit: m/s²
Non-uniform acceleration: a varies with time

5. Kinematic Equations (Constant Acceleration)

For motion with CONSTANT acceleration a, starting with initial velocity u:

v = u + at (velocity-time relation)
s = ut + ½at² (position-time relation)
v² = u² + 2as (velocity-position relation)

Free Fall (a special case)

a = +g (downward) or a = —g (upward)
g (acceleration due to gravity) ≈ 9.8 m/s² NEAR Earth's surface
KEY: All objects fall with the SAME acceleration (ignoring air resistance)

Worked Problem

Q: A ball is thrown upward with initial speed 20 m/s. Find: (a) maximum height, (b) time to reach top, (c) total time in air. Take g = 10 m/s². Solution: (a) At top, v = 0. v² = u² + 2as → 0 = 400 + 2(-10)h → h = 20 m (b) v = u + at → 0 = 20 + (-10)t → t = 2 s (c) Total time = 2 × time to top = 4 s (symmetry of projectile motion)

6. Graphical Analysis

Position-Time (x-t) Graphs

Slope of x-t graph = velocity
Straight line = uniform velocity
Parabola = constant acceleration

Velocity-Time (v-t) Graphs

Slope of v-t graph = acceleration
Area under v-t graph = displacement
Horizontal line = constant velocity

Worked Problem

Q: A car accelerates from rest at 2 m/s² for 5 s, travels at constant speed for 10 s, then brakes to stop in 3 s. Draw v-t graph and find total distance. Solution:

Phase 1: v = at = 2×5 = 10 m/s. s₁ = ½×2×25 = 25 m
Phase 2: v = 10 m/s. s₂ = 10×10 = 100 m
Phase 3: a = (0-10)/3 = -3.33 m/s². s₃ = ½×10×3 = 15 m (area of triangle)
Total distance = 25 + 100 + 15 = 140 m

7. Relative Velocity

Definition: Velocity of ONE object as observed from ANOTHER moving object
v_AB = v_A — v_B (velocity of A relative to B)
If two objects move in the SAME direction: v_rel = v₁ — v₂
If in OPPOSITE directions: v_rel = v₁ + v₂

Applications

Crossing a moving walkway
Two trains passing each other
Overtaking problems

8. Common Mistakes

Confusing distance and displacement: Displacement can be ZERO while distance is non-zero (round trip)
Sign convention errors: Always define a positive direction and stick to it
Using v = u + at when a is not constant: These equations ONLY work for uniform acceleration
Area under v-t is displacement, NOT distance: If velocity changes sign, area under the axis is negative

9. CBSE Exam Focus

Derivation of equations of motion using calculus (v = dx/dt, a = dv/dt)
Numerical problems on free fall (5-mark)
Relative velocity problems
Area under v-t graph for displacement
Stopping distance and reaction time problems

10. Key Formulas

v = u + at
s = ut + ½at²
v² = u² + 2as
sₙ = u + a(n — ½) (distance in nth second)
v_rel = v_A — v_B

11. Self-Test (5+ Q&A)

Q1: A particle moves along x-axis with v = 10 + 2t m/s. Find displacement from t = 0 to t = 5 s. A: s = ∫vdt = ∫(10+2t)dt = 10t + t² from 0 to 5 = 50 + 25 = 75 m.

Q2: A car moving at 72 km/h stops in 50 m. Find deceleration. A: u = 72 km/h = 20 m/s. v² = u² + 2as → 0 = 400 + 2a(50) → a = -4 m/s².

Q3: Two trains 100 m each run on parallel tracks at 54 km/h and 72 km/h in opposite directions. Find crossing time. A: v_rel = 54 + 72 = 126 km/h = 35 m/s. Distance = 200 m. Time = 200/35 = 5.71 s.

Q4: A stone is dropped from a tower of height 80 m. Find time to reach ground and speed on impact (g = 10 m/s²). A: s = ½gt² → 80 = 5t² → t = 4 s. v = gt = 40 m/s.

Q5: Can an object have non-zero acceleration but zero velocity? Give an example. A: YES — at the highest point of a vertically thrown ball, v = 0 but a = g = 9.8 m/s² downward.

12. Conclusion

Motion in a straight line is the SIMPLEST form of motion, but the concepts you learn here — displacement vs. distance, instantaneous velocity, acceleration, kinematic equations — form the FOUNDATION for all of mechanics. Master the sign convention and graphical analysis, and you will find more complex motion problems much easier.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Equations of motion (constant a)

v = u + at; s = ut + (1/2)at^2; v^2 = u^2 + 2as

Valid only for uniform acceleration.

Distance in nth second

s_n = u + a(n - 1/2)

Distance travelled during the nth second of motion.

Instantaneous velocity and acceleration

v = dx/dt; a = dv/dt = d^2x/dt^2

Slope of x-t graph is v; slope of v-t graph is a.

Relative velocity

v_AB = v_A - v_B

Same direction: subtract; opposite directions: add.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing distance and displacement

✓ Displacement is a vector and can be zero (round trip) even when distance travelled is large.

WATCH OUT

✗ Using v = u + at when acceleration is not constant

✓ The kinematic equations apply only to uniform acceleration; use calculus (integration) otherwise.

WATCH OUT

✗ Treating area under v-t graph as distance

✓ Area under a v-t graph is displacement; if velocity changes sign, area below the axis is negative.

WATCH OUT

✗ Ignoring sign convention

✓ Choose a positive direction at the start and apply it consistently to u, v, a, and s.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Position, velocity, acceleration

Definitions, instantaneous quantities, graphs

Questions

Equations of motion and free fall

Numerical problems on uniform acceleration and gravity

Questions

Relative velocity

Trains, overtaking, crossing problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1MEDIUM· Calculus

A particle moves with v = 10 + 2t m/s. Find the displacement from t = 0 to t = 5 s.

▶ Show solution

s = integral of v dt = integral(10 + 2t)dt = 10t + t^2, evaluated 0 to 5 = 50 + 25 = 75 m.

Q2MEDIUM· Kinematics

A car moving at 72 km/h stops in 50 m. Find the deceleration.

▶ Show solution

u = 20 m/s. v^2 = u^2 + 2as: 0 = 400 + 2a(50), so a = -4 m/s^2.

Q3MEDIUM· Relative Velocity

Two 100 m trains move at 54 km/h and 72 km/h in opposite directions on parallel tracks. Find the crossing time.

▶ Show solution

Relative speed = 54 + 72 = 126 km/h = 35 m/s. Total distance = 200 m. Time = 200/35 = 5.71 s.

Q4MEDIUM· Free Fall

A stone is dropped from an 80 m tower. Find the time to reach the ground and the impact speed (g = 10 m/s^2).

▶ Show solution

s = (1/2)g t^2: 80 = 5 t^2, so t = 4 s. Impact speed v = g t = 40 m/s.

Q5EASY· Concept

Can an object have zero velocity but non-zero acceleration? Give an example.

▶ Show solution

Yes. At the highest point of a ball thrown vertically upward, the velocity is momentarily zero, but the acceleration is g = 9.8 m/s^2 downward.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Distance is scalar and path-dependent; displacement is a vector and path-independent.
•v = dx/dt; a = dv/dt; slope of x-t is v, slope of v-t is a.
•Equations of motion (v = u + at, s = ut + (1/2)at^2, v^2 = u^2 + 2as) need constant acceleration.
•Free fall: a = g = 9.8 m/s^2; all objects fall equally ignoring air resistance.
•Area under a v-t graph gives displacement.
•Relative velocity v_AB = v_A - v_B; add speeds for opposite directions.
•An object can have zero velocity with non-zero acceleration.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-7 marks across the chapter

Question type	Marks each	Typical count	What it tests
Free-fall / kinematics numericals	3-5	1	Equations of motion and gravity
Graphs	2-3	1	x-t and v-t graph interpretation
Relative velocity	2-3	1	Trains and overtaking

Prep strategy

Memorise and practise the three equations of motion
Interpret slopes and areas of motion graphs
Set sign conventions before solving
Use calculus when acceleration varies

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Vehicle safety

Stopping-distance and reaction-time calculations underpin road-safety rules and braking-system design.

Sports analysis

Tracking velocity and acceleration helps analyse sprinters, jumpers, and projectiles.

Free-fall and gravity

Understanding free fall is the basis for skydiving, drop towers, and measuring g experimentally.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Define positive direction and list u, v, a, s, t before solving
Choose the kinematic equation that omits the unknown you don't need
Use graph slopes and areas for graphical questions
Exploit symmetry in projectile/free-fall timing

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Solve motion problems with variable acceleration using integration and differentiation.
Analyse multi-stage motion (acceleration, cruise, braking) with combined v-t graphs.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Kinematics)High

NEET PhysicsHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

During ascent gravity decelerates the ball uniformly until its velocity is zero at the top; during descent the same gravity accelerates it through the same distance. Because the magnitude of acceleration and the distance are identical, the time taken to rise equals the time taken to fall, and the ball returns with the same speed it was thrown.

The slope (gradient) of a velocity-time graph at any point equals the acceleration at that instant. A straight upward-sloping line means constant positive acceleration, a horizontal line means zero acceleration (constant velocity), and a downward slope means deceleration. The area under the graph gives the displacement.

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