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Updated on 23 Jun 2026, 11:36 IST
Motion in a Straight Line Class 11 Notes help students understand the basic ideas of kinematics, including position, displacement, speed, velocity, acceleration, equations of motion, graphs, motion under gravity, and relative velocity. This chapter is one of the most important starting points in CBSE Class 11 Physics because it builds the foundation for Laws of Motion, Work Energy and Power, Rotational Motion, Gravitation, Oscillations, and Waves.
In the latest NCERT Books, this topic is studied as Class 11 Physics Chapter 2: Motion in a Straight Line. In some older books and search queries, students may still search it as Class 11 Physics Chapter 2 Notes or Kinematics Class 11 Notes. This page is designed to help both CBSE board students and JEE/NEET aspirants revise the chapter in a clear and exam-focused way.
On this page, students can download Motion in a Straight Line Class 11 Notes PDF, revise formulas, understand graph-based questions, learn derivations of equations of motion, and practice important questions.
Also Check: Class 11 Physics Notes Chapter Wise
Motion in a Straight Line is the branch of kinematics that studies the motion of an object along a straight path. In this type of motion, only one coordinate changes with time. For example, a car moving on a straight road, a train moving on a straight track, or a stone falling vertically downward are examples of motion in a straight line.
This chapter mainly answers questions like:
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This chapter is important because it introduces the mathematical language of motion. Students learn how to use formulas, graphs, and calculus-based ideas to describe real motion. The concepts from this chapter are used repeatedly in later chapters of mechanics.
After completing this chapter, students should be able to:
CBSE board students should focus on the latest NCERT syllabus, while JEE and NEET aspirants should also study the full kinematics foundation, including distance, displacement, speed, velocity, and relative velocity.
Some basic topics may be reduced or rationalized in newer NCERT textbooks for school-level exams. However, these topics are still very important for JEE Main, JEE Advanced, NEET, and other entrance exams.
| Topic | CBSE Board Focus | JEE/NEET Focus |
| Rest and motion | Important | Important |
| Distance and displacement | Basic understanding | Very important |
| Speed and velocity | Important | Very important |
| Average and instantaneous velocity | Important | Very important |
| Acceleration | Very important | Very important |
| Equations of motion | Very important | Very important |
| Graphs of motion | Very important | Very important |
| Motion under gravity | Very important | Very important |
| Relative velocity | May be limited in board focus | Highly important |
A frame of reference is a coordinate system or point of view from which the position and motion of an object are observed.
Rest and motion are relative terms. An object may be at rest for one observer but in motion for another observer.
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A body is said to be at rest if its position does not change with time with respect to a chosen frame of reference.
Example: A student sitting inside a moving bus is at rest with respect to the bus, but the same student is in motion with respect to a person standing on the road.
A body is said to be in motion if its position changes with time with respect to a chosen frame of reference.
Example: A train moving on a track is in motion with respect to the railway platform.
Students often think rest and motion are absolute. In Physics, rest and motion depend on the observer or frame of reference.
Rectilinear motion is motion along a straight line. It is also called one-dimensional motion because only one coordinate is needed to describe the position of the object.
Examples of rectilinear motion:
In one-dimensional motion, the position of a particle can be represented using one coordinate, usually x.
If the particle moves along the x-axis, its position at any time t is written as x(t).
Position tells where an object is located with respect to a reference point or origin.
For motion along a straight line, the position of a particle is represented using positive or negative coordinates.
Example:
Path length is the actual length of the path travelled by an object.
Path length is always positive because it represents the total distance covered.
Distance is the total actual path length travelled by a body, while displacement is the shortest distance between the initial and final positions with direction.
Distance and displacement are very important for understanding speed and velocity.
| Feature | Distance | Displacement |
| Definition | Total actual path travelled | Shortest distance between initial and final position |
| Quantity type | Scalar | Vector |
| Direction | No direction | Has direction |
| Value | Always positive or zero | Can be positive, negative, or zero |
| Path dependence | Depends on actual path | Depends only on initial and final position |
| For round trip | Not zero | Zero |
| Formula idea | Sum of actual path lengths | Final position - Initial position |
Suppose a student walks 3 m east and then 4 m west.
Total distance = 3 m + 4 m = 7 m
Displacement = Final position - Initial position
Displacement = 1 m west
No, displacement can never be greater than distance. The magnitude of displacement is always less than or equal to the distance travelled because displacement is the shortest distance between two points.
Speed is the rate of change of distance, while velocity is the rate of change of displacement.
Speed is a scalar quantity, while velocity is a vector quantity.
Speed tells how fast an object is moving.
Formula:
Speed = Distance / Time
SI unit of speed:
m s-1
Average speed is the ratio of total distance travelled to total time taken.
Formula:
Average Speed = Total Distance / Total Time
If a car travels 100 km in 2 hours:
Average speed = 100 / 2 = 50 km h-1
Velocity is the rate of change of displacement with time.
Formula:
Velocity = Displacement / Time
SI unit of velocity:
m s-1
Velocity has both magnitude and direction.
Average velocity is the ratio of total displacement to total time taken.
Formula:
vavg = Δx / Δt
or
vavg = (x2 - x1) / (t2 - t1)
Where:
Instantaneous velocity is the velocity of an object at a particular instant of time.
In calculus form:
v = dx / dt
This means instantaneous velocity is the derivative of position with respect to time.
| Feature | Speed | Velocity |
| Meaning | Rate of change of distance | Rate of change of displacement |
| Quantity type | Scalar | Vector |
| Direction | No direction | Has direction |
| Can be negative? | No | Yes |
| Formula | Distance / Time | Displacement / Time |
| SI unit | m s-1 | m s-1 |
Yes, an object can have constant speed but variable velocity if its direction of motion changes.
Example: In uniform circular motion, the speed of the object may remain constant, but velocity changes continuously because direction changes at every point.
Acceleration is the rate of change of velocity with respect to time.
Formula:
a = Δv / Δt
or
a = dv / dt
SI unit of acceleration:
m s-2
Acceleration tells how quickly velocity changes.
Average acceleration is the change in velocity divided by the time interval.
Formula:
aavg = (v - u) / t
Where:
Instantaneous acceleration is the acceleration of an object at a particular instant of time.
Formula:
a = dv / dt
Since velocity is the derivative of position:
a = d2x / dt2
Uniform acceleration means velocity changes by equal amounts in equal intervals of time.
Example: A freely falling body near Earth has approximately uniform acceleration equal to g = 9.8 m s-2.
Non-uniform acceleration means velocity changes by unequal amounts in equal intervals of time.
Example: A vehicle moving in traffic usually has non-uniform acceleration.
Retardation, also called deceleration, is acceleration opposite to the direction of motion.
If a body is moving in the positive direction and slowing down, its acceleration is negative.
Example: A car applying brakes has retardation.
Students often think negative acceleration always means slowing down. Actually, negative acceleration means acceleration is in the negative direction. Whether the object slows down depends on the direction of velocity.
Equations of motion are formulas used to describe the motion of an object moving with constant acceleration.
The three main equations of motion are:
Where:
sn = u + a(2n - 1) / 2
Where:
The first equation of motion relates final velocity, initial velocity, acceleration, and time.
Starting with acceleration:
a = (v - u) / t
Multiplying both sides by t:
at = v - u
Rearranging:
v = u + at
v = u + at
A car starts with velocity 5 m s-1 and accelerates at 2 m s-2 for 4 s.
v = u + at
v = 5 + 2 × 4
v = 13 m s-1
The second equation of motion gives displacement in terms of initial velocity, acceleration, and time.
Average velocity for uniform acceleration is:
Average velocity = (u + v) / 2
Displacement is:
s = Average velocity × Time
So:
s = [(u + v) / 2] t
Using:
v = u + at
Substitute v:
s = [(u + u + at) / 2] t
s = [(2u + at) / 2] t
s = ut + 1/2 at2
s = ut + 1/2 at2
The third equation of motion relates final velocity, initial velocity, acceleration, and displacement.
We know:
v = u + at
So:
t = (v - u) / a
Also:
s = [(u + v) / 2] t
Substitute t:
s = [(u + v) / 2] [(v - u) / a]
s = (v2 - u2) / 2a
Therefore:
2as = v2 - u2
Rearranging:
v2 = u2 + 2as
v2 = u2 + 2as
This section is very useful for JEE and NEET aspirants.
Acceleration is the rate of change of velocity:
a = dv / dt
So:
dv = a dt
Integrating from initial velocity u to final velocity v, and from time 0 to t:
∫ dv = ∫ a dt
For constant acceleration:
v - u = at
Therefore:
v = u + at
Velocity is the rate of change of displacement:
v = ds / dt
So:
ds = v dt
Using:
v = u + at
We get:
ds = (u + at) dt
Integrating from displacement 0 to s, and time 0 to t:
∫ ds = ∫ (u + at) dt
s = ut + 1/2 at2
Acceleration can be written as:
a = dv / dt
Also:
v = ds / dt
So:
a = dv / dt = (dv / ds)(ds / dt)
Since:
ds / dt = v
Therefore:
a = v(dv / ds)
So:
a ds = v dv
Integrating:
∫ a ds = ∫ v dv
For constant acceleration:
as = (v2 - u2) / 2
Therefore:
v2 = u2 + 2as
A position-time graph shows how the position of an object changes with time.
In a position-time graph:
The slope of a position-time graph represents velocity.
Formula:
Velocity = Change in position / Change in time
v = Δx / Δt
| Type of Motion | Shape of Position-Time Graph | Meaning |
| Object at rest | Horizontal straight line | Position does not change |
| Uniform velocity | Inclined straight line | Constant velocity |
| Uniform acceleration | Curved graph | Velocity changes with time |
| Uniform retardation | Curve bending downward | Velocity decreases with time |
A velocity-time graph shows how velocity changes with time.
In a velocity-time graph:
The slope of a velocity-time graph represents acceleration.
Formula:
Acceleration = Change in velocity / Change in time
a = Δv / Δt
The area under a velocity-time graph represents displacement.
If the graph is above the time axis, displacement is positive. If the graph is below the time axis, displacement is negative.
| Type of Motion | Shape of Velocity-Time Graph | Meaning |
| Constant velocity | Horizontal line | Acceleration is zero |
| Uniform acceleration | Straight line with positive slope | Velocity increases uniformly |
| Uniform retardation | Straight line with negative slope | Velocity decreases uniformly |
| Non-uniform acceleration | Curved line | Acceleration changes with time |
Motion under gravity is motion in which the object moves only under the influence of Earth’s gravitational acceleration.
Near Earth’s surface, acceleration due to gravity is approximately:
g = 9.8 m s-2
For quick calculations, many problems use:
g = 10 m s-2
If upward direction is taken as positive:
If downward direction is taken as positive:
When a body is thrown vertically upward with initial velocity u:
At maximum height:
v = 0
Using:
v2 = u2 + 2as
For upward direction positive:
0 = u2 - 2gH
Therefore:
H = u2 / 2g
Using:
v = u + at
At highest point:
v = 0 and a = -g
So:
0 = u - gt
Therefore:
tup = u / g
For vertical upward projection returning to the same level:
T = 2u / g
At the highest point, velocity is zero, but acceleration is not zero. Acceleration is still g downward.
At the highest point of vertical motion, the velocity is zero, but acceleration is still g downward.
This is a very common conceptual question. Students often think acceleration becomes zero at the top because velocity is zero. That is wrong. Gravity continues to act on the body throughout the motion.
Velocity at highest point = 0
Acceleration at highest point = g downward = 9.8 m s-2
Free fall is the motion of an object under the influence of gravity alone.
If air resistance is neglected, all objects fall with the same acceleration near Earth’s surface.
If the object starts from rest:
Using equations of motion:
v = gt
s = 1/2 gt2
v2 = 2gs
Relative velocity is the velocity of one object as observed from another moving object.
If object A has velocity vA and object B has velocity vB, then velocity of A relative to B is:
vAB = vA - vB
If two bodies move in the same direction:
Relative velocity = Difference of velocities
Example:
If car A moves at 20 m s-1 and car B moves at 12 m s-1 in the same direction:
vAB = 20 - 12 = 8 m s-1
If two bodies move in opposite directions:
Relative speed = Sum of speeds
Example:
If two cars move toward each other with speeds 20 m s-1 and 12 m s-1:
Relative speed = 20 + 12 = 32 m s-1
| Concept | Formula |
| Average speed | Total distance / Total time |
| Average velocity | Total displacement / Total time |
| Average acceleration | (v - u) / t |
| Instantaneous velocity | v = dx / dt |
| Instantaneous acceleration | a = dv / dt |
| First equation of motion | v = u + at |
| Second equation of motion | s = ut + 1/2 at2 |
| Third equation of motion | v2 = u2 + 2as |
| Displacement in nth second | sn = u + a(2n - 1) / 2 |
| Maximum height | H = u2 / 2g |
| Time to reach maximum height | tup = u / g |
| Total time of flight | T = 2u / g |
| Relative velocity | vAB = vA - vB |
Question: A car starts from rest and accelerates uniformly at 3 m s-2 for 5 s. Find its final velocity.
Given:
u = 0
a = 3 m s-2
t = 5 s
Formula:
v = u + at
Solution:
v = 0 + 3 × 5
v = 15 m s-1
Answer: Final velocity = 15 m s-1
Question: A body starts from rest and moves with uniform acceleration of 2 m s-2 for 10 s. Find the displacement.
Given:
u = 0
a = 2 m s-2
t = 10 s
Formula:
s = ut + 1/2 at2
Solution:
s = 0 × 10 + 1/2 × 2 × 102
s = 100 m
Answer: Displacement = 100 m
Question: A ball is thrown vertically upward with speed 20 m s-1. Find the maximum height reached. Take g = 10 m s-2.
Given:
u = 20 m s-1
v = 0 at maximum height
g = 10 m s-2
Formula:
H = u2 / 2g
Solution:
H = 202 / (2 × 10)
H = 400 / 20
H = 20 m
Answer: Maximum height = 20 m
Question: Two trains move in the same direction with speeds 30 m s-1 and 20 m s-1. Find the relative velocity of the first train with respect to the second.
Given:
vA = 30 m s-1
vB = 20 m s-1
Formula:
vAB = vA - vB
Solution:
vAB = 30 - 20
vAB = 10 m s-1
Answer: Relative velocity = 10 m s-1
| Mistake | Correct Approach |
| Confusing distance and displacement | Distance is path length; displacement depends on endpoints |
| Treating speed and velocity as same | Speed is scalar; velocity is vector |
| Forgetting sign convention | Choose positive and negative direction before solving |
| Using equations of motion for non-uniform acceleration | These equations apply only for constant acceleration |
| Thinking acceleration is zero at highest point | Acceleration is g downward at the highest point |
| Forgetting area under v-t graph | Area under v-t graph gives displacement |
| Forgetting slope of x-t graph | Slope of x-t graph gives velocity |
| Forgetting slope of v-t graph | Slope of v-t graph gives acceleration |
Also Check: CBSE Class 11 Physics Formula Sheets
Students can download the Motion in a Straight Line Class 11 Notes PDF for quick revision before school exams, JEE, NEET, and other entrance exams.
The PDF includes:
Students can also explore complete Class 11 Physics courses on Infinity Learn for concept clarity, guided practice, and exam-focused preparation.
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Motion in a Straight Line is one of the most important chapters in Class 11 Physics. It introduces students to one-dimensional motion, position, distance, displacement, speed, velocity, acceleration, equations of motion, graphs, motion under gravity, and relative velocity.
Students should focus on:
For better preparation, students should revise the formulas daily, solve NCERT questions, practice graph-based questions, and download the Motion in a Straight Line Class 11 Notes PDF for quick exam revision.
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Motion in a Straight Line is the study of the motion of an object along a straight path. It is also called one-dimensional motion or rectilinear motion.
Distance is the total path travelled by an object, while displacement is the shortest distance between initial and final positions with direction. Distance is scalar, while displacement is vector.
No, displacement can never be greater than distance. The magnitude of displacement is always less than or equal to the total distance travelled.
Yes, an object can have constant speed but variable velocity if its direction changes. Uniform circular motion is a common example.
The slope of a position-time graph represents velocity. A steeper slope means greater velocity.
The area under a velocity-time graph represents displacement. If the graph lies below the time axis, the displacement is negative.
At the highest point, velocity becomes zero, but acceleration remains g downward. The acceleration is 9.8 m s-2 downward.
The three equations of motion for constant acceleration are:
Yes, relative velocity is very important for JEE and NEET. It is used in train problems, meeting problems, overtaking problems, and advanced motion questions.