Understanding Motion through Experience Class 9 Question Answer Advanced Science Chapter 2

Students can use NCERT Class 9 Advanced Science Solutions Chapter 2 Understanding Motion through Experience Question Answer to understand complex concepts with ease.

Understanding Motion through Experience Class 9 Questions and Answers

Understanding Motion through Experience Question Answer Class 9

Check Your Understanding

Question 1.
Define a frame of reference in your own words.
Answer:
A frame of reference is a fixed point or a set of objects (like a building, a tree, or a room) that we use as a “viewpoint” to decide if something else is moving. To describe the position or motion of an object, we need this reference point; without it, we cannot tell how fast or in which direction an object is travelling.

Question 2.
Give two real-life examples where motion depends on the observer.
Answer:

• A Passenger in a Moving Train: If we are sitting in a moving train, you appear to be at rest to a fellow passenger sitting next to us (because our position isn’t changing relative to them). However, to a person standing on the railway platform, we are in motion along with the train.
• The Movement of Earth: To us standing on the ground, the buildings and trees around us appear to be at rest. However, to an astronaut in space (the observer), everything on Earth—including us and the buildings, trees, etc., are in rapid motion because the Earth is constantly rotating and orbiting the Sun.

Question 3.
Why does a person sitting in a moving train appear at rest to another passenger?
Answer:
A person appears at rest to a fellow passenger because they share the same frame of reference. Since both individuals are moving at the same speed and in the same direction as the train, the distance and position between them do not change over time. In physics, motion is defined as a change in position relative to an observer; since there is no such change between the two passengers, they perceive each other as being at rest.

Question 4.
Classify the following as scalar or vector quantities: speed, velocity, displacement, distance, acceleration and mass.
Answer:

Scalar Quantities (Magnitude only)	Vector Quantities (Magnitude + Direction)
Mass	Displacement
Distance	Velocity
Speed	Acceleration

Question 5.
Explain the difference between distance and displacement with an activity diagram.
Answer:
A point is marked as point A. Moved 4 steps forward, then 3 steps to the right, and marked the final point as B.

• The distance is the total path walked (4 + 3 = 7 steps).
• The displacement is the straight-line distance from A to B (used a ruler to measure). It becomes equal to \(\sqrt{\left(4^2+3^2\right)}\) = 5 steps.
  
• The path shown (A → B shown by dotted line) represents distance.
• The straight line from A to B represents displacement.

Question 6.
Give two everyday examples of vector quantities.
Answer:
Two everyday examples of vector quantities are:

• Velocity: For example, a car moving at 60 km/h towards the north (it has both magnitude and direction).
• Force: For example, pushing a door towards the inside (force has magnitude and direction).

Question 7.
Draw two vectors of 4 units east and 3 units north and find the resultant using the triangle method.
Answer:

Since East and North are perpendicular (90-degree angle), the vectors form a right-angled triangle. We apply the Pythagoras’ theorem:
R = \(\sqrt{\left(A^2+B^2\right)}\)
R = \(\sqrt{\left(4^2+3^2\right)}\)
R= \(\sqrt{(16+9)}\)
R = \(\sqrt{(25)}\)
R = 5 units

Question 8.
Explain how vector subtraction is performed graphically.
Answer:
Vector subtraction is essentially the addition of a negative vector. To perform \(\vec{A}-\vec{B}=\vec{A}+(-\vec{B}) .\).
The Step-by-Step Process

1. Reversed the direction: Taken the vector being subtracted (\(\vec{B}\))and flipped it 180° to create its opposite (-\(\vec{B}\)). It keeps the same magnitude but points the other way.
2. Used the triangle method: Placed the tail of the flipped vector (-\(\vec{B}\)) at the head of the first vector (\(\vec{A}\)).
3. Drawn the resultant: Drawn a new arrow headed line from the tail of \(\vec{A}\) to the head of –\(\vec{B}\). This new vector represents \(\vec{A}-\vec{B}\).

Question 9.
Draw two opposite vectors of equal magnitude. Calculate its resultant.
Answer:
The resultant of two opposite vectors of equal magnitude is a null vector (or zero vector), meaning its magnitude is 0.

Vector Visualisation
When adding two opposite vectors of the same size, the second vector brings the path exactly back to the starting point.

To find the resultant \(\vec{R}\) of two vectors \(\vec{A}\) and \(\vec{B}\) those are opposite and equal in magnitude (|\(\vec{A}\)| = |\(\vec{B}\)| = x):
1. Define Direction: Let East be positive (+) and West be negative (-).

2. Assign Values:
(i) |\(\vec{A}\)| = + x
(ii) |\(\vec{B}\)| = -x
Sum of Vectors:
\(\vec{R}=\vec{A}+\vec{B}\)
\(\vec{R}=\overrightarrow{0}\)

Question 10.
A body starts from rest and accelerates at 4 m/s². Find the distance travelled in the 6^th second.
Answer:
To find the distance travelled in the 6^th second, we use the formula for the distance covered in the n^th second of motion:
S_n = u + \(\frac{a}{2}\)(2n – 1)
Given, u = 0
a = 4 m/s²
t = n = 6^th second
Now,
S₆ = 0 + \(\frac{4}{2}\)(2 × 6 – 1)
S₆ = 22 m
The distance travelled by the body in the 6^th second is 22 metres.

Question 11.
A car with an initial velocity of 8 m/s accelerates at 2 m/ s². Find the distance covered in the 5^th second.
Answer:
Given, u = 8 m/s
a = 2 m/s²
t = n = 5^th second
Now,
S_n = u + \(\frac{a}{2}\)(2n – 1)
S₅ = 8 + \(\frac{2}{2}\)(2 × 5 – 1)
= 8 + 10 – 1
S₅ = 17 m
The distance covered by the car in the 5^th second is 17 metre.

Reflect and Discuss

Question 1.
Why is specifying a reference frame necessary to describe motion?
Answer:
Specifying a reference frame is necessary because motion is not absolute; it is relative. Without a fixed viewpoint or reference point, it is impossible to determine whether an object has changed its position.
Here is why it is essential:

• Defining State of Rest or Motion: An object can be at rest and in motion at the same time, depending on the observer. For example, a person in a moving car is at rest relative to the car’s seat but in motion relative to a person standing on the road.
• Tracking Position: To say something is “5 kilometres away,” we must have a starting point (the origin). Without a reference frame, “5 kilometres” has no meaning because we don’t know from where the distance is being measured.
• Measuring Velocity and Direction: Speed and direction are measured relative to a background. A reference frame provides the coordinate system (like North, South, or x, y, z axes) needed to calculate how fast and in what direction an object is travelling.
• Consistency: It provides a common ground for observers to agree on a measurement. If two people use different reference frames without stating them, their descriptions of the same event will conflict.

Question 2.
How do direction and magnitude together describe displacement?
Answer:
In physics, displacement is defined as the shortest straight-line distance between the initial and final positions of an object. To fully describe it, both magnitude and direction are required because displacement is a vector quantity.
Here is how they work together:
The Magnitude (The “How Much”): The magnitude tells us the numerical value of the shortest distance. It represents the “gap” between the starting and ending points, regardless of the actual path taken. For example, if we walk around a block and end up 10 metres from where we started, the magnitude is 10 metres.

The Direction (The “Where To”): The direction specifies the orientation of that change in position. Without direction, we only know how far you moved, but not where we ended up. Saying “I moved 10 metres” is a distance; saying “I moved 10 metres North” is a displacement.

Why Both Are Essential
If we only have one without the other, the description of motion is incomplete:

Multiple Results for the Same Magnitude: If we move 5 metre from a point, we could end up anywhere on a circle with a 5-metre radius. Only by adding direction (e.g., “East”) we can pinpoint our exact final location.

The “Zero Displacement” Rule: Magnitude and direction allow us to distinguish displacement from distance. If we run 400 metres around a track and return to our start, our distance covered is 400 m, but since our direction resulted in our returning to the origin, our displacement is 0.

Question 3.
Which daily activities around you involve accelerated motion?
Answer:
Accelerated motion is very common in our daily lives because almost any change in speed or direction counts as acceleration. Here are several activities around us that involve it:

• Commuting to School or Work: When a school bus or car pulls away from a stop sign, it is speeding up (positive acceleration). When it approaches a red light and slows down, it is decelerating (negative acceleration).
• Walking or Running: When you start to walk from a standing position, we are accelerating. Even if we run at a constant speed but turn a corner, we are accelerating because our direction is changing.
• Using an Elevator: When we press a button, and the elevator starts to move, you feel a slight “heavy” or “light” sensation. This is because the elevator is accelerating to reach its cruising speed or decelerating as it arrives at our floor.
• Dropping an Object: If we accidentally drop a pen or a ball, it undergoes acceleration due to gravity. It starts at zero speed and moves faster and faster as it falls toward the ground.
• Playing Sports: Swinging a cricket bat, kicking a football, or throwing a basketball involves rapid acceleration. The ball changes its speed and direction the moment it is hit or thrown.
• Ceiling Fans: When we first switch on a fan, it undergoes angular acceleration as it speeds up from rest to its full rotating speed.

Project-Based Learning

Design a simple experiment using everyday materials to measure the speed of a moving object (using a bicycle, or a walking student). Present your method, observations, calculations and conclusions to the class.
Answer:
Objective: To calculate the average speed of a student walking/cycling a fixed distance using everyday tools and the formula: Speed = Distance/Time

Materials Required
Measuring Tape (or a 1-metre ruler).
Stopwatch (a smartphone works perfectly).
Chalk or Tape (to mark start and end lines).
A flat, straight path (like a school corridor or playground).

Method (Procedure)

1. Track is marked: The measuring tape is used to measure a straight path of 10 metres. “Start Line” and “Finish Line” are clearly with chalk.
2. Roles assigned: One student acts as the walker cyclist.
   • One student acts as the Timer (standing at the Finish Line).
   • One student acts as the Recorder to write down the results.
3. The walking/cycling: On the signal “GO,” the walker/ cyclist begins walking/cycling at a normal, steady pace from the Start Line.
4. Timing: The Timer starts the stopwatch when the moment the Start Line is crossed and stops it when the Finish Line is crossed.
5. Repeat: Repeated the process three times to ensure accuracy and calculated an average time.

Observations

Total Distance (d) = 10 metres

Trial	Time Taken (seconds)
Trial 1	8.2 s
Trial 2	7.9 s
Trial 3	8.5 s
Average Time (t)	8.2 second

Calculations
Using the average time recorded:
Distance (d): 10 m
Average Time (t): 8.2 s
Speed = 10 m/8.2 s = 1.22 m/s

Understanding Motion through Experience Class 9 Extra Questions and Answers

Short Answer Type Questions

Question 1.
Why is it impossible to define whether an object is completely at rest or in motion without specifying a reference point?
Answer:
Motion is not absolute; it is a relative concept that depends entirely on the observer’s position. A reference point provides the observation point necessary to track position changes; without it, there is no baseline to determine if the distance between objects is changing or remaining constant.

Question 2.
When two cars move down a highway in the same direction at the exact same speed, what is their state of motion relative to each other? Explain.
Answer:
The two cars are at relative rest with respect to each other. Because they travel in the same direction at identical velocities, the physical distance between them never changes, causing each car to appear stationary from the viewpoint of each other.

Question 3.
Differentiate between an inertial and a non-inertial frame of reference based on Newton’s laws of motion.
Answer:
An inertial frame is either stationary or moving at a constant velocity, allowing Newton’s laws to work perfectly without modifications. In contrast, a non- inertial frame is actively accelerating, meaning Newton’s laws do not hold true unless imaginary correction forces, known as pseudo-forces, are applied.

Question 4.
State the mathematical relationship between speed and time when distance is a controlled, constant variable, and clarify how a faster object behaves under this rule.
Answer:
When distance is constant, speed is inversely proportional to time (s = d/t). This means that a faster moving object will naturally take a significantly smaller amount of time to cover the designated baseline distance.

Question 5.
Differentiate between scalar and vector quantity.
Answer:

Feature	Scalar Quantity	Vector Quantity
Definition	A physical quantity that has only magnitude (numerical value and unit) but no specific direction.	A physical quantity that requires both magnitude and direction to be completely described.
Change	Changes only when its numerical value changes.	Changes if either its magnitude or its direction or both change.
Math Rules	Follows simple rules of ordinary algebra for addition and subtraction (e.g., 5m + 2m = 7m.	Requires special rules of vector algebra for addition and subtraction.
Examples	Distance, Speed, Mass, Time, Temperature.	Displacement, Velocity, Acceleration, Force.

Question 6.
Using the example of two passengers sitting next to each other in a car moving at 60 km/h, explain how an object can simultaneously be at rest and in motion?
Answer:
Whether an object is at rest or in motion depends entirely on the chosen frame of reference.

• From the perspective of a passenger inside the car, the person next to him appears completely still (at relative rest) because both are moving with exactly same velocity, and the distance between them never changes.
• However, from the perspective of an outside observer standing on the road, both passengers are moving at 60 km/h because their positions are rapidly changing relative to the ground.

This proves that motion is a relative concept, allowing both states to exist at the same time under different frames of reference.

Question 7.
A student walks across a classroom with a classmate. Explain how changing the frame of reference from the classmate to the classroom floor alters the description of the student’s motion.
Answer:
The chosen frame of reference dictates whether motion is detected:

• From the frame of reference of the classmate walking alongside at the exact same speed and direction, there is no motion observed because their relative positions remain constant.
• If the frame of reference shifts to the classroom floor, clearly there is a measurable motion because the walking student’s coordinates are continuously changing relative to the stationary floor.
• This demonstrates that an object’s state of motion cannot be defined in isolation; it must always be specified relative to a chosen frame of reference.

Question 8.
A student walks 5 metres east from point A to Point B, and then walks 5 metres west back to point A. Using the concepts of scalar and vector quantities, calculate the total distance covered and the net displacement. Justify why these two values are different.
Answer:

1. Calculations:
   • Distance Covered: 5 m (towards east) + 5 m (towards west) = 10 metre.
   • Displacement: 5 m (East) -5m (West) = 0 metre.
2. Justification: Distance is a scalar quantity that represents the total path length travelled; it accumulates continuously because it is independent of direction.

Conversely, displacement is a vector quantity defined as the shortest straight-line distance between the initial and final positions. Since the student returned precisely to his starting point, his final position has not changed relative to his start, resulting in zero net displacement.

Question 9.
An automobile travelling along a straight highway at a uniform speed of 25 m/s detects an obstruction ahead. The driver applies the brakes immediately, producing a uniform retardation (negative acceleration) of 5.0 m/s2 until the vehicle comes to a complete halt. Determine (a) the total time required for the automobile to stop, and (b) the braking distance covered before stopping. -Q
Answer:
Given parametres:

• Initial velocity (u) = 25 m/s
• Final velocity (υ) = 0 m/s (Since the vehicle comes to a complete halt)
• Acceleration (a) = -5.0 m/s² (Negative sign signifies braking/retardation)

Part (a): Finding time taken to stop (t)
Using the first equation of motion:
υ = u + at
0 = 25 + (-5.0 × t)
5.1 = 25
t = 5.0 s

Part (b): Finding braking distance (s)
Using the third equation of motion to verify independently:
υ² = u² + 2as
(0)² = (25)² + [2 × (-5.0) × s]
0 = 625 – 10s
10s = 625
s = 62.5 metre

Question 5.
A high-speed train starts from a complete state of rest at a station and moves down a straight track with a uniform acceleration of 2.0 m/s². Calculate (a) the final velocity attained by the train after exactly 15 seconds of travel, and (b) the total distance covered by the train during this time interval. 0
Answer:
Given parametres:

• Initial velocity (u) = 0 m/s (Since the train starts from a state of rest)
• Uniform acceleration (a) = 2.0 m/s²
• Time elapsed (f) = 15 s

Part (a): Finding final velocity (υ)
Using the first equation of motion:
υ = u + at
υ = 0 + (2.0 m/s² × 15 s)
υ = 30 m/s

Part (b): Finding distance covered (s)
Using the second equation of motion:
s = ut + 0.5at²
s = (0 × 15) + [0.5 × 2.0 m/s² × (15 s)²]
s = 0 + [1.0 × 225]
s = 225 metre

Long Answer Type Questions

Question 1.
Four major milestones along a straight-line city grid are mapped out on a coordinate plane where all grid values are measured in kilometres (km). The coordinates are defined as follows:
Point A (2, 2): Rahul’s house
Point B (5, 2): Local bus station
Point C (5, 6): Central library junction
Point D (8, 6): Science academy campus
In the morning, Rahul leaves his house and walks on foot along a straight path from point A to point B.
At the bus station, he boards an electric shuttle bus that travels along a straight path from point B to point D, via point C.
Based on the coordinates provided, calculate the following:
(a) The total distance travelled by Rahul on foot.
(b) The total distance covered by Rahul while riding the electric shuttle bus.
(c) The net magnitude of Rahul’s displacement from his house directly to the Science Academy campus.
Answer:
(a) Distance Travelled on foot (From A to B)
Formula: Using the coordinate distance formula:
d = \(\sqrt{\left[\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\right]}\)
Substitution: For points A(2, 2) and B(5, 2).
AB = \(\sqrt{\left[(5-2)^2+(2-2)^2\right]}\)
AB = \(\sqrt{\left[3^2+0^2\right]}=\sqrt{9}\) = 3 km

(b) Distance travelled by shuttle bus (From B to D via C)
Step 1 (Path BC): For points B(5, 2) and C(5, 6).
BC = \(\sqrt{\left[(5-5)^2+(6-2)^2\right]}\)
= \(\sqrt{\left[0+4^2\right]}\)
= \(\sqrt{16}\)
= 4 km

Step 2 (Path CD): For points C(5, 6) and D(8, 6).
CD = \(\sqrt{\left[(8-5)^2+(6-6)^2\right]}\)
= \(\sqrt{\left[3^2+0\right]}\)
= \(\sqrt{9}\)
= 3 km

Step 3 (Total Bus Distance): Summing the individual straight paths:
Total Bus Distance = BC + CD = 4 km + 3 km = 7 km

(c) Total displacement from house to Science Academy.
Concept: Displacement is a vector quantity
representing the shortest straight-line path connecting the initial position A(2, 2) directly to the final destination D(8, 6).
Substitution:
AD = \(\sqrt{\left[(8-2)^2+(6-2)^2\right]}\)
AD = \(\sqrt{\left[6^2-4^2\right]}\)
AD = \(\sqrt{[36+16]}\) = \(\sqrt{52} \mathrm{~km}\)
= \(\sqrt{(4 \times 13)}=2 \sqrt{13} \mathrm{~km}\)
(or approximately 7.21 km)

Question 2.
(a) Derive the second equation of motion, s = ut + 1/2at², algebraically for an object moving along a straight line path with a constant uniform acceleration. Define all mathematical symbols used during the derivation.
(b) A vehicle starting from a state of rest accelerates uniformly down a straight highway at a constant rate of 4.0 m/s>sup>2 for exactly 6.0 seconds. Calculate the total linear distance covered by the vehicle during this period using your derived expression.
Answer:
(a) 1. Parametres
Let an object begins its straight-line motion with an initial velocity u. Let it undergo a constant uniform acceleration a over a total elapsed time interval t, attaining a final velocity v. During this time, the total linear distance or net displacement covered by the object is represented as s.

2. Derivation:
When an object moves with uniform linear acceleration, its rate of change of velocity is constant.
Therefore, the average velocity (υ_avg) over the designated arithmetic mean of its initial and final velocities:
Average Velocity = \(\frac{(\text { Initial Velocity }+ \text { Final Velocity })}{2}\)
υ_avg = \(\frac{(u+v)}{2}\) ………… (1)
By fundamental definition, the total distance (s) covered is equal to the product of average velocity and the total elapsed time interval (t):
Distance (s) = Average Velocity × Time (t)
S = \(\) × t ………….. (2)
From the first equation of motion, we know that final velocity is expressed as:
υ = u + at ………. (3)
Substituting the algebraic expression for υ from equation 3 into equation 2:
S = \(\left[\frac{(u+(u+a t))}{2}\right]\) × t
S = \(\left[\frac{(2 u+a t)}{2}\right]\) × t
S = \(\left[\left(\frac{2 u}{2}\right)+\left(\frac{a t}{2}\right)\right]\) × t
S = \(\left[u+\frac{1}{2} a t\right]\) × t
S = ut + \(\frac{1}{2}\)at²

(b) Given:

• Initial velocity (u) = 0 m/s (Since the vehicle starts from rest).
• Constant acceleration (a) = 4.0 m/s²
• Time interval (t) = 6.0 seconds

Substituting the vales directly into the equation:
S = ut + \(\frac{1}{2}\)at²
S = (0 × 6.0) + [\(\frac{1}{2}\) × 4.0 m/s² × (6.0 s)²]
S = 0 + [ 2.0 × 36.0 ]
S = 72 metre
The vehicle covers a total linear braking/travel distance of exactly 72 metre over the designated 6.0 second.

Case-Based Questions

I. An electric delivery van is idling at a red traffic light on a straight city avenue. The moment the light switches to green, the van’s automated system initiates a smooth forward launch, executing a constant uniform acceleration of 2.5 m/s². The vehicle maintains this steady rate of acceleration for exactly 6.0 seconds to safely clear a wide intersection and merge into the main traffic flow. To predict and monitor the van’s real-time position and speed during this launch, the vehicle’s onboard processing computer relies on classical kinematic equations of motion. These equations allow the system to map the vehicle’s state from a complete rest to its cruising speed, calculating its exact linear tracking displacement and velocity variations step- by-step.

Question 1.
Which of the following parametres represents the initial velocity (u) of the delivery van based on the situational context?
(A) 2.5 m/s
(B) 6.0 m/s
(C) 0.0 m/s
(D) 15.0 m/s
Answer:
Option (C) is correct.
Explanation: The passage states that the van is initially idling at a red traffic light; which implies it is completely stationary. In classical physics, an object starting from a complete state of rest has an initial velocity (u) of exactly 0.0 m/s.

Question 2.
What final velocity (υ) does the delivery van attain at the end of the 6.0 second acceleration window?
(A) 8.5 m/s
(B) 15.0 m/s
(C) 45.0 m/s
(D) 90.0 m/s
Answer:
Option (B) is correct.
Explanation: Using the first equation of motion (υ = u + at),
where u = 0.0 m/s, a = 2.5 m/s², and t = 6.0 s:
υ = 0 + (2.5 × 6.0) = 15.0 m/s.

Question 3.
What is the total linear displacement (s) covered by the vehicle during its 6.0 second launch across the intersection?
(A) 15 metres
(B) 30 metres
(C) 45 metres
(D) 90 metres
Answer:
Option (C) is correct
Explanation: Using the second equation of motion [s = ut + 1/2at²], where u = 0.0 m/s, a = 2.5 m/s², and t = 6.0 s:
S = (0 × 6.0) + [1/2 × 2.5 × (6.0)²] = 0 + [1.25 × 36] = 45 metre.
Alternatively, this can be cross-verified using the third equation of motion: υ² = u² + 2as ⇒ 15² = 0² + 2(2.5)s ⇒ 225 = 5s ⇒ S = 45 m.

Question 4.
If the van had to stop suddenly at the intersection instead, what mathematical adjustment would the processing computer make to the acceleration parametre (a)?
(A) It would double its numerical magnitude to represent an increase in kinetic energy.
(B) It would change its sign to a negative value to signify retardation (braking).
(C) It would change the parametre to zero because the van is travelling in a straight line.
(D) It would convert the value into a vector that points upward to counter gravity.
Answer:
Option (B) is correct
Explanation: When a moving vehicle brakes or slows down to stop, its final velocity is lower than its initial velocity. This yields a negative rate of change in velocity over time, which is defined as a negative acceleration or retardation. The internal software maps this mechanical state by assigning a negative sign to the acceleration , value.

Understanding Motion through Experience Class 9 MCQ

Question 1.
Which of the following statements correctly identifies the mathematical relationship between speed and time when the total distance covered remains constant?
(A) Speed is directly proportional to the square of time.
(B) Speed is directly proportional to time.
(C) Speed is inversely proportional to time.
(D) Speed is completely independent of time.
Answer:
Option (C) is correct.
Explanation: When the distance is held constant (at 5 metres), a shorter time interval results in a higher calculated speed. Mathematically, speed (s = d/t) shares an inverse relationship with time (f) under constant distance (d), meaning the motion that takes the least amount of time is inherently the fastest.

Question 2.
Two professional sprinters are running a race down a straight track side-by-side at a matching, uniform speed of 8m/s. From the perspective of Sprinter A, what is the state of motion of Sprinter B?
(A) Sprinter B is accelerating forward at a rate of 8 m/s².
(B) Sprinter B appears to be at relative rest (stationary).
(C) Sprinter B is moving backwards at a constant velocity of 8 m/s.
(D) Sprinter B is moving in a non-inertial frame of reference.
Answer:
Option (B) is correct.
Explanation: When two objects move in the exact same direction at the exact same velocity, they are at relative rest to one another. Because the physical distance between the two sprinters never changes as they move down the track, Sprinter B will appear stationary from Sprinter A’s chosen frame of reference.

Question 3.
A scientist conducting an experiment inside a vehicle moving with a constant velocity observes that Newton’s laws of motion hold perfectly true without needing any mathematical corrections. What type of reference frame is this vehicle operating in?
(A) Non-inertial Frame
(B) Accelerating Frame
(C) Pseudo Frame
(D) Inertial Frame
Answer:
Option (D) is correct.
Explanation: By definition, an inertial frame of reference is a framework that is either entirely at rest or moving with a constant velocity. In these balanced, non-accelerating environments, Newton’s laws of motion work perfectly without requiring special engineering or mathematical corrections like pseudo-forces.

Question 4.
While sitting perfectly still on a stationary passenger train, the train on the parallel track next to you begins to roll slowly forward. For a split second, your brain mistakenly perceives that your own train is moving backward. What does this common sensory illusion demonstrate?
(A) Motion is an absolute concept that can be defined without an observer.
(B) The human brain can automatically calculate pseudo-forces accurately.
(C) Motion is relative, and the brain is struggling to establish a fixed frame of reference.
(D) Your train has instantly entered an accelerating, non- inertial reference frame.
Answer:
Option (C) is correct.
Explanation: This sensory phenomenon highlights that motion is relative rather than absolute. Because your brain lacks an immediate point of certainty, it struggles to decide whether the moving train next to you or your own train should be treated as the “fixed” reference point, leading to a temporary misinterpretation of relative position.

Question 5.
During a science laboratory activity, a student records that it takes exactly 2.5 seconds to run across a fixed distance of 10 metres. If the student then slows down to a walking pace that takes 8.0 seconds to cover the exact same distance, what is the difference between their running speed and walking speed? 0
(A) 1.25 m/s
(B) 2.75 m/s
(C) 4.00 m/s
(D) 5.50 m/s
Answer:
Option (B) is correct.
Explanation: First, calculate the running speed by dividing the distance by the running time:
\(\frac{10 \mathrm{~m}}{2.5 \mathrm{~s}}\) = 4.0 m/s.
Next, calculate the walking speed by dividing the same distance by the walking time:
\(\frac{10 \mathrm{~m}}{8.0 \mathrm{~s}}\) = 1.25 m/s.
Finally, find the difference between the two speeds:
4. m/s – 1.25 m/s = 2.75 m/s.

Assertion-Reason Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): A passenger reading a book inside a train travelling at 60 km/h is simultaneously at rest and in motion.
Reason (R): Motion is a relative concept; an object’s state of motion depends entirely on the chosen observer’s frame of reference.
Answer:
Option (A) is correct.
Explanation: Both statements are scientifically accurate. The passenger is at relative rest with respect to the train (as the distance between them is unchanging) but is simultaneously in motion with respect to an outside observer standing on the stationary platform. Since motion is relative and not absolute, reason (R) perfectly explains the dual state highlighted in assertion (A).

Question 2.
Assertion (A): In a 5-metre space, a student who completes the run in 1.5 second is faster than a student who walks it in 4.0 second.
Reason (R): When the total distance travelled is held constant, the speed of an object is directly proportional to the time taken. 0
Answer:
Option (C) is correct.
Explanation: The Assertion is true because a lower time interval over the same distance yields a higher calculated speed (3.33 m/s running vs. 1.25 m/s walking). However, the Reason is completely false. Mathematically, speed (s = d/t) is inversely proportional to time when distance (d) is constant, meaning the motion taking the least amount of time is the fastest.

Question 3.
Assertion (A): Special mathematical corrections called “pseudo-forces” must be introduced to accurately describe an object’s motion inside an accelerating car.
Reason (R): An environment that is undergoing acceleration is classified as a non-inertial frame of reference, where Newton’s standard laws of motion do not work perfectly on their own.
Answer:
Option (A) is correct.
Explanation: Both assertion (A) and reason (R) are true and directly connected. Inside an accelerating vehicle (a non-inertial frame), objects appear to move or shift without any visible physical contact pushing them. To correct this framework and apply classical mechanics accurately, scientists introduce imaginary adjustments called pseudo-forces. Thus, reason (R) provides the exact theoretical reason why the action in assertion (A) is mandatory.