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Edited By Safeer PP | Updated on Aug 09, 2022 03:00 PM IST

NCERT Exemplar Class 11 Physics solutions chapter 3 introduces Motion's basic concept and connects it to our daily activities. NCERT Exemplar Class 11 Physics chapter 3 solutions help understand how Motion is crucial to everything in the universe, be it the acceleration of a car or the earth's rotation. This chapter of NCERT Class 11 Physics Solutions revolves around Motion characteristics like position and displacement, introductory concepts of instantaneous velocity, and kinematic equations to analyse Motion from a fresh perspective of velocity.

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This Story also Contains

- NCERT Exemplar Class 11 Physics Solutions Chapter 3 MCQI
- NCERT Exemplar Class 11 Physics Solutions Chapter 3 MCQII
- NCERT Exemplar Class 11 Physics Solutions Chapter 3 Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 3 Long Answer
- Class 11 Physics NCERT Exemplar Solutions Chapter 3 Includes The Following Topics:
- What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion In A Straight Line?
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- Important Topics To Cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 3
- NCERT Exemplar Class 11 Solutions

Question:1

Answer:

The correct answer is the option (b)Let us draw a parallel line from A to the time axis at . It intersects the graph at B & the change in displacement time is zero. So, the displacement from A to & hence the average velocity of the body also vanishes to 0.

Question:2

Answer:

The correct answer is the option (a) x < 0, v < 0, a > 0Velocity is downward, i.e., negative, thus, v<0.

The lift retard before reaching the 4th floor and hence the acceleration will be upwards, i.e., a>0.

Question:3

In one dimensional motion, instantaneous speed v satisfies

(a) The displacement in time T must always take non-negative values.

(b) The displacement x in time T satisfies –

(c) The acceleration is always a non-negative number.

(d) The motion has no turning points.

Answer:

The correct answer is the option (b) The displacement x in time T satisfies -v

Thus, v

Thus, -

Question:5

The displacement of a particle is given by where x is in metres and t is seconds. The distance covered by the particle in first 4 seconds is

a) 4 m

b) 8 m

c) 12 m

d) 16 m

Answer:

The answer is the option (b) 8mNow, we know that,

&

Now,

Now, distance is equal to the area between the time-axis graph and the (v-t) graph,

Hence, option (b).

Question:7

The variation of quantity A with quantity B, plotted in Fig. 3.2 describes the motion of a particle in a straight line.

(a) Quantity B may represent time.

(b) Quantity A is velocity if motion is uniform.

(c) Quantity A is displacement if motion is uniform.

(d) Quantity A is velocity if motion is uniformly accelerated.

Answer:

The correct answer is the option:(a) Quantity B may represent time.

(c) Quantity A is displacement if the motion is uniform.

(d) Quantity A is velocity if the motion is uniformly accelerated.

Verification of opt (a) & (d)

If the quantity B would have represented velocity instead of time, then the graph would’ve become a straight line, viz., uniformly accelerated motion, hence the motion is not uniform.

Verification of opt (c)

If A represents displacement and B represents time, then the graph will be a straight line which would represent uniform motion.

Question:8

A graph of x versus t is shown in Fig. 3.3. Choose correct alternatives from below.

(a) The particle was released from rest at t = 0.

(b) At B, the acceleration a > 0.

(c) At C, the velocity and the acceleration vanish.

(d) Average velocity for the motion between A and D is positive.

(e) The speed at D exceeds that at E.

Answer:

The correct answer is the option:(a) The particle was released from rest at t = 0.

(c) At C, the velocity and acceleration vanish.

(e) The speed at D exceeds that at E.

Explanation: Now, we know that,

Slope of the x-t graph gives us

Verification of opt(a)-

is zero or particle is at rest at A since graph (x-t) is parallel to the time axis.

Slope increases after A and hence velocity also increases.

Verifying option (c) and rejecting opt (b)-

Now, or v = 0 since the tangent at B & C is graph (x-t), viz., parallel to the time axis. Hence, acceleration = 0.

Verifying opt (e)-

Speed at D is greater than speed at E since the slope at D is greater at D than at E.

Rejecting opt (d)-

Average velocity at A is zero as graph (x-t) is parallel to time axis, also displacement is negative at D, which makes it clear that the velocity at D is also negative.

Question:9

For the one-dimensional motion, describe by

Answer:

The correct answer is the option:Explanation: Now,

We know that

Now,

v

. v

Thus, it is clear that v lies between 0 to 2 and option (d) is verified.

Now,

Thus, sin t lies between 1 and -1 for all t > 0.

Thus, x will always be positive and option (a) is verified.

Now,

&

Hereby opt (b) is discarded.

Now,

&

Thus, acceleration can be negative as well, and hence opt (c) is also discarded here.

Question:10

A spring with one end attached to a mass and the other to a rigid support is stretched and released.

a) magnitude of acceleration, when just released is maximum

b) magnitude of acceleration, when at equilibrium position is maximum

c) speed is maximum when mass is at equilibrium position

d) magnitude of displacement is always maximum whenever speed is minimum

Answer:

The correct answer is the option:(a) Magnitude of acceleration, when just released is maximum.

(c) Speed is maximum when mass is at the equilibrium position.

Now let us stretch the spring to a displacement x by force F,

Now, P.E. at

Since restoring force is proportional to x, Simple Harmonic Motion is executed here.

Therefore,

Or

&

Thus, when spring is released, the magnitude will be maximum. Hence, opt(a) is verified here.

At x = 0, the speed of mass is maximum.

Hence opt (c) is also verified.

At x = 0, magnitude of a = 0.

Hence, opt (b) is discarded.

The speed of mass may or may not be zero when it is at its maximum displacement.

Hence, opt (d) is also discarded here.

Question:11

A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,

a) the direction of motion of the ball changes every 10 seconds

b) speed of ball changes every 10 seconds

c) average speed of ball over any 20 seconds intervals is fixed

d) the acceleration of ball is the same as from the train

Answer:

The correct answer is the option:(b) The speed of the ball changes every 10 seconds.

(c) average speed of the ball over any 20 seconds interval is fixed.

(d) the acceleration of the ball is the same as from the train.

Explanation: If we observe the motion from the ground, we will see that the ball strikes with the wall after every 10 seconds. The direction of the ball is the same since it is moving at a very small speed in the moving train, therefore, it will not change w.r.t. observer from the earth.

The speed of ball can change after a collision, hence, option (a) will be discarded and opt (b) is verified.

Average speed of the ball at any time remain same or is 1 m/s, i.e., it is uniform.

Hence opt (c) is also verified.

When the ball strikes to the wall, initial speed of the ball will be in the direction of the moving train w.r.t. the ground as well as its speed will also change (vTG)

Thus,

The speed of the ball after collision with a side of the train is in the opposite direction of the train

Thus, the magnitude of acceleration on both the walls of the compartment will be the same, but in opposite direction. Hence, opt (b), (c) & (d) are verified here.

NCERT Exemplar Class 11 Physics Solutions Chapter 3 very short answer

Question:12

Refer to the graphs below and match the following:

Graph | Characteristics |

a) | i) has and throughout |

b) | ii) has throughout and has a point with and a point with |

c) | iii) has a point with zero displacement for |

d) | iv) has and |

Answer:

(i) From the graph (d), it is indicated that the slope is always positive between to (tan ?).Hence, (i) (d)

(ii) At point A, v = 0 & a = 0 as the slope is zero; thus, the graph always lies in +x direction.

Hence, (ii) (b).

(iii) There is zero displacement only in graph (a), where y = 0.

Hence, (iii) (a).

(iv) In the graph (c), since v < 0, the slope is negative here.

Hence, (iv) (c).

Question:13

Answer:

When we hit a ball with a bat, the acceleration of the ball decreases, till its velocity becomes zero. Hence, the acceleration will be in the backward directionAfter the velocity of the ball has been decreased to zero, it increases in the forward direction. Thus, the acceleration will be negative in the forward direction

Question:14

Give examples of a one-dimensional motion where

a) the particle moving along positive x-direction comes to rest periodically and moves forward

b) the particle moving along positive x-direction comes to rest periodically and moves backward

Answer:

(a) Let us consider a motion whereThus,

&

(ii) Let us consider a function of motion where,

Thus the displacement of the particle is in negative direction and it comes to rest periodically.

Thus,

is a periodic function.

Now,

After zero displacement, velocity changes periodically.

Thus, is the function required.

(i) Now, let us consider a function

Thus, the particle moves in a positive direction, periodically with zero displacements.

Hence is the required function.

Question:15

Give example of a motion where at a particular instant.

Answer:

Let x(t) be the function of motion,Here, & A are constant & B is the amplitude.

At time t the displacement is x(t),

Here

Thus,

Now,

Thus, we get,

x > 0, i.e., x is always positive, since A>B

v<0, i.e., v is always negative, since v<0

& a>0, i.e., a is always positive.

The value of varies from 0 to

Question:17

A ball is dropped and its displacement vs time graph is as shown in the figure where displacement x is from the ground and all quantities are positive upwards.

a) Plot qualitatively velocity vs time graph

b) Plot qualitatively acceleration vs time graph

Answer:

If we observe the graph we know that the displacement (x) is always positive. The velocity of the body keeps on increasing till the displacement becomes zero, after that the velocity decreases to zero in the opposite direction till the maximum value of x is reached, viz., smaller than earlier. When the body reaches towards x=0, the velocity increases and acceleration is in the downward direction. And when the body’s displacement is , i.e., the body moves upwards, the direction will be downwards and velocity will decrease, i.e., .(a) Velocity time graph

(b) Acceleration time graph.

Question:18

A particle executes the motion described by where

a) Where does the particles start and with what velocity?

b) Find maximum and minimum values of . Show that and increase with time and decreases with time.

Answer:

Here,So,

&

+

Thus, is the starting point of the particle and its velocity is

(b)

Maximum at since

Minimum at since at ,

v(t) is,

maximum at since

minimum at since, ,

a(t) is,

maximum at since at ,

minimum at since at ,

Question:22

The velocity-displacement graph of a particle is shown in the figure.

a) Write the relation between v and x.

b) Obtain the relation between acceleration and displacement and plot it.

Answer:

a) Consider the point P(x,v) at any time t on the graph such that angle ABO is such thatWhen the velocity decreases from to zero during the displacement, the acceleration becomes negative.

is the relation between v and x.

x=x

The points are

and

Question:23

It is a common observation that rain clouds can be at about a kilometre altitude above the ground.

a) If a raindrop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h

b) A typical raindrop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.

c) Estimate the time required to flatten the drop.

d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.

e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two raindrops is 5 cm.

Answer:

Thus,

(a) Let's find out the velocity of raindrop on the ground’

Thus,

(b) Momentum of the raindrop when it touches the ground

mass of drop(m) = Vol. × density

Now, density of water

Thus,

Now, we know that Momentum (p) = mv

(c) Time required for a drop to be flattened-

(d) Now, we know that,

(e) Here,

Radius of umbrella

Thus, Area of umbrella

Now, the square area covered by one drop

Therefore, no. of drops falling on the umbrella

Therefore 314 drops fell on the umbrella.

Thus, the net force on the umbrella

Question:24

Answer:

Given : (for truck)We know that,

i.e.,

thus,

Given : (for car)

Again,

A human takes atleast 0.5 seconds to respond, thus time taken by the car driver to respond is sec …. (car takes t time to stop)

There is no responding time for the truck driver so he applies breaks with passing signal to car back side, hence,

From (i) & (ii),

Thus,

Thus,

Now the distance covered by the car & the truck in sec will be,

…….

First 0.5 seconds the car moves with uniform speed but after responding brakes are applied for 0.5 sec and the retarding motion of the car starts.

Thus,

Therefore, the car must be 1.25 m behind the truck to avoid bumping into it.

Question:25

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by and for on m/s. It repeats this cycle till it reaches the height of 20 m.

a) At what time is its velocity maximum?

b) At what time is its average velocity maximum?

c) At what times is its acceleration maximum in magnitude?

d) How many cycles are required to reach the top?

Answer:

(a) for the velocity to be maximum,Thus,

(b)Now, average velocity

Let us integrate both the sides from 0 to 3

Thus,

Now, average velocity

Thus, the average velocity is maximum at 2.36 seconds.

(c) When the body returns at its mean position or changes direction in periodic motion, time for acceleration is maximum.

Thus, the acceleration is maximum at t = 3 sec.

(d)Now, for 3 to 6 sec

Integrate from 3 to 6 s

.............because distance is in downward direction

Thus, net distance

Thus, in three cycle

Remaining height will be

The monkey can climb up to 9m without slipping but in the 4^{th }cycle it will slip and the height remaining to climb will be 6.5 m.

Net no. of cycle = 4.

Question:26

Answer:

Let be the speed of the 1& be the speed of the second ball,

Let be the height of the two balls before coming to rest & be the height covered by 1 ball before coming to rest.

Now, we know that,

Thus,

and

Now,

Thus,

Calculating time for 1

Thus,

Now, calculating time for second ball,

This,

Thus, time intervals between these two balls will be,

To help comprehend the concepts well, students can take the help of NCERT Exemplar Class 11 Physics Solutions Chapter 3 PDF Download, which could be put to download by differrent web page download options

Well-versed academicians prepare these solutions to provide the students an insight into a creative learning process that would help them perform better in exams.

**Also, check NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

**Also, Read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

3 Motion in a Straight Line

- 3.1 Introduction
- 3.2 Position, path length, and displacement
- 3.3 Average velocity and average speed
- 3.4 Instantaneous velocity and speed
- 3.5 Acceleration
- 3.6 Kinematic equations for uniformly accelerated Motion
- 3.7 Relative velocity

- Students will be able to throw light on the significance of Motion in their typical activities.
- They would be able to utilise the concept of Motion to explain and observe the working of everything that surrounds them.
- NCERT Exemplar Class 11 Physics chapter 3 solutions enlighten the students by providing logical explanations behind situations.
- NCERT Exemplar Class 11 Physics solutions chapter 3 helps get a clear idea of average speed and velocity, which is a crucial part of our lives.
- Students are also introduced to numerical interpretations of these concepts with the help of kinematic expressions.

· NCERT Exemplar Class 11 Physics solutions chapter 3 covers the properties and terms related to Motion, such as position, path length, the frame of reference, and a detailed description of displacement. This chapter provides an in-depth analysis of the study of the Motion of objects along a straight line.

· This chapter also helps students to differentiate between speed and velocity and know the terms like average speed, average velocity, instantaneous speed, and instantaneous velocity. It also offers a graphical representation using diverse examples of velocity and speed. It dives deep into the study of acceleration to understand how an object gains or loses speed.

· The mathematical and graphical analysis of Motion and the relative terms are introduced in this chapter through Kinematic equations for uniform accelerated Motion. NCERT Exemplar Class 11 Physics chapter 3 solutions would also help highlight the critical feature of Motion in terms of frames of reference, by the concept of relative velocity.

Chapter 1 | Physical world |

Chapter 2 | Units and Measurement |

Chapter 3 | Motion in a straight line |

Chapter 4 | Motion in a Plane |

Chapter 5 | Laws of Motion |

Chapter 6 | Work, Energy and Power |

Chapter 7 | System of Particles and Rotational motion |

Chapter 8 | Gravitation |

Chapter 9 | Mechanical Properties of Solids |

Chapter 10 | Mechanical Properties of Fluids |

Chapter 11 | Thermal Properties of Matter |

Chapter 12 | Thermodynamics |

Chapter 13 | Kinetic Theory |

Chapter 14 | Oscillations |

Chapter 15 | Waves |

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Download EBook1. How can these solutions help?

NCERT Exemplar Class 11 Physics Solutions Chapter 3 can help you to understand the applications of motions in real life and help you perform better in exams.

2. What are the essential topics of this chapter?

Essential topics of NCERT Exemplar Class 11 Physics Solutions Chapter 3 are the Motion in a Straight Line, position, path length, displacement, average velocity and average speed, Kinematic equations, relative velocity, instantaneous velocity and speed.

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