NCERT Solutions For Class 11 Physics Chapter 3: Motion in a Straight Line

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NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line deals with the concepts of Position, path length and displacement, acceleration, kinematic equations, and relative velocity. Motion is the change in the position of an object over a specific period of time with respect to its surroundings. According to Newton’s first law of motion, a body without any net force on it continues to move in a straight line with a permanent velocity until and unless the body is subjected to a net force.

Class 11 Physics Chapter 3 Motion in a Straight Line belongs to Unit 2 Kinematics. Unit 2 along with Unit 1 Physical World and Measurement and Unit 3 Laws of Motion has a weightage of 23 marks. Class 11 Physics Chapter 3 NCERT Solutions covers position-time graphs, calculation of velocity, acceleration, etc.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 3

NCERT Solutions for Class 11 Physics Chapter 3

Read More: Motion in a Straight Line MCQ

Class 11 Physics Chapter 3 – Concepts Covered

An object is in motion if its position changes with time

Based on the position coordinates, motion can be classified as –

(i) One-dimensional motion: A particle in motion along a straight line or a path is said to be in one dimensional motion.
(ii) Two-dimensional motion: A particle in motion in a plane is said to be in two dimensional motion.
(iii) Three-dimensional motion: A particle that is moving in space is said to undergo three dimensional motion.

The total length of the actual path travelled by a particle between its initial and final positions is referred to as the distance travelled by the particle.
Displacement of a particle is the change in the position of a particle in a particular direction during a period of time.

- Displacement has direction while distance does not.

- The magnitude of displacement can be both negative and positive whereas distance is always positive.

Speed is the time rate at which a given object is moving along a path.

An object is said to move with a uniform speed if it covers equal distances in equal intervals of time, howsoever small these intervals may be.

Velocity is the rate and direction of the movement of an object.

An object is said to move with uniform velocity if it covers equal displacements in equal intervals of time, howsoever small these intervals may be.

Class 11 Physics Chapter 3 Related Guides:

Speed and Velocity	Derivation of equation of motion	Frames of Reference
Relative Speed	Uniform Acceleration	Difference between Kinetics and kinematics
Acceleration Time Graph	Path Length	Uniform motion
Average Acceleration Formula	Speed Distance Time Formula	Displacement Formula
Average velocity formula	Average Speed Formula	V-T graphs

Class 11 Physics Study Guides:

NCERT Solutions for Class 11 Physics	Physics Constants	Relation between articles in Physics
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NCERT Solutions For Class 11 Physics Chapter 3: Motion in a Straight Line

NCERT Solutions for Class 11 Physics Chapter 3

Class 11 Physics Chapter 3 – Concepts Covered

CBSE CLASS XII Related Questions

1.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

4.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

5.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Similar Physics Concepts

Comments

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NCERT Solutions For Class 11 Physics Chapter 3: Motion in a Straight Line

NCERT Solutions for Class 11 Physics Chapter 3

Class 11 Physics Chapter 3 – Concepts Covered

CBSE CLASS XII Related Questions

1.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

4. If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

5.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

4.
If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain.

5.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: