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Motion in a Straight Line NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 2 Motion in a Straight Line PDF Download (2026) + Important Questions Welcome to one of the most foundational chapters in your Class 11 Physics journey! If you want to understand how the universe moves, you have to start with the basics: Motion in a Straight Line . Whether you are watching a car speed down a highway or dropping a ball from a rooftop, kinematics (the study of motion) is everywhere. This chapter is incredibly important because the concepts of velocity, acceleration, and graphs build the base for almost everything you will study in Mechanics. For students targeting the 2026 Board Exams, JEE, or NEET, mastering this chapter is absolutely non-negotiable! In this comprehensive guide, we will break down the chapter into simple, easy-to-digest concepts, provide step-by-step NCERT solutions, and equip you with high-probability board questions. Let's get moving!
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ChapterMotion in a Straight Line
SubjectPhysics
Class11 Physics Chapter 2 is all about: Feature Details
BoardCBSE
DifficultyModerate
Exam WeightageHighly weighted (Combined with Motion in a Plane, usually 7-10 marks)
Learning Objectives
- Differentiate between distance and displacement, and speed and velocity.
- Understand average, relative, and instantaneous velocity.
- Derive the kinematic equations of motion using graphical and calculus methods.
- Analyze position-time (x-t) and velocity-time (v-t) graphs accurately.
- Solve complex numerical problems related to objects in free fall.
Key Concepts, Definitions and Formulas
- Point Object: An object whose size is negligible compared to the distance it travels
- Displacement: The shortest straight-line distance between the initial and final positions. It is a vector quantity (can be positive, negative, or zero)
- Instantaneous Velocity: The velocity of an object at a specific instant of time. Mathematically, it is the derivative of position with respect to time: $v = \frac{dx}{dt}$
- Acceleration: The rate of change of velocity. If a car goes from 0 to 60 km/h, the rate at which its speed increases is acceleration. $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$
- Kinematic Equations (for uniformly accelerated motion): $v = u + at$ $s = ut + \frac{1}{2}at^2$ $v^2 = u^2 + 2as$ (Where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time)
- Free Fall: When an object falls solely under the influence of Earth's gravity, its acceleration is constant ($g = 9.8 \text{ m/s}^2$ downward)
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Full NCERT Solutions
NCERT Q2.1: 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.
Answer: (a) Yes. The size of the carriage is very small compared to the distance between the two stations. (b) Yes. The size of the monkey is extremely small compared to the circumference of the circular track. (c) No. The ball's size is not negligible compared to the distance it covers while turning sharply. Its spin involves the motion of its different parts. (d) No. The size of the beaker is not negligible compared to the height of the table it falls from.
NCERT Q2.2: 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 given in the textbook. Choose the correct entries: (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.
Answer: Let's analyze the standard NCERT graph conceptually: (a) A lives closer to the school than B. (The position of A's home, P, is at a lower value on the y-axis than B's home, Q). (b) A starts from the school earlier than B. (A starts at t = 0 on the x-axis, whereas B starts at a later time t > 0). (c) B walks faster than A. (The slope of the x-t graph represents velocity. B's line is steeper than A's line, meaning B has a higher velocity). (d) A and B reach home at the same time. (Both lines end at the same time value on the t-axis).
NCERT Q2.3: A woman starts from her home at 9.00 am, walks with 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 an auto with a speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion.
Answer: Step-by-step explanation: 1. Home to Office: Distance = 2.5 km Speed = 5 km/h Time taken = Distance / Speed = 2.5 / 5 = 0.5 hours = 30 minutes. She reaches the office at 9:30 am. 2. At Office: She stays at the office (x = 2.5 km) from 9:30 am to 5:00 pm. During this time, the graph will be a straight horizontal line parallel to the time axis. 3. Office to Home: Distance = 2.5 km Speed = 25 km/h Time taken = 2.5 / 25 = 0.1 hours = 6 minutes. She reaches home at 5:06 pm. (To plot the graph: X-axis represents time from 9:00 am to 5:10 pm. Y-axis represents distance from 0 to 3 km. Draw a diagonal line up to 9:30, a horizontal line till 5:00, and a steep diagonal line back to zero at 5:06).
NCERT Q2.4: A jet airplane 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?
Answer: Let the speed of the jet plane with respect to the ground be $v_j = 500 \text{ km/h}$. Let the speed of the combustion products with respect to the jet plane be $v_{cg} = -1500 \text{ km/h}$ (negative because the gas ejects in the opposite direction of the plane's motion). We know the relative velocity formula: $v_{cg} = v_c - v_j$ Where $v_c$ is the speed of combustion products with respect to the ground. $-1500 = v_c - 500$ $v_c = -1500 + 500$ $v_c = -1000 \text{ km/h}$ The speed of the combustion products relative to an observer on the ground is 1000 km/h in the direction opposite to the jet's motion.
NCERT Q2.5: A car moving along a straight highway with a speed of 126 km/h 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?
Answer: Step 1: Write given values and convert units. Initial velocity, $u = 126 \text{ km/h} = 126 \times \frac{5}{18} \text{ m/s} = 35 \text{ m/s}$. Final velocity, $v = 0 \text{ m/s}$ (brought to a stop). Distance, $s = 200 \text{ m}$. Step 2: Find acceleration (retardation). Using the 3rd equation of motion: $v^2 = u^2 + 2as$ $0^2 = (35)^2 + 2 \times a \times 200$ $0 = 1225 + 400a$ $400a = -1225$ $a = \frac{-1225}{400} = -3.06 \text{ m/s}^2$ The retardation is $3.06 \text{ m/s}^2$ . Step 3: Find time taken to stop. Using the 1st equation of motion: $v = u + at$ $0 = 35 + (-3.06)t$ $3.06t = 35$ $t = \frac{35}{3.06} \approx 11.4 \text{ s}$ The car takes 11.4 seconds to stop.
Extra Important Questions for 2026 Exams
Important Q1: The slope of a velocity-time graph gives: A) Distance B) Displacement C) Acceleration D) Speed
Answer: C) Acceleration Explanation: Acceleration is defined as the change in velocity over time, which is exactly what the slope ($dv/dt$) represents.
Important Q2: A body is thrown vertically upwards with a velocity $u$. The maximum height reached by the body is: A) $u/g$ B) $u^2/2g$ C) $u^2/g$ D) $u/2g$
Answer: B) $u^2/2g$ Explanation: At maximum height, final velocity $v = 0$. Using $v^2 = u^2 + 2as$ (where $a = -g$), we get $0 = u^2 - 2gs$, so $s = u^2/2g$.
Important Q3: If the displacement of an object is proportional to the square of time, then the object moves with: A) Uniform velocity B) Uniform acceleration C) Increasing acceleration D) Decreasing acceleration
Answer: B) Uniform acceleration Explanation: $x \propto t^2 \Rightarrow x = kt^2$. Velocity $v = dx/dt = 2kt$. Acceleration $a = dv/dt = 2k$ (which is a constant).
Important Q4: Can a body have zero velocity and still be accelerating? Give an example.
Answer: Yes. When a ball is thrown vertically upwards, at its highest point, its velocity becomes zero for a brief instant. However, the acceleration due to gravity ($9.8 \text{ m/s}^2$ downward) is still acting on it.
Important Q5: Distinguish between distance and displacement.
Answer: Distance: The total path length covered by a body. It is a scalar quantity and is always positive. Displacement: The shortest distance between the initial and final positions. It is a vector quantity and can be positive, negative, or zero.
Important Q6: A train travels from station A to station B at a speed of 40 km/h and returns at 60 km/h. What is its average speed?
Answer: Average speed for equal distances is given by $\frac{2v_1v_2}{v_1 + v_2}$. Average speed = $\frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48 \text{ km/h}$.
Important Q7: Define stopping distance. On what factors does it depend?
Answer: Stopping distance is the distance traveled by a moving vehicle before it comes to a complete halt after the brakes are applied. It depends on the initial velocity of the vehicle and the braking capacity (retardation or friction) of the road.
Important Q8: Derive the equation $s = ut + \frac{1}{2}at^2$ using the calculus method.
Answer: We know that velocity is the rate of change of position: $v = \frac{ds}{dt}$. And we know $v = u + at$. Substituting this in the first equation: $\frac{ds}{dt} = u + at \Rightarrow ds = (u + at)dt$ Integrating both sides. Assume at $t=0$, $s=0$ and at time $t$, position is $s$: $\int_0^s ds = \int_0^t (u + at)dt$ $s = \int_0^t u dt + \int_0^t at dt$ $s = u[t]_0^t + a[\frac{t^2}{2}]_0^t$ $s = ut + \frac{1}{2}at^2$
Important Q9: 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. (Take $g = 10 \text{ m/s}^2$).
Answer: First descent: $u = 0$, $s = 90 \text{ m}$, $a = 10 \text{ m/s}^2$. Time taken: $s = \frac{1}{2}at^2 \Rightarrow 90 = 5t^2 \Rightarrow t^2 = 18 \Rightarrow t \approx 4.24 \text{ s}$. Speed just before hitting the ground: $v = u + at = 10 \times 4.24 = 42.4 \text{ m/s}$. Rebound: It loses 1/10th speed. New speed $u' = \frac{9}{10} \times 42.4 = 38.16 \text{ m/s}$. Time to reach max height: $v = u' - gt \Rightarrow 0 = 38.16 - 10t \Rightarrow t = 3.81 \text{ s}$. Total time so far = 4.24 + 3.81 = 8.05 s. Second descent: Takes another 3.81 s to hit the ground. Total time = 11.86 s. Hits with speed 38.16 m/s. The graph will be a series of triangular shapes, peaking at 42.4, 38.16, etc.
Important Q10: What is relative velocity? Derive an expression for the relative velocity of object A with respect to object B moving in a straight line.
Answer: Relative velocity is the velocity of one object as observed from another moving object. Let $x_A(t)$ and $x_B(t)$ be the positions of A and B at time t. $x_A(t) = x_A(0) + v_A t$ $x_B(t) = x_B(0) + v_B t$ The relative displacement of A with respect to B is $x_{AB}(t) = x_A(t) - x_B(t)$ $x_{AB}(t) = (x_A(0) - x_B(0)) + (v_A - v_B)t$ Since displacement = initial relative position + relative velocity x time, we get $v_{AB} = v_A - v_B$. A boy stands at the edge of a cliff 20 m high and throws a stone vertically upwards with an initial speed of 15 m/s. (Take $g = 10 \text{ m/s}^2$).
Important Q11: What is the maximum height reached by the stone from the top of the cliff?
Answer: $v^2 = u^2 - 2gh \Rightarrow 0 = (15)^2 - 2(10)h \Rightarrow 20h = 225 \Rightarrow h = 11.25 \text{ m}$.
Important Q12: How much time does it take for the stone to reach its maximum height?
Answer: $v = u - gt \Rightarrow 0 = 15 - 10t \Rightarrow t = 1.5 \text{ s}$.
Important Q13: With what velocity does the stone hit the ground at the base of the cliff?
Answer: Consider the whole journey. Net displacement $s = -20 \text{ m}$ (downward from start). $v^2 = u^2 + 2as \Rightarrow v^2 = (15)^2 + 2(-10)(-20)$ $v^2 = 225 + 400 = 625$ $v = \sqrt{625} = 25 \text{ m/s}$ (downward direction). (Directions: A = Both true & R explains A; B = Both true but R doesn't explain A; C = A is true, R is false; D = A is false, R is true)
Important Q14: Assertion (A): The speedometer of a car measures the instantaneous speed of the car. Reason (R): Average speed is equal to total distance divided by total time.
Answer: B . Both statements are scientifically correct, but R is not the reason for A. The speedometer shows speed at a specific instant (instantaneous), which is a separate concept from average speed.
Important Q15: Assertion (A): It is possible to have an accelerated motion with a constant speed. Reason (R): In uniform circular motion, speed is constant, but the direction of velocity changes continuously.
Answer: A . Since velocity is a vector, changing direction means velocity changes, which implies acceleration, even if the magnitude (speed) remains constant.
FAQ Section
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.
(a) Yes. The size of the carriage is very small compared to the distance between the two stations. (b) Yes. The size of the monkey is extremely small compared to the circumference of the circular track. (c) No. The ball's size is not negligible compared to the distance it covers while turning sharply. Its spin involves the motion of its different parts. (d) No.
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 given in the textbook. Choose the correct entries: (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.
Let's analyze the standard NCERT graph conceptually: (a) A lives closer to the school than B. (The position of A's home, P, is at a lower value on the y-axis than B's home, Q). (b) A starts from the school earlier than B. (A starts at t = 0 on the x-axis, whereas B starts at a later time t > 0). (c) B walks faster than A. (The slope of the x-t graph represents velocity.
A woman starts from her home at 9.00 am, walks with 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 an auto with a speed of 25 km/h. Choose suitable scales and plot the x-t graph of her motion.
Step-by-step explanation: 1. Home to Office: Distance = 2.5 km Speed = 5 km/h Time taken = Distance / Speed = 2.5 / 5 = 0.5 hours = 30 minutes. She reaches the office at 9:30 am. 2. At Office: She stays at the office (x = 2.5 km) from 9:30 am to 5:00 pm. During this time, the graph will be a straight horizontal line parallel to the time axis. 3. Office to Home: Distance = 2.5 km Speed = 25 km/h Time taken = 2.
A jet airplane 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 the speed of the jet plane with respect to the ground be $v_j = 500 \text{ km/h}$. Let the speed of the combustion products with respect to the jet plane be $v_{cg} = -1500 \text{ km/h}$ (negative because the gas ejects in the opposite direction of the plane's motion). We know the relative velocity formula: $v_{cg} = v_c - v_j$ Where $v_c$ is the speed of combustion products with respect to the ground.
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves