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NCERT Solutions Class 11 Physics Chapter 3 Motion In A Straight Line – Here are all the NCERT solutions for Class 11 Physics Chapter 3. This solution contains questions, answers, images, explanations of the complete chapter 3 titled Of Motion In A Straight Line taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 3 Motion In A Straight Line After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 3 Motion In A Straight Line in one place.

NCERT Solutions Class 11 Physics Chapter 3 Motion In A Straight Line

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Physics for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 3 Motion In A Straight Line , Physics, Class 11.

Class 11
Subject Physics
Book Physics Part I
Chapter Number 3
Chapter Name

Motion In A Straight Line

NCERT Solutions Class 11 Physics chapter 3 Motion In A Straight Line

Class 11, Physics chapter 3, Motion In A Straight Line solutions are given below in PDF format. You can view them online or download PDF file for future use.

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

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Question & Answer

Q.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

Ans : (a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object. (b) The size of a monkey is very as compared to the size Of circular track. Therefore, the monkey can be considered as a point sized object on the track. (c) The size Of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object. (d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
Q.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 Figure. 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).

Ans : (a) A lives closer to school than B. (b) A starts from school earlier than B, (c) B walks faster than A. (d) A B reach home at the same time. (e) B overtakes A once on the road. Explanation: (a) In the given x-t graph, it can be observed that distance OP < OQ Hence, the distance Of school from the A's home is less than that from B's home. (b) In the given graph, it can be observed that for x = O, t = O for A, whereas for x = 0, t has some finite value for B. Thus, A starts his journey from school earlier than B. (c) In the given x-t graph, it can be observed that the slope of B is greater than that of A. Since the slope of the x—t gives the speed, a greater slope means that the speed of B is greater than the speed A. (d) It Is clear from the given graph that both A and B reach their respective homes at the same time. (e) B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.
Q.3: A woman starts from her home at 9.00 am, walks with a speed of 5 km \(\mathrm{h}^{-1}\) 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 \(\mathrm{h}^{-1}\). Choose suitable scales and plot the x-t graph of her motion.

Ans : Speed of the woman = 5 km/h Distance between her office and home Distance Time taken Speed - 2.5 km \(\begin{array}{l}{\text { Time taken }=\frac{\text { Distance }}{\text { Speed }}} \\ {=\frac{2.5}{5}=0.5 \mathrm{h}=30 \mathrm{min}}\end{array}\) It is given that she covers the same distance in the evening bv an auto. Now, speed of the auto = 25 km/h \(\begin{array}{l}{\text { Time taken }=\frac{\text { Distance }}{\text { Speed }}} \\ {=\frac{2.5}{25}=\frac{1}{10}=0.1 \mathrm{h}=6 \mathrm{min}}\end{array}\) The suitable x-t graph Of the motion Of the woman is shown in the given figure.
Q.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.

Ans : Distance covered with 1 step = 1 m Time taken = 1 s Time taken to move first 5 m forward = 5 s Time taken to move 3 m backward = 3 s Net distance covered = 5 - 3 = 2 m Net time taken to cover 2 m = 8 s Drunkard covers 2 m in 8 s. Drunkard covered 4 m in 16 s. Drunkard covered 5 m in 24 s. Drunkard covered 8 m in 32 s. In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and a falls into the pit. Net time taken by the drunkard to cover 13m = 32 + 5 = 37 s The x-t graph of the drunkard's motion can be shown as:
Q.5: A jet airplane travelling at the speed of 500 km \(\mathrm{h}^{-1}\) ejects its products of combustion at the speed of 1500 km \(\mathrm{h}^{-1}\) relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

Ans : Speed of the jet airplane, v jet = 500 km/h Relative speed of its products of combustion with respect to the plane, vsmoke = - 1500 km/h Speed of its products of combustion with respect to the ground = v'smoke Relative speed of its products of combustion with respect to the airplane, Vsmoke = v'smoke — vjet - 1500 = v'smoke - 500 v'srnoke = - 1ooo km/h The negative sign indicates that the direction Of its products Of combustion is opposite to the direction of motion of the jet airplane,

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