## NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight LineNCERT Solutions for Physics Chapter 3 Class XI is an important resource for exam preparation. They include essential questions from CBSE sample papers and previous year's question papers, as well as answers to the questions from the textbook. Solutions to model problems are provided to assist you in exam preparation and verify your degree of readiness. All objects in the universe experience motion. While we sleep, blood continues to flow through our veins and arteries, and air enters and exits our bodies. Motion is the term for a change in an object's position over time. A student's 11th-grade year is crucial to develop a solid understanding of fundamental ideas. You will benefit from the material you learn with the NCERT Solutions for Class 11 Physics Chapter 3 because they are updated to align with the latest Syllabus (2022-23). ## NCERT Solutions Class XI Physics Chapter 3
In which of the following examples of motion can the body be considered approximately a point object? - A railway carriage moving without jerks between two stations.
- A cap on top of a man cycling smoothly on a circular track.
- A spinning cricket ball that turns sharply on hitting the ground.
- A tumbling beaker that has slipped off the edge of a table.
- The carriage will be regarded as a point object because the distance between the two stations is quite great compared to the size of a carriage.
- The monkey will be regarded as a point object because it is much larger than a circular track compared to the size of the man it is sitting on.
- The ball makes a sharp turn over a relatively small distance. Cricket balls are not regarded as point objects since the size of the spinning cricket ball is comparable to the distance through which the ball abruptly turns when it hits the ground.
- The table's height is relatively small. Because a beaker's size is comparable to the height of a table, it will not be regarded as a point object.
The bodies will therefore be regarded as a point object in options (a) and (b).
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. Choose the correct entries in the brackets - (A/B) lives closer to the school than (B/A)
- (A/B) starts from the school earlier than (B/A)
- (A/B) walks faster than (B/A)
- A and B reach home at the (same/different) time
- (A/B) overtakes (B/A) on the road (once/twice).
- Since A has to cover a short distance than B therefore we can conclude that A lives close to school than B [OP < OQ].
- Since x=0 and t=0 for A therefore it has to start earlier from school than B. However for B there is some finite amount of time.
- B walks faster than A because slope of B is greater than A.
- A and B, both will reach their home at the same time.
- B overtakes A once on the road and this happens at the point of intersection.
A woman starts from her home at 9.00 am, walks at a speed of 5 km/h 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 auto at a speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion.
Distance between her office and home = 2.5 km. Speed of the woman while walking = 5 km/h Time taken to reach the office = (2.5/5 ) h=(1/2) h = 30 minutes Speed of auto = 25 km/h Time taken to reach home in auto = 2.5/25 = (1/10) h = 0.1 h = 6 minutes In the graph, O is taken as the origin of the distance and the time, then at t = 9.00 am, x = 0 and at t = 9.30 am, x = 2.5 km OA is the portion on the x-t graph that represents her walk from home to the office. AB represents her time of stay in the office from 9.30 to 5. Her return journey is represented by BC.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards, followed again by 5 steps forward and 3 steps backwards, 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.
In 1 second the drunkard walks 1 step. He moves 5m in forward direction in 5 seconds and also comes back 3m in the next 3 seconds. So in total the time period is 8 seconds and the man has covered a distance of 2m. Therefore we can say that in order to cover the distance of 8m the man will take 32 seconds. The remaining distance between man and pit is 5m and we know that in 5 seconds he covers 5m in forward direction. Therefore the man will take 32 + 5 = 37 seconds to fall into the pit which is at a distance of 13m.
A jet aeroplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Let v Let v Therefore combustion products speed with respect to observer on the ground is v v
A car moving along a straight highway with a speed of 126 km h
Car has an initial velocity of = u Final velocity = v Total distance covered by car before it comes to rest = 200 m Using equation of motion we get, Therefore time taken by the car to come to rest is 11.44 seconds.
Two trains, A and B, of length 400 m each, are moving on two parallel tracks with a uniform speed of 72 km h
Length of train A = 400 m Length of train B = 400 m Speed of both trains = 72 km/hr Thus the original distance between the two trains is s
On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars, B and C, approach car A in opposite directions with a speed of 54 km/h 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 is the minimum acceleration of car B required to avoid an accident?
Speed of car A = 36 km/h So car B needs to cover a total distance of 1000 + 400 = 1400 m. It means that if car B wants to overtake car A before car C then car B needs to cover 1400 m in 40 s. Using the equation of motion relation, Thus car B can avoid accident by minimum acceleration of 1 m s
Two towns, A and B, are connected by regular bus service, with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h
Let speed of each bus be = V Let speed of the cyclist be = V Thus relative velocity of bus while travelling in the direction of motion of cyclist is V The bus passes every 18 minutes when the bus is in the direction of motion of cyclist, thus the time is (18/60) s Thus relative velocity of bus while travelling in the opposite direction of motion of cyclist is V The bus passes every 6 minutes when the bus is in the opposite direction of motion of cyclist, thus the time is (6/60) s
A player throws a ball upwards with an initial speed of 29.4 m/s. - What is the direction of acceleration during the upward motion of the ball?
- What are the velocity and acceleration of the ball at the highest point of its motion?
- Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of the x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
- To what height does the ball rise, and after how long does the ball return to the player's hands? (Take g = 9.8 m s
^{-2}and neglect air resistance).
(a) No matter in what direction the body is moving the acceleration due to gravity always acts on the body in downwards direction towards the centre of earth. If the body is moving along the direction of gravity its speed increases on the other hand if it is moving in opposite direction of g its speed decreases. (b) Acceleration due to gravity is a fixed quantity and it always acts on body in downward direction towards center of earth. So no matter what it will always be 9.8 m s (c) If we consider the highest point of ball motion as x = 0, t = 0, and vertically downward direction to be +ve direction of the x-axis, then - During the upward motion of the ball before reaching the highest point position, x = +ve, velocity, v = -ve and acceleration, a = +ve.
- During the downward motion of the ball after reaching the highest point position, velocity and acceleration, all three quantities are positive.
(d) Initial speed of the ball, u= -29.4 m/s final velocity of the ball, v = 0 m/s Acceleration due to gravity = 9.8 m/s From equation of motion, Thus the maximum height reached by the ball -44.1 m and the negative sign indicates that the movement of the ball is in upward direction that is against g. From equation of motion, Thus the total time taken by the ball to reach back players hand is 3 + 3 = 6 seconds.
Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion - with zero speed at an instant may have non-zero acceleration at that instant
- with zero speed may have non-zero velocity
- with constant speed must have zero acceleration
- with positive value of acceleration must be speeding up
- When an object is thrown upward, its speed is zero at its highest point, but it has a constant acceleration throughout the flight. When the ball is at its highest point, its acceleration is equal to the acceleration caused by gravity. Consequently, the statement is true.
- If an item has any velocity, it can never have zero speed. The speed of the particle is equal to its own velocity because it is travelling in a one-dimensional space. Consequently, the statement is false.
- A rate of change in velocity is called acceleration. If the velocity is constant, then there will be no acceleration and no change in velocity. Consequently, the statement is true.
- The ball is accelerated upwards when it is thrown by gravity, which is a positive acceleration, but the speed is continuously reduced. Consequently, the statement is false.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Ball is dropped from the height of = 90 m While dropping the ball is stationary initially therefore initial velocity of the ball is = 0 m/s Let us consider v as the final velocity of the ball. Using the equation of motion we get,
Provide clear explanations and examples to distinguish between: - The total length of a path covered by a particle and the magnitude of displacement over the same interval of time.
- The magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval].
In (a) and (b), compare and find which of the two quantities is greater. When can the given quantities be equal? [For simplicity, consider one-dimensional motion only].
(a) Let us take an example of two players playing pass-pass with football. Suppose one of them is A and the other one is B. Now A hit the ball towards B and again B passed it back to A. Thus, the initial position and the final position of the ball is same as a result of which net magnitude of displacement is 0. But the distance covered by the ball is two times distance between the player A and B. (b) Let us consider the same example as above where the ball takes t seconds of time to cover the total distance. Therefore the magnitude of the average velocity of football will be given by [y is the distance between the player A and B. It is 2 times because the football covered y distance from A to B and same distance again from B to A] Thus, the second quantity is greater than the first quantity. The above quantity are equal if the ball only moves from one player to another in a single direction (considering one-dimensional motion).
A man walks on a straight road from his home to a market 2.5 km away at a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h - Magnitude of average velocity, and
- Average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time and not as the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Distance between home and market is 2.5 km = 2500m Speed of the man while walking towards market is 5 km/hr (a) Magnitude of the average velocity is 0 because the main started from his home which is his initial position and came back home which is his final position also. So the initial and the final position is same therefore no displacement and no average velocity (b)
In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and the magnitude of average velocity. No such distinction is necessary when we consider the instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
The speed of an object at a specific moment or instant is its instantaneous velocity. Since the gap is so narrow, it is assumed that the object's direction of motion is remaining unchanged, causing the displacement and distance to become equal. Instantaneous speed and instantaneous velocity are therefore equivalent.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent the one-dimensional motion of a particle?
None of the four graphs shows a one-dimensional motion. One-dimensional motion is not represented by the graph (a). This is due to the fact that the item is in two places at a given time t, which is impractical for one-dimensional motion. One-dimensional motion is not represented by the graph (b). This is due to the particle having two opposing movements at a given time t. It is an illustration of two-dimensional motion or plane motion. Because it depicts a particle moving at a negative speed, graph (c) is incorrect. An object's speed cannot be negative. The particle's total route cannot shorten with time, hence the graph (d), which depicts the lengthening and shortening of the total path with time, cannot describe one-dimensional motion.
The figure shows the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that shows that the particle moves in a straight line for t > 0 and on a parabolic path for t >0? If not, suggest a suitable physical context for this graph.
It is not correct to say that the particle moves in a straight line for t > 0 (i.e., -ve) and on a parabolic path for t > 0 (i.e., + ve) because the x-t graph does not represent the path of the particle. A suitable physical context for the graph can be the particle is dropped from the top of a tower at t =0.
A police van moving on a highway with a speed of 30 km h
Suggest a suitable physical situation for each of the following graphs.
The velocity in the opposite direction suddenly increases in the graph (a) before dropping off. It resembles when a football player kicks a ball that is stationary in front of a vertical wall. The ball's velocity quickly rises when it is kicked. The ball bounces back in the opposite direction with a decreasing velocity after hitting the wall. The ball eventually comes to rest. The velocity in the graph (b) alternates between positive and negative signs, and each time, its magnitude falls. It is an illustration of a ball that freely falls from a height, bounces several times, and finally comes to rest. The acceleration suddenly increases over a brief period of time as seen in graph (c). A batter hitting a ball with a bat is an illustration of it.
The following figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, - 1.2 s.
Negative, Negative, Positive (at t = 0.3 s) Positive, Positive, Negative (at t = 1.2 s) Negative, Positive, Positive (at t = -1.2 s) Acceleration of a particle exhibiting Simple Harmonic Motion (SHM) is given by the realtion, a = -? For t = 0.3 s When time is 0.3 seconds then x is negative. So the slope of the x-t plot will also be negative as a result of which velocity and position both will be negative. However acceleration will be positive from equation (i). For t = 1.2 s When time is 1.2 seconds then x is negative. So the slope of the x-t plot will also be positive as a result of which velocity and position both will be positive. However acceleration will be negative from equation (i). For t = - 1.2 s When time is -1.2 seconds then x is negative. So the slope of the x-t plot will also be negative. Since x and t both are negative the velocity becomes positive. However acceleration will be positive from equation (i).
The figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
Interval 3 (Greatest), Interval 2 (Least) Positive (Intervals 1 & 2), Negative (Interval 3) The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time. The graph clearly shows that the slope is at its highest and lowest points at intervals 3 and 2, respectively. As a result, interval 3 has the highest average particle speed, whereas interval 2 has the lowest. Since the slope is positive in intervals 1 and 2, so is the average velocity in these intervals. In contrast, it is negative in interval 3 since this interval has a negative slope.
The following figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion gives the signs of v and a in the three intervals. What are the accelerations at points A, B, C and D?
The maximum rate at which the speed changes over time is in interval 2. As a result, interval 2 experiences the largest average acceleration. In interval 3, the average speed is at its highest. The velocity in intervals 1, 2, and 3 is positive. The slope affects acceleration. Because the slope is positive, the acceleration is positive at intervals 1 and 3. Due to the negative slope in interval 2, the acceleration is also negative. Since the slope is parallel to the time axis at points A, B, C, and D, there is no acceleration at these locations.
A three-wheeler starts from rest, accelerates uniformly with 1 m s
Distance covered by a body in nth-second in a straight line is, Where, a is the acceleration u is the initial velocity n stands for time interval where n = 1, 2, 3, . . ., n Thus in the above case Acceleration is 1m/s From above equation we can relate that, S Substituting different values of n in equation (ii) we get,
Thus the graph plotted will be a straight line.
A boy, standing on a stationary lift (open from above), throws a ball upwards with the maximum initial speed he can, equal to 49 m s
The initial velocity of the ball, u = 49 m/s Case: I When the lift is still, the boy throws the ball upward. The upward direction is in vertical motion so it is taken as positive. The ball has no or zero displacement at all. From the equation of motion we know that, Case: II The ball will first move at a speed of (49 m/s + 5 m/s) = 54 m/s, when the lift begins to move at a speed of 5 m/s. Thus the ball will have a displacement of s = 5t Therefore the time taken is Both situations will involve the same amount of time.
On a long, horizontally moving belt figure, a child runs to and fro with a speed 9 km h - speed of the child running in the direction of motion of the belt?
- speed of the child running opposite to the direction of motion of the belt?
- time taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?
Speed of child = 9 km/h Speed of belt = 4 km/h - When the boy runs in the direction of the belt's motion, the stationary observer will measure his speed as (9 + 4) km/hr, or 13 km/hr.
- If the boy runs against the belt's direction of motion, his speed as measured by a stationary observer is (9 - 4) km/hr, or 5 km/hr.
50 metres separate the child's parents. The child's speed in either direction as viewed by the parents will remain the same, i.e., 9 km/h = 2.5 m/s, while both parents are standing on the moving belt. As a result, the child needs 50/2. 5 = 20 seconds to travel toward one of his parents. Since the child and his parents are both standing on the same belt, the belt's motion affects them both equally. As a result, the child continues to move at the same speed of 9 km/h for both parents (regardless of the direction of mobility). This leads to the conclusion that the child still needs the same amount of time to get to any one of his parents.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s
For the first stone: Given details are, Acceleration, a = -g = - 10 m/s2 Initial velocity, u Now, we know We know that the stone was thrown during the time interval t = 0 seconds therefore it cannot be negative as a result of which t = 8 seconds is correct. For the second stone: Given details are, Acceleration, a = - g = - 10 m/s Initial velocity, u Now, we know >
The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between - t = 0 s to 10 s,
- t = 2 s to 6 s.
(a) During the first instance t = 0 s and t = 10 s, so distance traversed by the particle during this time period is, Average speed of the particle is given as 60m/10 s which is equal to 6 m/s (b) During the second instance t = 2 s and t = 6 s, so distance traversed by the Let S Let S For S From equation of motion we know that, Distance covered from 2 to 5 s, S From equation of motion we know that, For the second instant that is when t = 5 sec to 10 sec, u = 12 m/s and a = -2.4 m/s Since t = 5 sec to t = 6 sec means n = 1 for this motion Distance covered in the 6 the sec is
The velocity-time graph of a particle in one-dimensional motion is shown in the figure. - x (t
_{2}) = x (t_{1}) + v (t_{1}) (t_{2}- t_{1}) + (1/2) a(t_{2}- t_{1})2 - v(t
_{2}) = v(t_{1}) + a(t_{2}- t_{1}) - V
_{average}= [ x(t_{2}) - x (t_{1})] /(t_{2}- t_{1}) - a
_{average}= [ v(t_{2}) - v (t_{1})] /(t_{2}- t_{1}) - x (t
_{2}) = x (t_{1}) + v_{av}(t_{2}- t_{1}) + (1/2) a_{av}(t_{2}- t_{1})2 - x(t
_{2}) - x (t1) = Area under the v-t curve bounded by t-axis and the dotted lines.
The equations are given as, - x (t
_{2}) = x (t_{1}) + v (t_{1}) (t_{2}- t_{1}) + (1/2) a(t_{2}- t_{1})2 - v(t
_{2}) = v(t_{1}) + a(t_{2}- t_{1}) - V
_{average}= [ x(t_{2}) - x (t_{1})] /(t_{2}- t_{1} - a
_{average}= [ v(t_{2}) - v (t_{1})] /(t_{2}- t_{1}) - x (t
_{2}) = x (t_{1}) + v_{av}(t_{2}- t_{1}) + (1/2) a_{av}(t_{2}- t_{1})2 - x(t
_{2}) - x (t1) = Area under the v-t curve bounded by t- axis and the dotted lines.
Option (a), (b) and (e) are equations of a straight line. Also from the graph it is clear that graph between t The ratio of the displacement at two different positions to the time interval difference is known as the average velocity. Thus average velocity between the time interval t The ratio of the velocity difference between two points to the time interval difference is known as the average acceleration. The body's displacement between two time intervals is shown by the area under the velocity-time graph. This is so because displacement is the result of adding the corresponding time and velocity. Therefore the displacement between time interval t x(t Therefore, Options (c), (d) and (f) are the correct answers for this question. Next TopicClass 11 Physics Chapter 4 |