Chapter 2 Motion in a Straight Line

NCERT Solutions Class 11 Physics Chapter 2 – Motion in a Straight Line

NCERT solutions are a comprehensive guide to solving problems mathematically using various algebraic and calculus techniques and provide a critical resource for students preparing for competitive exams such as the JEE and NEET. Motions along a straight path have some important characteristics we can analyze mathematically. Kinematics is an area of study that helps us to describe the behavior of objects in motion without any consideration or reference to the cause(s) of that motion. Thus in this chapter, the student will gain a better understanding of motion based on the following (general) concepts; position - path-length - displacement - speed - and velocity.

ALLEN’s NCERT Solutions for Class 11 Physics (Motion in a Straight Line) are prepared by expert faculty to simplify kinematics with clear explanations and step-wise numerical solutions. This approach is based on conceptual clarity and application based problem solving needed for JEE and NEET.

These ideas of rectilinear motion apply to all kinds of motion, whether it be objects falling through the air because of the force of gravity, or cars decelerating down the highway because their brakes were suddenly pulled on.

1.0Class 11 Physics Chapter 2 : Key Concepts

This chapter focuses on the motion of point objects along a single axis. Key lessons include:

Position, Path Length, and Displacement: Understand the difference between position, path length, and displacement, including distance travelled and change in position.
Average Speed and Average Velocity: Learn how to calculate average speed and average velocity using distance, displacement, and time.
Instantaneous Velocity and Speed: Using the concept of limits and derivatives to find velocity at a specific moment: $v = \frac{d x}{d t}$
Acceleration: Understanding the rate of change of velocity: $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$
Kinematic Graphs:

Position-Time (x-t) Graphs: The slope represents velocity.
Velocity-Time (v-t) Graphs: The slope represents acceleration, and the area under the curve represents displacement.

Kinematic Equations for Uniformly Accelerated Motion: Mastering the three fundamental equations:

v = u + at
$s = u t + \frac{1}{2} a t^{2}$
$v^{2} = u^{2} + 2 a s$

Free Fall: Analyzing motion under the influence of gravity where a = -g (approximately - $9.8 m/s^{2}) .$
Relative Velocity: Calculating the velocity of one object with respect to another in a straight line: $v_{B A} = v_{B} - v_{A}$

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 2

SOLVED EXAMPLES

The position of an object moving along x -axis is given by $x = a + bt^{2}$ where $a = 8.5 m, b = 2.5 ms^{- 2}$ and t is measured in seconds. What is its velocity at $t = 0 s$ and $t = 2.0 s$ What is the average velocity between $t = 2.0 s$ and $t = 4.0 s$ ? Sol. In notation of differential calculus, the velocity is

$v = \frac{dx}{dt} = \frac{d}{dt} (a + bt^{2}) = 2 b t = 5.0 tm s^{- 1} .$

At $t = 0 s, v = 0 ms^{- 1}$ and

$at t = 2.0 s, v = 10 ms^{- 1} .$

Average velocity $= \frac{x _{2} - x _{1}}{t _{2} - t _{1}}$

$= \frac{x ( 4.0 ) - x ( 2.0 )}{4.0 - 2.0} = \frac{a + 16 b - a - 4 b}{2.0} = \frac{2 a - 12 b}{2} = 17 - 30 = 12 \times b = 12 \times 2.5 = 30 ms^{- 1} .$

Uniform motion, velocity is the same as the average velocity at all instants.

Obtain equations of motion for constant acceleration using method of calculus. Sol. By definition, $a = \frac{dv}{dt}$

$d v = a d t$

Integrating both sides

$\int_{v_{0}}^{v} dv = \int_{0}^{t} adt = a \int_{0}^{t} dt (a is constant)$

$v - v_{0} = at$

$v = v_{0} + at$

Further,

$v = \frac{dx}{dt} dx = v dt$

Integrating both sides

$\int_{x_{0}}^{x} d x = \int_{0}^{t} v d t = \int_{0}^{t} (v_{0} + a t) d t x - x_{0} = v_{0} t + \frac{1}{2} a t^{2} x = x_{0} + v_{0} t + \frac{1}{2} a t^{2} \dots . (ii)$

We can write

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} ∵ \frac{d x}{d t} = v$

Integrating both sides,

$\int_{v_{0}}^{v} v d v = \int_{x_{0}}^{x} a d x \Rightarrow [\frac{v ^{2}}{2}]_{v_{0}}^{v} = a [x]_{x_{0}}^{x} \frac{v ^{2} - v _{0}^{2}}{2} = a (x - x_{0}) v^{2} = v_{0}^{2} + 2 a (x - x_{0}) \dots . (iii)$

The advantage of this method is that it can be used for motion with non-uniform acceleration also.

A ball is thrown vertically upwards with a velocity of $20 m s^{- 1}$ from the top of a multi storey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take $g = 10 m s^{- 2}$ . Sol. (a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Figure.

Now $v_{0} = + 20 m s^{- 1}$

$a = - g = - 10 m s^{- 2}, v = 0 m s^{- 1}$

If the ball rises to height y from the point of launch, then using the equation

$v^{2} = v_{0}^{2} + 2 a (y - y_{0})$

we get

$0 = (20)^{2} + 2 (- 10) (y - y_{0})$

Solving, we get, $(y - y_{0}) = 20 m$ .

(b) We can solve this part of the problem in two ways. Note carefully the methods used.

First Method : In the first method, we split the path in two parts : the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken $t_{1}$ and $t_{2}$ . Since the velocity at B is zero, we have :

$v = v_{0} + at 0 = 20 - 10 t_{1} t_{1} = 2 s$

This is the time in going from A to B. From B, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation.

$y = y_{0} + v_{0} t + \frac{1}{2} at^{2}$

We have, $y_{0} = 45 m, y = 0, v_{0} = 0$ ,

$a = - g = - 10 ms^{- 2} 0 = 45 + \frac{1}{2} (- 10) t_{2}^{2} \Rightarrow 5 t_{2}^{2} = 45$

Solving, we get $t_{2} = 3 s$ Therefore, the total time taken by the ball before if hits the ground

$= t_{1} + t_{2} = 2 s + 3 s = 5 s .$

Second Method : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation

$y = y_{0} + v_{0} t + \frac{1}{2} at^{2}$

Now $y_{0} = 25 m, y = 0 m$

$v_{0} = 20 ms^{- 1}, a = - 10 m s^{- 2}, t = ? 0 = 25 + 20 t + (\frac{1}{2}) (- 10) t^{2}$

Or, $5 t^{2} - 20 t - 25 = 0$ Solving this quadratic equation for t, we get $t = 5 s$ we do not have to worry about the path of the motion as the motion is under constant acceleration.

Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Sol. An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth's radius, g can be taken to be constant. equal to $9.8 m s^{- 2}$ . Free fall is thus a case of motion with uniform acceleration.

We assume that the motion is in y-direction. more correctly in -y-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward. it is in the negative direction and we have $a = - g = - 9.8 ms^{- 2}$ The object is released from rest at $y = 0$ Therefore, $v_{0} = 0$ and the equations of motion become:

$v = 0 - gt = - 9.8 t ms^{- 1} y = 0 - \frac{1}{2} gt^{2} = - 4.9 t^{2} m v^{2} = 0 - 2 gy = - 19.6 y m^{2} s^{- 2}$

These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration, velocity, and distance, with time have been plotted in Figure.

Motion of an object under free fall.

(a) variation of acceleration with time (b) Variation of velocity with time (c) Variation of distance with time

Galileo's law of odd numbers: The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely. 1 : 3 : 5 : 7......]." Prove it.

Sol. Let us divide the time Interval of motion of an object under free fall into many equal intervals $τ$ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have

$y = - \frac{1}{2} gt^{2}$

Using this equation, we can calculate the position of the object after different time Intervals, $0, τ .2 τ .3 τ$ . First time interval, $y_{1} - y_{0} = - \frac{1}{2} g τ^{2}$ Second time interval, $y_{2} - y_{1} = - \frac{3}{2} g τ^{2}$ Third time interval, $y_{3} - y_{2} = - \frac{5}{2} g τ^{2}$

We find that the distances are In the simple ratio 1 : 3 : 5 : 7 : 9 : 11... as shown In the last column.

This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.

Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_{0})$ and the braking capacity, or deceleration, -a that is caused by the braking. Derive an expression for stopping distance of a vehicle In terms of $v_{0}$ and a. Sol. Let the distance travelled by the vehicle before it stops be $d_{s}$ . Then, using equation of motion $v^{2} = v_{0}^{2} + 2 ax$ and noting that $v = 0$ we have the stopping distance

$d_{s} = \frac{- v _{0}^{2}}{2 a}$

Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).

Stopping distance is an important factor considered in setting speed limits, for example, in school zones.

Reaction time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Figure). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

Sol. The ruler drops under free fall. Therefore, $v_{0} = 0$ and $a = - g = - 9.8 ms^{- 2}$ . The distance travelled d and the reaction time $t_{r}$ are related by

$d = \frac{1}{2} gt_{r}^{2}$

or, $t_{r} = \frac{2 d}{g} s$ Given $d = 21.0 cm$ and $g = 9.8 ms^{- 2}$ the reaction time is

$t_{r} = \frac{2 \times 0.21}{9.8} s ≅ 0.2 s .$

EXERCISE QUESTIONS WITH SOLUTIONS

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. Sol. (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 small as compared to the size of a 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.
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).

Sol. (a) As OP < OQ, A lives closer to the school than B.

(b) For $x = 0, t = 0$ for A; while t has some finite value for B. Therefore, A starts from the school earlier than B. (c) Since the velocity is equal to slope of x-t graph in case of uniform motion and slope of x-t graph for B is greater than that for A, hence B walks faster than 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.

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.

Sol. Speed of the woman $= 5 km / h$ Distance between her office and home $= 2.5 km$ Time taken $= \frac{Distance}{Speed} = \frac{2.5}{5} = 0.5 h = 30 min$ It is given that she covers the same distance in the evening by an auto.

Now, speed of the auto 25 km/h Time taken $= \frac{Distance}{Speed}$ $= \frac{2.5}{25} = \frac{1}{10} = 0.1 h = 6 min$ The suitable $x - t$ graph of the motion of the woman is shown in the given figure.

ques-3-exer-ques-chap-2-class-11-physics

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.

Sol. 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 6 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 falls into the pit.

Net time taken by the drunkard to cover

$13 m = 32 + 5 = 37 s$

The $x - t$ graph of the drunkard's motion can be shown as:

5. 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? Sol. Initial velocity of the car,

$u = 126 \frac{km}{h} = 35 \frac{m}{s}$

Final velocity of the car, $v = 0$ Distance covered by the car before coming to rest, $s = 200 m$ Retardation produced in the car $= a$ From third equation of motion, 'a' can be calculated as:

$v^{2} - u^{2} = 2 a s 0 - (35)^{2} = 2 \times a \times 200 \Rightarrow a = \frac{35 \times 35}{2 \times 200} = 3.06 m / s^{2}$

From first equation of motion, time (t) taken by the car to stop can be obtained as:

$v = u + at t = \frac{( v - u )}{a} = \frac{( - 35 )}{( - 3.06 )} = 11.44 s$

A player throws a ball upwards with an initial speed of $29.4 ms^{- 1}$ . (a) What is the direction of acceleration during the upward motion of the ball? (b) What are the velocity and acceleration of the ball at the highest point of its motion? (c) Choose the $x = 0 m$ and $t = 0$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player's hands? (Take $g = 9.8 ms^{- 1}$ and neglect air resistance). Sol. (a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth. (b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., $9.8 m / s^{2}$ , (c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive. (d) Initial velocity of the ball, $u = 29.4 m / s$ Final velocity of the ball, $v = 0$ (At maximum height, the velocity of the ball becomes zero) Acceleration $a = - g = - 9.8 m / s^{2}$

From third equation of motion, height (s) can be calculated as:

$v^{2} - u^{2} = 2 g s s = \frac{( v ^{2} - u ^{2} )}{2 g} = \frac{( 0 ) ^{2} - ( 29.4 ) ^{2}}{2 \times ( - 9.8 )} = 3 s$

Time of ascent = Time of descent Hence, the total time taken by the ball to return to the player's hands $= 3 + 3 = 6 s$ .

Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

(a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration particle must be speeding up.

Sol. (a) True, when an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.

(b) False, Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero. (c) True, A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero. (d) This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards. This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.

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 $t = 0$ to12 s.

Sol. Ball is dropped from a height, $s = 90 m$ Initial velocity of the ball, $u = 0$ Acceleration, $a = g = 9.8 m / s^{2}$ Final velocity of the ball = v From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:

$s = ut + \frac{1}{2} at^{2} 90 = 0 + \frac{1}{2} \times 9.8 t^{2} t = 18.38 = 4.29 s$

From first equation of motion, final velocity is given as:

$v = u + at = 0 + 9.8 \times 4.29 = 42.04 m / s$

Rebound velocity of the ball, $u_{r} = \frac{9 v}{10}$

$= \frac{9 \times 42.04}{10} = 37.84 m / s$

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

$v = u_{r} + at^{'} 0 = 37.84 + (- 9.8) t^{'} t^{'} = \frac{- 37.84}{- 9.8} = 3.86 s$

Total time taken by the ball $= t + t^{'}$

$= 4.29 + 3.86 = 8.15 s$

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time. The velocity with which the ball rebounds from the floor $= \frac{9 \times 37.84}{10} = 34.05 m / s$ Total time taken by the ball for second rebound $= 8.15 + 3.86 = 12.01 s$

The speed-time graph of the ball is represented in the given figure as:

ques-8-pysics-exercise-ques-chap-2-class-11

Explain clearly, with examples, the distinction between: (a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval. (b) 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]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider onedimensional motion only]. Sol. (a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.

chap-2-exercise-ques-class-11-physics-ques-9

Whereas, total path length $= AB + BC$ It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other. (b) Average velocity $= \frac{Displacement}{Time int erval}$ For the given particle.

$Average velocity = \frac{A C}{t}$

Average speed $= \frac{Total path length}{Time int erval}$

$= \frac{( AB + BC )}{t}$

Since $(AB + BC) > AC$ , average speed is greater than the magnitude of average velocity.

The two quantities will be equal if the particle continues to move along a straight line.

A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 kmh^{- 1}$ . Finding the market closed, he instantly turns and walks back home with a speed of $7.5 km h^{- 1}$ . What is the (a) magnitude of average velocity, and (b) 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 magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]

Sol. Time taken by the man to reach the market from home, $t_{1} = \frac{2.5}{5} = \frac{1}{2} h = 30 min$ Time taken by the man to reach home from the market, $t_{2} = \frac{2.5}{7.5} = \frac{1}{3} h = 20 min$ Total time taken in the whole journey $30 + 20 = 50 min$ (i) 0 to 30 min

$Average velocity = \frac{Displacement}{Time} = \frac{2.5}{1/2} = 5 kmh^{- 1} Average speed = \frac{Distan ce}{Time} = \frac{2.5}{1/2} = 5 kmh^{- 1}$

(ii) 0 to 50 min

$Time = 50 min = \frac{50}{60} = \frac{5}{6} h Net displacement = 0 Total distance = 2.5 + 2.5 = 5 km Average velocity = \frac{Displacement}{Time} = 0 Average speed = \frac{Distan ce}{Time} = \frac{5}{5/6} = 6 km / h$

(iii) 0 to 40 min Speed of the man $= 7.5 km / h$ Distance travelled in first 30min

$= 2.5 km$

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

$10 min = \frac{7.5 \times 10}{60} = 1.25 km$

Net displacement $= 2.5 - 1.25$

$= 1.25 km$

Total distance travelled $= 2.5 + 1.25$

$= 3.75 km$

$Average velocity = \frac{Displacement}{Time} = \frac{1.25}{( 40/60 )} = 1.875 km / h Average speed = \frac{Distan ce}{Time} = \frac{3.75}{( 40/60 )} = 5.625 km / h$

In Exercises 9 and 10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? Sol. Instantaneous velocity is given by the first derivative of distance with respect to time i.e. Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time. Therefore, instantaneous speed is always equal to instantaneous velocity.
Look at the graphs (a) to (d) (Figure) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

exercise-ques-12-physics-class-11-chap-2

Sol. (a) The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time. (b) The given v-t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time. (c) The given v-t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative. (d) The given v-t graph. shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.

Figure shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph 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.

ques-13-physics-class-11-exercise-ques-chap-2

Sol. No, because the x-t graph does not represent the trajectory of the path followed by a particle. From the graph, it is noted that at $t = 0, x = 0$

A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 kmh^{- 1}$ . If the muzzle speed of the bullet is $150 ms^{- 1}$ , with what speed does the bullet hit the thief's car? (Note: Obtain that speed which is relevant for damaging the thief's car). Sol. Speed of the police van,

$v_{p} = 30 kmh^{- 1} = 8.33 m / s$

Muzzle speed of the bullet. $v_{b} = 150 m / s$ Speed of the thief's car,

$v_{t} = 192 km / h = 53.33 m / s$

Since the bullet is fired from a moving van, its resultant speed can be obtained as:

$= 150 + 8.33 = 158.33 m / s$

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief's car can be obtained as:

$v_{b t} = v_{b} - v_{t} = 158.33 - 53.33 = 105 m / s$

Suggest a suitable physical situation for each of the following graphs (Figure)

physics-class-11-chap-2-exercise-ques-15

Sol. (a) A ball at rest on a smooth floor is kicked, it rebounds from a wall with reduced speed and moves to the opposite wall, where it stops. (b) A ball thrown up with same initial velocity, rebounds from floor with reduced speed after each hit. (c) A uniformly moving cricket ball turned back by hitting it with a bat for a very short time interval.

16. 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 $t = 0.3 s, 1.2 s, - 1.2 s$ .

chap-2-class-11-physics-exercise-ques-16

Sol. Negative, Negative, Positive (at $t = 0.3 s$ ) Positive, Positive, Negative (at $t = 1.2 s$ ) Negative, Positive, Positive (at $t = - 1.2 s$ ) For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation: $a = - ω^{2} x$ $ω =$ angular frequency $t = 0.3 s$ , In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive. $t = 1.2 s$ , In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive.

Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.

$t = - 1.2 s,$

In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

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.

exercise-ques-17-chap-2-class-11-physics

Sol. 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. It is clear from the graph that the slope is maximum and minimum respectively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.

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, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

physics-class-11-chap-2-exercise ques-18

Sol. Average acceleration is greatest in interval 2. Average speed is greatest in interval 3. v is positive in intervals 1, 2, and 3. a is positive in intervals 1 and 3 and negative in interval 2. $a = 0$ at A, B, C, D Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.

Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.

In interval 1: The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2: The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3: The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval. Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points $A, B, C$ , and D , acceleration of the particle is zero.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Access NCERT Solutions for Class 11 Physics chapter-wise with simple and detailed explanation for every question. Learn textbook concepts, solve questions step-wise and build your Physics fundamentals chapter-wise.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 2

Calculus-Based Understanding of Motion: Understand the derivation of equations of motion using differentiation and integration. Learn how velocity and acceleration are related to the rate of change of displacement and velocity.
Clear Graphical Interpretation: Learn to interpret displacement-time and velocity-time graphs with step-by-step explanations of slopes, areas under curves, displacement, velocity, and acceleration.
Correct Sign Convention in Kinematics: Understand how to select positive and negative directions for upward and downward motion. Clear examples help avoid sign errors in free-fall and other kinematics problems.
Step-by-Step Numerical Solutions: Get detailed solutions to NCERT numerical problems based on motion in a straight line, relative motion, free fall, pursuit problems, and parachutist motion, with formulas, calculations, units, and required steps.
Prepared by ALLEN Subject Experts: These NCERT Solutions for Class 11 Physics Chapter 2 are prepared by ALLEN subject experts, with accurate calculations, clear explanations, and alignment with the latest NCERT syllabus.
Simple and Concept-Based Explanations: Important concepts from Motion in a Straight Line are explained in simple language using formulas, graphs, examples, and logical steps for easy understanding.
Complete NCERT Question Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 2 examples and exercise questions, covering key concepts of kinematics for CBSE, JEE, and NEET preparation.

NCERT Solutions Class 11 Physics Chapter 2 – Motion in a Straight Line

1.0Class 11 Physics Chapter 2 : Key Concepts

This chapter focuses on the motion of point objects along a single axis. Key lessons include:

Position, Path Length, and Displacement: Understand the difference between position, path length, and displacement, including distance travelled and change in position.
Average Speed and Average Velocity: Learn how to calculate average speed and average velocity using distance, displacement, and time.
Instantaneous Velocity and Speed: Using the concept of limits and derivatives to find velocity at a specific moment: $v = \frac{d x}{d t}$
Acceleration: Understanding the rate of change of velocity: $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$
Kinematic Graphs:

Position-Time (x-t) Graphs: The slope represents velocity.
Velocity-Time (v-t) Graphs: The slope represents acceleration, and the area under the curve represents displacement.

Kinematic Equations for Uniformly Accelerated Motion: Mastering the three fundamental equations:

v = u + at
$s = u t + \frac{1}{2} a t^{2}$
$v^{2} = u^{2} + 2 a s$

Free Fall: Analyzing motion under the influence of gravity where a = -g (approximately - $9.8 m/s^{2}) .$
Relative Velocity: Calculating the velocity of one object with respect to another in a straight line: $v_{B A} = v_{B} - v_{A}$

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 2

SOLVED EXAMPLES

The position of an object moving along x -axis is given by $x = a + bt^{2}$ where $a = 8.5 m, b = 2.5 ms^{- 2}$ and t is measured in seconds. What is its velocity at $t = 0 s$ and $t = 2.0 s$ What is the average velocity between $t = 2.0 s$ and $t = 4.0 s$ ? Sol. In notation of differential calculus, the velocity is

$v = \frac{dx}{dt} = \frac{d}{dt} (a + bt^{2}) = 2 b t = 5.0 tm s^{- 1} .$

At $t = 0 s, v = 0 ms^{- 1}$ and

$at t = 2.0 s, v = 10 ms^{- 1} .$

Average velocity $= \frac{x _{2} - x _{1}}{t _{2} - t _{1}}$

$= \frac{x ( 4.0 ) - x ( 2.0 )}{4.0 - 2.0} = \frac{a + 16 b - a - 4 b}{2.0} = \frac{2 a - 12 b}{2} = 17 - 30 = 12 \times b = 12 \times 2.5 = 30 ms^{- 1} .$

Uniform motion, velocity is the same as the average velocity at all instants.

Obtain equations of motion for constant acceleration using method of calculus. Sol. By definition, $a = \frac{dv}{dt}$

$d v = a d t$

Integrating both sides

$\int_{v_{0}}^{v} dv = \int_{0}^{t} adt = a \int_{0}^{t} dt (a is constant)$

$v - v_{0} = at$

$v = v_{0} + at$

Further,

$v = \frac{dx}{dt} dx = v dt$

Integrating both sides

$\int_{x_{0}}^{x} d x = \int_{0}^{t} v d t = \int_{0}^{t} (v_{0} + a t) d t x - x_{0} = v_{0} t + \frac{1}{2} a t^{2} x = x_{0} + v_{0} t + \frac{1}{2} a t^{2} \dots . (ii)$

We can write

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} ∵ \frac{d x}{d t} = v$

Integrating both sides,

The advantage of this method is that it can be used for motion with non-uniform acceleration also.

A ball is thrown vertically upwards with a velocity of $20 m s^{- 1}$ from the top of a multi storey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take $g = 10 m s^{- 2}$ . Sol. (a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Figure.

Now $v_{0} = + 20 m s^{- 1}$

$a = - g = - 10 m s^{- 2}, v = 0 m s^{- 1}$

If the ball rises to height y from the point of launch, then using the equation

$v^{2} = v_{0}^{2} + 2 a (y - y_{0})$

we get

$0 = (20)^{2} + 2 (- 10) (y - y_{0})$

Solving, we get, $(y - y_{0}) = 20 m$ .

(b) We can solve this part of the problem in two ways. Note carefully the methods used.

$v = v_{0} + at 0 = 20 - 10 t_{1} t_{1} = 2 s$

$y = y_{0} + v_{0} t + \frac{1}{2} at^{2}$

We have, $y_{0} = 45 m, y = 0, v_{0} = 0$ ,

$a = - g = - 10 ms^{- 2} 0 = 45 + \frac{1}{2} (- 10) t_{2}^{2} \Rightarrow 5 t_{2}^{2} = 45$

Solving, we get $t_{2} = 3 s$ Therefore, the total time taken by the ball before if hits the ground

$= t_{1} + t_{2} = 2 s + 3 s = 5 s .$

Second Method : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation

$y = y_{0} + v_{0} t + \frac{1}{2} at^{2}$

Now $y_{0} = 25 m, y = 0 m$

$v_{0} = 20 ms^{- 1}, a = - 10 m s^{- 2}, t = ? 0 = 25 + 20 t + (\frac{1}{2}) (- 10) t^{2}$

Or, $5 t^{2} - 20 t - 25 = 0$ Solving this quadratic equation for t, we get $t = 5 s$ we do not have to worry about the path of the motion as the motion is under constant acceleration.

Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Sol. An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth's radius, g can be taken to be constant. equal to $9.8 m s^{- 2}$ . Free fall is thus a case of motion with uniform acceleration.

$v = 0 - gt = - 9.8 t ms^{- 1} y = 0 - \frac{1}{2} gt^{2} = - 4.9 t^{2} m v^{2} = 0 - 2 gy = - 19.6 y m^{2} s^{- 2}$

Motion of an object under free fall.

(a) variation of acceleration with time (b) Variation of velocity with time (c) Variation of distance with time

Galileo's law of odd numbers: The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely. 1 : 3 : 5 : 7......]." Prove it.

$y = - \frac{1}{2} gt^{2}$

We find that the distances are In the simple ratio 1 : 3 : 5 : 7 : 9 : 11... as shown In the last column.

This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall.

Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity $(v_{0})$ and the braking capacity, or deceleration, -a that is caused by the braking. Derive an expression for stopping distance of a vehicle In terms of $v_{0}$ and a. Sol. Let the distance travelled by the vehicle before it stops be $d_{s}$ . Then, using equation of motion $v^{2} = v_{0}^{2} + 2 ax$ and noting that $v = 0$ we have the stopping distance

$d_{s} = \frac{- v _{0}^{2}}{2 a}$

Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).

Stopping distance is an important factor considered in setting speed limits, for example, in school zones.

Reaction time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual. You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Figure). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.

Sol. The ruler drops under free fall. Therefore, $v_{0} = 0$ and $a = - g = - 9.8 ms^{- 2}$ . The distance travelled d and the reaction time $t_{r}$ are related by

$d = \frac{1}{2} gt_{r}^{2}$

or, $t_{r} = \frac{2 d}{g} s$ Given $d = 21.0 cm$ and $g = 9.8 ms^{- 2}$ the reaction time is

$t_{r} = \frac{2 \times 0.21}{9.8} s ≅ 0.2 s .$

EXERCISE QUESTIONS WITH SOLUTIONS

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. Sol. (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 small as compared to the size of a 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.
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).

Sol. (a) As OP < OQ, A lives closer to the school than B.

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.

Net time taken by the drunkard to cover

$13 m = 32 + 5 = 37 s$

The $x - t$ graph of the drunkard's motion can be shown as:

$u = 126 \frac{km}{h} = 35 \frac{m}{s}$

Final velocity of the car, $v = 0$ Distance covered by the car before coming to rest, $s = 200 m$ Retardation produced in the car $= a$ From third equation of motion, 'a' can be calculated as:

$v^{2} - u^{2} = 2 a s 0 - (35)^{2} = 2 \times a \times 200 \Rightarrow a = \frac{35 \times 35}{2 \times 200} = 3.06 m / s^{2}$

From first equation of motion, time (t) taken by the car to stop can be obtained as:

$v = u + at t = \frac{( v - u )}{a} = \frac{( - 35 )}{( - 3.06 )} = 11.44 s$

A player throws a ball upwards with an initial speed of $29.4 ms^{- 1}$ . (a) What is the direction of acceleration during the upward motion of the ball? (b) What are the velocity and acceleration of the ball at the highest point of its motion? (c) Choose the $x = 0 m$ and $t = 0$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player's hands? (Take $g = 9.8 ms^{- 1}$ and neglect air resistance). Sol. (a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth. (b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., $9.8 m / s^{2}$ , (c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive. (d) Initial velocity of the ball, $u = 29.4 m / s$ Final velocity of the ball, $v = 0$ (At maximum height, the velocity of the ball becomes zero) Acceleration $a = - g = - 9.8 m / s^{2}$

From third equation of motion, height (s) can be calculated as:

$v^{2} - u^{2} = 2 g s s = \frac{( v ^{2} - u ^{2} )}{2 g} = \frac{( 0 ) ^{2} - ( 29.4 ) ^{2}}{2 \times ( - 9.8 )} = 3 s$

Time of ascent = Time of descent Hence, the total time taken by the ball to return to the player's hands $= 3 + 3 = 6 s$ .

Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

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 $t = 0$ to12 s.

$s = ut + \frac{1}{2} at^{2} 90 = 0 + \frac{1}{2} \times 9.8 t^{2} t = 18.38 = 4.29 s$

From first equation of motion, final velocity is given as:

$v = u + at = 0 + 9.8 \times 4.29 = 42.04 m / s$

Rebound velocity of the ball, $u_{r} = \frac{9 v}{10}$

$= \frac{9 \times 42.04}{10} = 37.84 m / s$

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:

$v = u_{r} + at^{'} 0 = 37.84 + (- 9.8) t^{'} t^{'} = \frac{- 37.84}{- 9.8} = 3.86 s$

Total time taken by the ball $= t + t^{'}$

$= 4.29 + 3.86 = 8.15 s$

The speed-time graph of the ball is represented in the given figure as:

Explain clearly, with examples, the distinction between: (a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval. (b) 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]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider onedimensional motion only]. Sol. (a) The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle. The total path length of a particle is the actual path length covered by the particle in a given interval of time. For example, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.

$Average velocity = \frac{A C}{t}$

Average speed $= \frac{Total path length}{Time int erval}$

$= \frac{( AB + BC )}{t}$

Since $(AB + BC) > AC$ , average speed is greater than the magnitude of average velocity.

The two quantities will be equal if the particle continues to move along a straight line.

A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 kmh^{- 1}$ . Finding the market closed, he instantly turns and walks back home with a speed of $7.5 km h^{- 1}$ . What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time

$Average velocity = \frac{Displacement}{Time} = \frac{2.5}{1/2} = 5 kmh^{- 1} Average speed = \frac{Distan ce}{Time} = \frac{2.5}{1/2} = 5 kmh^{- 1}$

(ii) 0 to 50 min

(iii) 0 to 40 min Speed of the man $= 7.5 km / h$ Distance travelled in first 30min

$= 2.5 km$

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

$10 min = \frac{7.5 \times 10}{60} = 1.25 km$

Net displacement $= 2.5 - 1.25$

$= 1.25 km$

Total distance travelled $= 2.5 + 1.25$

$= 3.75 km$

$Average velocity = \frac{Displacement}{Time} = \frac{1.25}{( 40/60 )} = 1.875 km / h Average speed = \frac{Distan ce}{Time} = \frac{3.75}{( 40/60 )} = 5.625 km / h$

In Exercises 9 and 10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? Sol. Instantaneous velocity is given by the first derivative of distance with respect to time i.e. Here, the time interval is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time. Therefore, instantaneous speed is always equal to instantaneous velocity.
Look at the graphs (a) to (d) (Figure) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Figure shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph 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.

Sol. No, because the x-t graph does not represent the trajectory of the path followed by a particle. From the graph, it is noted that at $t = 0, x = 0$

A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 kmh^{- 1}$ . If the muzzle speed of the bullet is $150 ms^{- 1}$ , with what speed does the bullet hit the thief's car? (Note: Obtain that speed which is relevant for damaging the thief's car). Sol. Speed of the police van,

$v_{p} = 30 kmh^{- 1} = 8.33 m / s$

Muzzle speed of the bullet. $v_{b} = 150 m / s$ Speed of the thief's car,

$v_{t} = 192 km / h = 53.33 m / s$

Since the bullet is fired from a moving van, its resultant speed can be obtained as:

$= 150 + 8.33 = 158.33 m / s$

Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief's car can be obtained as:

$v_{b t} = v_{b} - v_{t} = 158.33 - 53.33 = 105 m / s$

Suggest a suitable physical situation for each of the following graphs (Figure)

Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.

$t = - 1.2 s,$

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.

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, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.

In interval 1: The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2: The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Units and Measurements
Chapter 3	Motion in a Plane
Chapter 4	Laws of Motion
Chapter 5	Work, Energy and Power
Chapter 6	System of Particles and Rotational Motion
Chapter 7	Gravitation
Chapter 8	Mechanical Properties of Solids
Chapter 9	Mechanical Properties of Fluids
Chapter 10	Thermal Properties of Matter
Chapter 11	Thermodynamics
Chapter 12	Kinetic Theory
Chapter 13	Oscillations
Chapter 14	Waves

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 2

Calculus-Based Understanding of Motion: Understand the derivation of equations of motion using differentiation and integration. Learn how velocity and acceleration are related to the rate of change of displacement and velocity.
Clear Graphical Interpretation: Learn to interpret displacement-time and velocity-time graphs with step-by-step explanations of slopes, areas under curves, displacement, velocity, and acceleration.
Correct Sign Convention in Kinematics: Understand how to select positive and negative directions for upward and downward motion. Clear examples help avoid sign errors in free-fall and other kinematics problems.
Step-by-Step Numerical Solutions: Get detailed solutions to NCERT numerical problems based on motion in a straight line, relative motion, free fall, pursuit problems, and parachutist motion, with formulas, calculations, units, and required steps.
Prepared by ALLEN Subject Experts: These NCERT Solutions for Class 11 Physics Chapter 2 are prepared by ALLEN subject experts, with accurate calculations, clear explanations, and alignment with the latest NCERT syllabus.
Simple and Concept-Based Explanations: Important concepts from Motion in a Straight Line are explained in simple language using formulas, graphs, examples, and logical steps for easy understanding.
Complete NCERT Question Coverage: Get detailed solutions to NCERT Class 11 Physics Chapter 2 examples and exercise questions, covering key concepts of kinematics for CBSE, JEE, and NEET preparation.

Frequently Asked Questions

Why are NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line important for students?

How do Class 11 Physics NCERT Solutions for Chapter 2 help in board, JEE, and NEET preparation?

What concepts are covered in NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line?

Are NCERT Solutions for Class 11 Physics Chapter 2 by ALLEN useful for competitive exams?

What is the significance of studying motion in a straight line in Physics?

What is difference between distance and displacement class 11 physics?

How are velocity-time graphs used in Motion in a Straight Line?

Join ALLEN!

NCERT Solutions Class 11 Physics Chapter 2 – Motion in a Straight Line

1.0Class 11 Physics Chapter 2 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 2

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 2

Table of Contents

Frequently Asked Questions

Why are NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line important for students?

How do Class 11 Physics NCERT Solutions for Chapter 2 help in board, JEE, and NEET preparation?

What concepts are covered in NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line?

Are NCERT Solutions for Class 11 Physics Chapter 2 by ALLEN useful for competitive exams?

What is the significance of studying motion in a straight line in Physics?

What is difference between distance and displacement class 11 physics?

How are velocity-time graphs used in Motion in a Straight Line?

Join ALLEN!

NCERT Solutions Class 11 Physics Chapter 2 – Motion in a Straight Line

1.0Class 11 Physics Chapter 2 : Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Physics Chapter 2

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0NCERT Class 11 Physics – Chapter-Wise Solutions & Explanations

4.0Key Features of NCERT Solutions for Class 11 Physics Chapter 2

Table of Contents