Motion in a Straight Line Class 11 Notes | CBSE Physics Chapter 2

Motion in a Straight Line is Chapter 2 of CBSE Class 11 Physics — your very first taste of kinematics, the language used to describe how things move. Before you can ever explain why objects move (that comes in Laws of Motion), you must learn to measure position, displacement, velocity, and acceleration along a single line.

By the end of these notes you will be able to tell displacement from distance, read position-time and velocity-time graphs at a glance, apply the three kinematic equations to any uniformly accelerated body, solve relative-velocity problems, and handle motion under gravity confidently. This is a high-scoring chapter carrying roughly 6–8 marks in boards and a guaranteed question in JEE and NEET.

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Table of Contents

• Key Concepts — Position, displacement, velocity, speed, acceleration, graphs, relative velocity, free fall
• Weightage in Board & Entrance Exams
• Important Definitions
• Solved Examples
• Important Questions for Board Exams
• Quick Revision Points

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Key Concepts

1. Position, Path Length and Displacement

To describe motion we first fix a reference point (origin) and a set of axes — together these form a frame of reference. The position of an object is its location measured from the origin along the chosen axis.

Path length (distance) is the total length of the actual path covered by the object. It is a scalar, is always positive, and can never be zero for a moving body.

Displacement (Δx) is the change in position: the straight-line distance from the start point to the end point, with direction. It is a vector and can be positive, negative, or zero.

Δx = x₂ − x₁

• If you walk 4 m east and 3 m back west, path length = 7 m but displacement = 1 m east.
• For a round trip, displacement is zero but path length is not.
• Magnitude of displacement ≤ path length, always.

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2. Average Velocity and Average Speed

Average velocity is the displacement divided by the time interval. It is a vector, in the direction of displacement.

v̄ = Δx/Δt = (x₂ − x₁)/(t₂ − t₁)

Average speed is the total path length divided by the total time taken. It is a scalar and is always positive for a moving body.

Average speed = total path length / total time

• SI unit: m/s for both.
• Average speed ≥ magnitude of average velocity (they are equal only when motion is along a straight line in one direction).

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3. Instantaneous Velocity and Speed

Instantaneous velocity is the velocity of a body at a particular instant of time. It is the limit of average velocity as the time interval approaches zero.

v = lim (Δt→0) Δx/Δt = dx/dt

• It is the slope of the tangent to the position-time graph at that instant.
• The magnitude of instantaneous velocity equals the instantaneous speed.
• The speedometer of a car shows instantaneous speed.

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4. Acceleration

Acceleration is the rate of change of velocity with time. It is a vector and tells us how quickly velocity changes.

Average acceleration ā = Δv/Δt and instantaneous acceleration a = dv/dt = d²x/dt²

• SI unit: m/s².
• If velocity and acceleration point the same way, the body speeds up; if opposite, it slows down (retardation/deceleration).
• Acceleration is the slope of the velocity-time graph.

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5. Kinematic Equations for Uniformly Accelerated Motion

When acceleration a is constant, the motion is called uniformly accelerated motion, and three standard equations connect initial velocity u, final velocity v, acceleration a, time t, and displacement s.

Equation	What it connects
v = u + at	velocity and time (no s)
s = ut + ½at²	displacement and time (no v)
v² = u² + 2as	velocity and displacement (no t)

A useful fourth result — the distance covered in the nth second:

sₙ = u + ½a(2n − 1)

Note: These equations are valid only when acceleration is constant. Take care with signs — choose one direction as positive and stick to it throughout the problem.

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6. Position-Time Graphs

A position-time (x–t) graph plots position on the y-axis against time on the x-axis. Its slope at any point gives the instantaneous velocity.

[DIAGRAM: x–t graph — a horizontal line means the body is at rest; a straight slanted line means uniform velocity; an upward curve (increasing slope) means acceleration.]

• Straight horizontal line: object at rest (velocity = 0).
• Straight slanted line: uniform velocity (constant slope).
• Curved line: non-uniform velocity — accelerated motion.
• A steeper slope means a greater speed.

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7. Velocity-Time Graphs

A velocity-time (v–t) graph plots velocity against time. It is one of the most powerful tools in kinematics because both its slope and its area carry meaning.

• Slope of the v–t graph = acceleration.
• Area under the v–t graph = displacement.
• A straight slanted line means uniform acceleration; a horizontal line means uniform velocity (zero acceleration).

[DIAGRAM: v–t graph for uniform acceleration — a straight line rising from u to v; the area under it (a trapezium) equals the displacement s = ut + ½at².]

The kinematic equations can actually be derived from the v–t graph: v = u + at comes from the slope, and s = ut + ½at² comes from the area.

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8. Relative Velocity in One Dimension

The relative velocity of object A with respect to object B is the velocity of A as seen by an observer moving with B.

v(AB) = v(A) − v(B)

• Same direction: relative velocity = v(A) − v(B) (small if speeds are close — two trains moving alongside seem slow relative to each other).
• Opposite directions: relative velocity = v(A) + v(B) (they approach quickly).
• The time to meet = relative displacement / relative velocity.

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9. Motion Under Gravity (Free Fall)

Near the Earth’s surface, every freely falling body has a constant downward acceleration called acceleration due to gravity (g ≈ 9.8 m/s²), independent of its mass.

The kinematic equations apply directly, replacing a with g and choosing a sign convention (usually downward positive for a dropped body):

• v = u + gt
• h = ut + ½gt²
• v² = u² + 2gh

For a body thrown vertically upward with speed u: it stops momentarily at the top (v = 0), so the maximum height is H = u²/2g, the time to reach the top is t = u/g, and the total time of flight is 2u/g.

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Weightage in Board & Entrance Exams

Exam	Typical Weightage	Most-Tested Areas
CBSE Board (Class 11)	6–8 marks	Distance vs displacement, kinematic equations, v–t graphs, free fall
JEE Main / Advanced	1–2 questions	Graphs, relative velocity, motion under gravity numericals
NEET	1–2 questions	Kinematic equations, average velocity vs speed, free fall

[TABLE: Question-type split — VSA (1 mark): definitions, scalar vs vector; SA (2–3 marks): graph interpretation, kinematic numericals; LA (5 marks): derivation of equations from v–t graph, free-fall problems.]

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Important Definitions

Term	Definition
Position	Location of an object relative to a chosen origin along an axis
Path length (distance)	Total length of the actual path covered; a scalar, always positive
Displacement	Change in position; shortest directed distance: Δx = x₂ − x₁ (a vector)
Average velocity	Displacement per unit time: v̄ = Δx/Δt (a vector)
Average speed	Total path length per unit time (a scalar)
Instantaneous velocity	Velocity at an instant: v = dx/dt
Acceleration	Rate of change of velocity: a = dv/dt = d²x/dt²
Uniform motion	Equal displacements in equal time intervals (constant velocity)
Relative velocity	Velocity of one body as seen from another: v(AB) = v(A) − v(B)
Acceleration due to gravity	Constant downward acceleration of a free-falling body: g ≈ 9.8 m/s²

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Solved Examples

Example 1

A car travels 60 km east in 1 hour, then 40 km west in the next hour. Find its average speed and average velocity.

Answer: Path length = 60 + 40 = 100 km in 2 h, so average speed = 100/2 = 50 km/h. Displacement = 60 − 40 = 20 km east, so average velocity = 20/2 = 10 km/h east.

Example 2

A body starts from rest and accelerates uniformly at 2 m/s² for 5 s. Find its final velocity and the distance covered.

Answer: v = u + at = 0 + 2 × 5 = 10 m/s. s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m.

Example 3

A car moving at 20 m/s is brought to rest in 50 m by applying brakes. Find the retardation.

Answer: Using v² = u² + 2as: 0 = 20² + 2a(50). a = −400/100 = −4 m/s² (retardation of 4 m/s²).

Example 4

A stone is dropped from a tower 80 m high. Find the time taken to reach the ground and its velocity on impact. (g = 10 m/s²)

Answer: h = ut + ½gt² → 80 = 0 + ½ × 10 × t², so t² = 16, t = 4 s. v = u + gt = 0 + 10 × 4 = 40 m/s.

Example 5

A ball is thrown vertically upward with a velocity of 20 m/s. Find the maximum height reached and the total time of flight. (g = 10 m/s²)

Answer: H = u²/2g = 20²/(2 × 10) = 400/20 = 20 m. Time of flight = 2u/g = (2 × 20)/10 = 4 s.

Example 6

Two trains A and B move on parallel tracks at 60 km/h and 40 km/h in the same direction. Find the velocity of A relative to B, and relative to B if B moves in the opposite direction.

Answer: Same direction: v(AB) = 60 − 40 = 20 km/h. Opposite directions: v(AB) = 60 + 40 = 100 km/h.

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Important Questions for Board Exams

1-Mark Questions (VSA)

1. Distinguish between distance and displacement in one line.
2. Can displacement be zero while distance is not? Give an example.
3. What does the slope of a position-time graph represent?
4. What does the area under a velocity-time graph represent?
5. Is acceleration due to gravity dependent on the mass of the falling body?

2–3-Mark Questions (SA)

1. Define average velocity and instantaneous velocity. How are they related?
2. A body covers equal distances in equal time intervals along a straight line. What can you say about its velocity and acceleration?
3. Draw and explain the velocity-time graph for a uniformly accelerated body starting from rest.
4. Two cars approach each other on a straight road. Explain how to find the time before they meet using relative velocity.

5-Mark Questions (LA)

1. Derive the three kinematic equations of uniformly accelerated motion using a velocity-time graph.
2. A ball is thrown vertically upward. Derive expressions for its maximum height, time of ascent, and total time of flight.
3. Explain position-time and velocity-time graphs for uniform and non-uniform motion, with sketches and interpretation of slope and area.

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Quick Revision Points

• Distance is a scalar (path length); displacement is a vector: Δx = x₂ − x₁
• |Displacement| ≤ distance; for a round trip displacement = 0
• Average velocity = Δx/Δt; average speed = path length/time
• Average speed ≥ |average velocity|
• Instantaneous velocity v = dx/dt = slope of x–t graph
• Acceleration a = dv/dt = slope of v–t graph
• Kinematic equations (constant a): v = u + at; s = ut + ½at²; v² = u² + 2as
• Distance in nth second: sₙ = u + ½a(2n − 1)
• Area under v–t graph = displacement
• Relative velocity (1-D): v(AB) = v(A) − v(B)
• Free fall: replace a with g (≈9.8 m/s²); max height H = u²/2g; time of flight = 2u/g

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Next Chapter: Chapter 3 — Motion in a Plane