# Important Questions Class 11 Physics Chapter 3

## Important Questions for CBSE Class 11 Physics Chapter 3

### Important Questions for CBSE Class 11 Physics Chapter 3 – Motion in a Straight Line

In Chapter 3, Motion in a Straight Line of Class 11 Physics, students will learn about relative velocity and its derivations. The chapter also discusses acceleration and how it relates to speed and velocity. After that, students will look at kinematic equations and uniformly accelerated motion.

Extramarks Important Questions for Class 11 Physics Chapter 3 makes it easier to study for students who prefer learning in a question-answer format. These questions are compiled by subject matter experts and are written in clear, simple language.

CBSE Class 11 Physics Chapter 3 Important Questions

Study Important Questions for Class 11 Physics Chapter 3 – Motion in A Straight Line

Some of the 1-mark, 2-marks, 3-marks, 4-marks, and 5-marks Class 11 Physics Chapter 3 Important Questions are discussed below. Students can access the link on the Extramarks website to review additional questions.

1 Mark Answers and Questions

Q1. Two balls of varying masses are launched vertically upward at the same initial speed. Which will reach the highest point?

A1. If two balls of different masses are thrown vertically upward at the same initial speed, they will both rise to a greater height.

Q2. Can an object’s speed be negative? Justify.

A2. An object’s speed can never be negative. This is due to the fact that the distance is always positive.

### 2 Marks Answers and Questions

Q1. Give the characteristics of displacement.

A1. The characteristics of displacement are as follows.

(1) Displacement is a vector quantity that has magnitude as well as direction.

(2) A given body’s displacement can be positive, negative, or zero.

Q2. What causes a particle’s velocity to vary?

A2. The velocity of a particle varies when

(1) changes in velocity magnitude.

(2) the motion’s direction shifts.

### 3 Marks Answers and Questions

Q1. A stone is dropped from the top of a cliff and found to travel 44.1m before diving at the last second. What is the cliff’s height? (g = 9.8m/s2)

A1. Consider the height of the cliff to be h.

u = 0m/s,

a = g = 9.8m/s2.

If n is the total time taken by the stone while falling,

Snth = u+a2(2n−1)

44.1 = 0+9.82(2n−1)

N = 102 = 5s

Now,

h = ut+12at2

h = 12(9.8)(5)2 = 122.5m, which is the required height.

Q2. A woman leaves her house at 9.00 a.m., walks at a speed of 5 km/hr on a straight road to her office, which is 2.5 km away, works there until 5.00 p.m., and then returns home by auto at a speed of 25 km/hr. Select appropriate scales and draw the x-t graph of her motion.

A2.  It is given that:

Speed of the woman  = 5 km/h

Distance between her office and home  = 2.5 km

Time taken = Distance / Speed

= 2.5 / 5 = 0.5 h = 30 min

It is given that she covers the same distance in the evening by auto.

Now, the speed of the auto = 25 km/h

Time taken = Distance / Speed

= 2.5 / 25 = 1 / 10 = 0.1 h = 6 min

### 4 Marks Answers and Questions

Q1. Regular bus service connects two towns, A and B, with a bus leaving in either direction every T minutes. A man cycling at 20 km/hr from A to B notices that a bus passes him every 18 minutes in the direction of his motion and every 6 minutes in the opposite direction. What is the bus service period T, and at what speed (assumed constant) do the buses travel on the road?

A1.

Consider V to be the speed of the bus running between towns A and B.

In the question, it is given that:

Speed of the cyclist is  v=20 km/hr

The relative speed of the bus moving in the direction of the cyclist will be V−v = (V−20)m/s.

The bus went past the cyclist every 18 min, i.e.,  18 / 60 h  (when he moves in the direction of the bus).

Hence, distance covered by the bus  = (V − 20) × 18 / 60 km …. (i)

As one bus leaves after every T minutes,

the distance travelled by the bus will be  = V × T / 60 …. (ii)

As equations (i) and (ii) are equal,

(V − 20) × 18 / 60 = V × T / 60 ….. (iii)

The relative speed of the bus moving in the opposite direction of the cyclist will be (V + 20) km/h.

Thus, the time taken by the bus to go past the cyclist  = 6 min = 6 / 60 hr.

⇒(V + 20)×6 / 60 = V × T / 60 …. (iv)

From (iii) and (iv), we get

(V + 20) × 6 / 60 = (V − 20) × 18 / 60

V + 20 = 3V  60

2V = 80

V = 40 km/h, which is the required speed.

Substituting the value of V in equation (iv),

(40 + 20) × 6 / 60 = 40 T / 60

T = 360 / 40 = 9 min, which is the required time period.

### 5 Marks Answers and Questions

Q1. A man walks from his house to a market 2.5 kilometres away at a speed of 5 kilometres per hour on a straight road. When he discovers that the market is closed, he immediately turns around and walks back home at a speed of 7.5 km/h. What exactly is the

(a) the average velocity magnitude, and

(b) the average speed of the man over the time intervals?

(i) 0 to 30 minutes

(ii) 0 to 50 minutes

(iii) 0 to 40 minutes

(Note: This exercise will show you why it is better to define average speed as total path length divided by time rather than average velocity magnitude. You don’t want to tell the tired man on his way home that his average speed was 0.)

A1. In the question, it is given that,

Time taken by the man to reach the market from home is t1 = 2.5/5 = 1/2hr = 30min.

Time taken by the man to reach home from the market is t2 = 2.5/7.5 = 1/3hr = 20min.

Total time taken in the whole journey  = 30 + 20 = 50 min

1. i) 0 to 30 min

Average velocity = Displacement/Time

Average speed = Distance/Time

1. ii) 0 to 50 min

Time = 50 min = 50/60 = 5/6 h

Net displacement = 0

Total distance = 2.5 + 2.5 = 5 km

Average velocity = Displacement / Time = 0

Average speed = Distance / Time = 5/(5/6) = 6 km/h

iii) 0 to 40 min

Speed of the man = 7.5 km/h

Distance travelled in first 30 min = 2.5 km

Distance travelled by the man (from the market to home) in the next 10 min

= 7.5 × 10/60 = 1.25 km

Net displacement = 2.5  1.25 = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km

Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h

Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h

### Important Questions for Chapter 13 Physics Class 11

The chapter is full of theories and concepts that require clear understanding to grasp them and score the best possible marks. Studying and practising are two of the best ways to learn this chapter. Students should take the time to answer the important questions, as this will help them get good grades. They must also learn the theories and practise the chapter’s practical aspects. Through Class 11 Physics Chapter 3 Important Questions, students can test their understanding of the chapter. This will encourage them to write efficient answers in their exams.

### Class 11 Physics Chapter 3 Important Questions

Some of the topics that will be covered in this chapter are as follows.

Motion is one of the most important topics in the chapter on motion in a straight line.

Everything in the solar system is in motion. An object is said to be in motion if its position changes from one location to another over time. Motion is a relative concept, and when a body changes positions, it is always in relative motion.

Types of Motion: The type of motion is an important topic that is  part of the motion in a straight line. It is critical to understand an object’s precise location. We need to know the object’s coordinate positions. The number of coordinate positions varies depending on the type of object. There are occasionally two or three coordinates.

Displacement: It is an important topic related to motion in a straight line. Displacement is the position of an object at a given time or the change in the direction of the object during that time. It is a vector that is drawn from the beginning to the end.

Uniform Speed and Uniform Velocity: An object is said to move at a uniform speed if it travels the same distance in the same amount of time. The time intervals can be small or large, but they must be equal.

When an object moves with uniform velocity, it covers the same amount of distance in equal intervals of time. The time intervals can be small or large, but they must be equal.

Variable Speed and Variable Velocity: An object moves with variable speed when it travels unequal distances in equal amounts of time. The time span can be short or long.

When an object moves with variable velocity, it covers an unequal amount of distance in equal intervals of time. Time intervals can be short or long.

### Important Questions for Motion in a Straight Line Class 11

Some of the questions that are most likely to appear in the board examinations from Class 12 Physics Chapter 3 are as follows.

• When is the condition of the relationship s = ut correct?
• If two balls of different masses are thrown vertically upward with alike initial speeds, which one will rise to a higher height?
• Under what circumstances is the relative velocity of two bodies equal?

To access the complete set of Extramarks Important Questions Class 11 Physics Chapter 3, visit the link provided below.

Benefits of Important Questions For Class 11 Science Chapter 3

These important questions will assist students in achieving high marks and excelling in their board exams. Some of the key benefits of Class 11 Physics Chapter 3 Important Questions include the following.

• Before compiling a list of Important Questions, the subject matter experts at Extramarks conduct thorough research to ensure that the questions on the list have the highest likelihood of appearing in the exams.
• The Important Questions for Class 11 Physics Chapter 3 are in accordance with the CBSE board format, and the answers to these Important Questions are given in such a detailed and explanatory manner that all students can understand them comprehensively.

### Conclusion

Students who want  help with extra study material should refer to Extramarks Class 11 Physics Chapter 3 Important Questions. It will be beneficial to the students, and they will be able to achieve the highest possible marks in the board exams. Referring to Important Questions for Class 11 Physics Chapter 3 will boost students’ confidence in passing the exam.

Q.1 Read the assertion and reason carefully to mark the correct option out of the options given below. Assertion: The displacement-time graph for the objects motion with negative acceleration is curved downward. Reason: The displacement of the object is proportional to the square of the time interval.
Assertion is true but reason is false.
Assertion and reason both are false.
Both assertion and reason are true and the reason is the correct explanation of the assertion.
Both assertion and reason are true but reason is not the correct explanation of the assertion.
Marks:1
Ans
The displacement of the object moving in a straight line with negative acceleration can be represented by using second equation of motion.
s=ut12at2

According to this equation, if initial speed of the ball is zero then displacement of the object is proportion to the square of the time interval, which is the equation of parabola. The shape of the curve also depends on the sign of acceleration. For positive acceleration, the curve is upward whereas for negative acceleration, the curve is downward.

Q.2 A ball is dropped downwards, after 1 second another ball is dropped downwards from the same point. The distance between them after 3 seconds will be
20 m
25 m
50 m
98 m
Marks:1
Ans

Q.3 A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. To have more than two balls in the sky at any time the speed of throw should be (g = 9.8 m/s2)
19.6 m/s
any speed less than 19.6 m/s
more than 19.6 m/s
9.8 m/s
Marks:1
Ans
For two balls in air if one is just thrown and other is at the highest point. Then, the time interval between them is t = 2 seconds
Now v = u – g t
0 = u – g × 2 so, u = 19.6 m/s
Hence, for more than two balls in air, the speed must be greater than 19.6 m/s.

Q.4 What will be the nature of velocity-time graph for a uniform motion?

Marks:1
Ans
Velocity-time graph will be parallel to time axis.
Q.5 A particle is moving along X axis the position is given by x = K + jt2 where K = 8m and j = 4 m/s2 t is time.  Find velocity of particle at t = 0, t= 3 sec.
Marks:3
Ans
X = k + Jt2

K = 8m, j = 4 m/s2

X = 8 + 4t2
To get velocity differenciate both sides v =  =  (8 + 4t2)

v  4 x 2 t = 8t

so  = v = velocity

when t = 0, v = 0

when t = 3 sec. v = 8 X 3

= 24 m/s.

## FAQs (Frequently Asked Questions)

### 1. If an object has a one-dimensional motion with a positive acceleration value and negative velocity, is the object speeding up?

No, the object at the starting point is not speeding up in this case. We can see that the particle is slowing down because its velocity is negative. This type of scenario occurs only when an object is thrown into the air. If both the acceleration and the velocity of the object are positive, we can say the object is accelerating, or is falling from a great height.

### 2. What is the difference between distance and displacement?

Distance denotes the actual path taken by the object, whereas displacement denotes the difference between the object’s initial and final positions. The former is known as a scalar quantity, whereas the latter is known as a vector quantity. An object’s distance travelled is always positive; it can never be negative or zero. An object’s displacement, on the other hand, can be positive, negative, or even zero. The object’s distance is determined by the path it follows, but its displacement is not.

### 3. What are some of the important questions of Chapter 3 of the Physics textbook for Class 11 students, "Motion in a Straight Line"?

The following are a few important questions for Class 11 Physics Chapter 3: Motion in a Straight Line:

• What is the relative velocity of two bodies with similar velocities?
• Is it possible for an object’s speed to be negative? Justify.
• How is the displacement time graph for uniform linear motion shaped?