## NCERT Class 11 Physics Chapter 3 Exercise Solutions : Motion in a Straight Line, NCERT Class 11 Physics Chapter 3

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## NCERT Class 11 Physics Chapter 3 Exercise Solutions

NCERT Class 11 Physics Chapter 3, titled “Motion in a Straight Line,” deals with the concepts of motion, distance, and displacement, and their relationship with time. The chapter also covers topics such as average speed, average velocity, and instantaneous velocity, as well as the graphical representation of motion. The exercise at the end of the chapter consists of a set of questions and problems designed to test the student’s understanding of these concepts. The solutions to these questions and problems, which are known as the NCERT Class 11 Physics Chapter 3 Exercise Solutions, provide step-by-step explanations and work through each problem to help students better understand the material and improve their problem-solving skills.

## Qu.1 : In which of the following examples of motion, can the body be considered approximately a point object:

• (a) a railway carriage moving without jerks between two stations.
• (b) a monkey sitting on top of a man cycling smoothly on a circular track.
• (c) a spinning cricket ball that turns sharply on hitting the ground.
• (d) a tumbling beaker that has slipped off the edge of a table.

In the examples given, the body that can be considered approximately a point object is the cricket ball in example (c).

A point object is an object that is so small, or has such a small mass, that it can be considered to have all its mass concentrated at a single point. This means that the object behaves as if all its mass is located at its center of mass, and its motion can be described using basic principles of physics such as Newton’s laws of motion.

In example (a), the railway carriage is not a point object because it has a large mass and physical dimensions, and its motion cannot be accurately described using basic principles of physics. Similarly, in example (b), the monkey and the man cycling on a circular track are not point objects because they both have large masses and physical dimensions.

In example (d), the tumbling beaker is not a point object because it has a physical size and shape and its motion cannot be accurately described using basic principles of physics.

However, in example (c), the spinning cricket ball can be considered a point object because it has a small mass and physical size, and its motion can be accurately described using basic principles of physics such as Newton’s laws of motion.

## Qu.2 : The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below ;

• (a) (A/B) lives closer to the school than (B/A)
• (b) (A/B) starts from the school earlier than (B/A)
• (c) (A/B) walks faster than (B/A)
• (d) A and B reach home at the (same/different) time
• (e) (A/B) overtakes (B/A) on the road (once/twice).

From the position-time graph in Fig. 3.19, the correct entries in the brackets are:

• (a) B lives closer to the school than A.
• (b) A starts from the school earlier than B.
• (c) B walks faster than A.
• (d) A and B reach home at the same time.
• (e) A overtakes B on the road once.

To determine these answers, we can analyze the position-time graph for each child.

For child A, the position-time graph is a straight line with a positive slope, which indicates that A is moving with a constant velocity in the positive direction (toward home). The slope of the line is equal to the velocity of A.

For child B, the position-time graph is also a straight line with a positive slope, but the slope is steeper than that of A’s line, indicating that B is moving with a higher velocity than A in the positive direction (toward home).

From the positions of the two lines on the graph, we can see that B starts from the school later than A and reaches home at the same time as A, which means that B is walking faster than A. We can also see that A overtakes B on the road once, as A’s line is above B’s line for a portion of the graph.

The fact that B’s line is steeper than A’s line also indicates that B lives closer to the school than A, as B takes less time to cover the same distance as A.

## Qu. 3 : A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.

To plot the x-t graph of the woman’s motion, we first need to determine the distance and time intervals for which the graph should be plotted.

The woman starts from her home at 9.00 am and returns home at 5.00 pm, which is a total time interval of 8 hours. We can set the time interval on the x-axis of the graph as 0-8 hours, with a scale of 1 hour per division.

The woman walks 2.5 km to her office and returns home by auto, covering a total distance of 2.5 km + 2.5 km = 5 km. We can set the distance interval on the y-axis of the graph as 0-5 km, with a scale of 1 km per division.

With these scales, we can now plot the x-t graph of the woman’s motion as follows:

1. From 9.00 am to 11.00 am (2 hours), the woman walks from her home to her office at a speed of 5 km/h. This is represented by a straight line with a slope of 5 km/h (velocity) on the graph, starting at the point (9,0) and ending at the point (11,2.5).
2. From 11.00 am to 5.00 pm (6 hours), the woman stays at her office and does not move. This is represented by a horizontal line on the graph, starting at the point (11,2.5) and ending at the point (5,2.5).
3. From 5.00 pm to 9.00 pm (4 hours), the woman returns home by auto at a speed of 25 km/h. This is represented by a straight line with a slope of 25 km/h (velocity) on the graph, starting at the point (5,2.5) and ending at the point (9,5).

The final x-t graph of the woman’s motion should look like the following:

[Insert graph here]

## Qu. 4 : A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

To plot the x-t graph of the drunkard’s motion, we first need to determine the distance and time intervals for which the graph should be plotted.

The drunkard starts at a distance of 0 m from the pit and ends at a distance of 13 m from the pit, which is a total distance of 13 m. We can set the distance interval on the y-axis of the graph as 0-13 m, with a scale of 1 m per division.

The drunkard takes 5 steps forward and 3 steps backward repeatedly, with each step taking 1 s. This means that he takes a total time of 5 steps * 1 s/step + 3 steps * 1 s/step = 8 s for each cycle. We can set the time interval on the x-axis of the graph as 0-80 s, with a scale of 8 s per division.

With these scales, we can now plot the x-t graph of the drunkard’s motion as follows:

1. From 0 s to 8 s (8 s), the drunkard takes 5 steps forward and 3 steps backward. This is represented by a line with a slope of 1 m/s (velocity) on the graph, starting at the point (0,0) and ending at the point (8,8).
2. From 8 s to 16 s (8 s), the drunkard takes 5 steps forward and 3 steps backward again. This is represented by a line with a slope of 1 m/s (velocity) on the graph, starting at the point (8,8) and ending at the point (16,16).
3. From 16 s to 24 s (8 s), the drunkard takes 5 steps forward and 3 steps backward again. This is represented by a line with a slope of 1 m/s (velocity) on the graph, starting at the point (16,16) and ending at the point (24,24).
4. From 24 s to 32 s (8 s), the drunkard takes 5 steps forward and 3 steps backward again. This is represented by a line with a slope of 1 m/s (velocity) on the graph, starting at the point (24,24) and ending at the point (32,32).

The final x-t graph of the drunkard’s motion should look like the following:

[Insert graph here]

From the x-t graph, we can see that the drunkard takes 32 s to fall in the pit 13 m away from the start. This can also be calculated using the formula for distance traveled, which is d = vt, where d is the distance, v is the velocity, and t is the time. In this case, the velocity of the drunkard is 1 m/s, and the time is 32 s, so the distance traveled is d = 1 m/s * 32 s = 32 m. This is equal to the distance from the start to the pit, so the drunkard takes 32 s to fall in the pit.

## Qu. 5: A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

The speed of the products of combustion relative to the ground can be calculated using the formula for the addition of velocities in one dimension, which is v = v1 + v2, where v is the final velocity, v1 is the initial velocity, and v2 is the added velocity.

In this case, the initial velocity of the products of combustion is 500 km/h relative to the ground, and the added velocity is 1500 km/h relative to the jet plane. Therefore, the final velocity of the products of combustion relative to the ground is v = 500 km/h + 1500 km/h = 2000 km/h.

This means that the speed of the jet plane with respect to an observer on the ground is 2000 km/h – 500 km/h = 1500 km/h.

## Qu. 6: A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?

The retardation of a moving object is the rate at which it slows down or decelerates. It is typically measured in units of acceleration, such as meters per second squared (m/s^2).

To calculate the retardation of the car, we can use the formula a = (vf – vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.

In this case, the initial velocity of the car is 126 km/h, the final velocity is 0 km/h (since the car comes to a stop), and the time is unknown. The distance traveled during this time is 200 m.

We can convert the initial velocity and distance to meters per second and meters, respectively, using the following conversions: 1 km/h = 1000 m/3600 s = 1000/3600 m/s = 5/18 m/s, and 1 km = 1000 m. This gives us an initial velocity of 126 km/h * 5/18 m/s/km/h = 35/9 m/s, and a distance of 200 m.

We can now use the formula a = (vf – vi) / t to calculate the acceleration (retardation) of the car. Substituting the values, we get a = (0 m/s – 35/9 m/s) / t = -35/9 m/s / t.

To solve for the time t, we can use the formula d = (vi + vf) / 2 * t, where d is the distance traveled. Substituting the values, we get 200 m = (35/9 m/s + 0 m/s) / 2 * t = 35/9 m/s * t. Solving for t, we get t = 200 m / (35/9 m/s) = 540/35 s = 15.43 s.

Therefore, the retardation of the car is a = -35/9 m/s / t = -35/9 m/s / 15.43 s = -2.29 m/s^2, and it takes 15.43 s for the car to come to a stop.

## Qu. 7:

Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ?

To solve this problem, we can use the formula d = vt + (1/2)at^2, where d is the distance traveled, v is the initial velocity, t is the time, and a is the acceleration.

In this case, the initial velocity of train B is 72 km/h, the time is 50 s, and the acceleration is 1 m/s^2. The distance traveled by train B during this time can be calculated as follows:

d = 72 km/h * 50 s + (1/2) * 1 m/s^2 * 50 s^2 = 3600 m + 25 m = 3625 m

Since the guard of train B just brushes past the driver of train A after 50 s, this means that the original distance between the two trains was equal to the length of train A, which is 400 m.

Therefore, the original distance between the two trains was 400 m.

## Qu. 8: On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?

To solve this problem, we can use the formula a = (vf – vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.

In this case, the initial velocity of car B is 54 km/h, the final velocity is unknown (since we don’t know how fast B has to go to overtake A), and the time is also unknown (since we don’t know how long it takes for B to overtake A).

Since the distance AB is equal to AC at a certain instant, this means that the distance between car A and car B is equal to the distance between car A and car C. Therefore, the distance between car B and car C is also equal to 1 km.

We can use this information to determine the time it takes for car B to overtake car A. Let’s call this time t. During this time, car A travels a distance of 36 km/h * t = 36t km. At the same time, car B travels a distance of 54 km/h * t = 54t km. Since the distance between car B and car C is also 1 km, this means that the total distance traveled by car B is 54t km + 1 km = 55t km.

We can now use the formula a = (vf – vi) / t to calculate the minimum acceleration of car B. Substituting the values, we get a = (vf – 54 km/h) / t. We can rearrange this formula to solve for vf as follows: vf = a * t + 54 km/h.

We can now use this formula to calculate the final velocity of car B. Substituting the values, we get vf = a * t + 54 km/h = a * (55t km – 36t km) / (54 km/h) + 54 km/h.

Solving for a, we get a = (vf – 54 km/h) / (t * 54 km/h / (55t km – 36t km)).

Since we are looking for the minimum acceleration of car B, we can set vf to the maximum speed allowed on the road, which is typically around 100 km/h. Substituting this value for vf, we get a = (100 km/h – 54 km/h) / (t * 54 km/h / (55t km – 36t km)).

Solving for a, we get a = (100 km/h – 54 km/h) / (t * 54 km/h / (55t km – 36t km)) = 46 km/h / (t * 54 km/h / (55t km – 36t km)).

Since t and (55t km – 36t km) are both unknown, we cannot solve for a directly. However, we can use this formula to calculate the minimum acceleration of car B for different values of t. For example, if t = 10 s, we get a = 46 km/h / (10 s * 54 km/h / (550 km – 360 km