Case Study Questions for Class 9 Science Chapter 4 Describing Motion Around Us (Exploration Book) 2026-27

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Home CBSE Class 9 Science Case Study Qs Case Study Questions for Class 9 Science Chapter 4 Describing Motion Around Us (Exploration Book) 2026-27

This page provides Case Study Questions for Class 9 Science Chapter 4 – Describing Motion Around Us from the latest NCERT Exploration textbook. Each set contains a reading passage followed by 4 objective/short-answer questions, exactly as expected in the CBSE Board examination pattern.

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Exam Tip: Case Study questions carry 4 marks each in the CBSE Class 9 Science paper. Read the passage carefully before attempting the MCQs or short answers.

1

Case Study: Braking Distance and Safe Driving

Read the passage carefully, then answer all four questions
Ch 4 · Describing Motion Around Us

When brakes are applied to a moving vehicle, it moves some distance before coming to a complete stop. This is called the braking distance. The total distance travelled during this stopping phase depends upon several factors: the initial velocity of the vehicle when the brakes are applied, the condition of the road surface (wet, dry, etc.), the braking capacity of the vehicle (which determines its negative acceleration), and the driver’s reaction time.

During the driver’s reaction time—the brief moment before the brakes are actually pressed—the vehicle continues to move forward at its initial constant velocity. Once the brakes are fully engaged, the vehicle decelerates uniformly until it comes to rest. Understanding the physics behind these factors is crucial for road safety and explains why it is vital to maintain a safe following distance from the vehicle ahead.

1
What does a negative sign in the calculated acceleration of a braking car indicate?
a The car is physically moving in the reverse direction.
b The acceleration is acting in a direction opposite to the vehicle’s velocity.
c The car has immediately reached zero velocity.
d The acceleration of the vehicle is increasing continuously.
Correct Answer (b) The acceleration is acting in a direction opposite to the vehicle’s velocity. ✅
Explanation

When an object is slowing down, the magnitude of its velocity decreases. Mathematically, this is represented by an average acceleration that has a negative sign relative to the positive direction of motion, meaning the acceleration acts opposite to the direction of velocity.

2
A car is moving on a highway at a velocity of 20 m s⁻¹. The driver applies the brakes to produce a constant negative acceleration of -4 m s⁻². What is the braking distance (distance travelled after brakes are applied)?
a 25 m
b 40 m
c 50 m
d 80 m
Correct Answer (c) 50 m ✅
Explanation

Using the third kinematic equation, v² = u² + 2as, where final velocity (v) is 0, initial velocity (u) is 20 m s⁻¹, and acceleration (a) is -4 m s⁻². Substituting the values: 0 = (20)² + 2(-4)s. This simplifies to 0 = 400 – 8s. Solving for s gives s = 50 m.

3
Which of the following statements about a vehicle coming to a stop is INCORRECT?
a A wet road surface generally reduces braking capacity, increasing the stopping distance.
b During the driver’s reaction time, the acceleration of the vehicle is zero.
c A vehicle moving at double the initial velocity will have exactly double the braking distance, assuming the same deceleration.
d The total stopping distance includes both the distance travelled during the reaction time and the actual braking distance.
Correct Answer (c) A vehicle moving at double the initial velocity will have exactly double the braking distance, assuming the same deceleration. ✅
Explanation

This statement is incorrect. Based on the equation s = u² / (2a) (derived from v² = u² + 2as with v = 0), the stopping distance is proportional to the square of the initial velocity. If the velocity is doubled, the braking distance increases by a factor of four, not two.

4
Assertion (A): It is necessary to maintain a significantly larger safe following distance from the vehicle ahead when driving at high speeds.

Reason (R): The total stopping distance of a vehicle is independent of the driver’s reaction time.
a Both A and R are true, and R is the correct explanation of A.
b Both A and R are true, but R is not the correct explanation of A.
c A is true, but R is false.
d A is false, but R is true.
Correct Answer (c) A is true, but R is false. ✅
Explanation

The assertion is true; higher speeds require vastly larger following distances due to the physics of kinetic energy and braking. However, the reason is false. The total stopping distance is highly dependent on the driver’s reaction time, as the car continues to travel at its initial high velocity before the brakes are even applied.

2

Case Study: Understanding Motion through Graphs

Read the passage carefully, then answer all four questions
Ch 4 · Describing Motion Around Us

Graphical representations provide a powerful visual method to describe how physical quantities like position and velocity change with time. In a position-time graph, a straight line indicates that the object is moving with a constant velocity. Geometrically, the steepness or “slope” of this line gives the magnitude of the velocity. A steeper line indicates a higher velocity, while a horizontal line indicates the object is at rest.

Similarly, a velocity-time graph provides deep insights into an object’s motion. The slope of the line on a velocity-time graph reveals how fast the velocity is changing with time, which is the acceleration. Furthermore, if you calculate the area enclosed by the velocity-time graph and the time axis for a specific time interval, that mathematical area equals the total displacement of the object during that interval. These graphical tools are essential for analyzing complex real-world linear motions.

1
If the position-time graph of an object is a straight horizontal line parallel to the time axis, what does it indicate about the object’s state?
a The object is moving with a uniform velocity.
b The object is stationary (at rest).
c The object is moving with constant acceleration.
d The object is returning to its starting point.
Correct Answer (b) The object is stationary (at rest). ✅
Explanation

A straight line parallel to the time (x) axis on a position-time graph means that the position value on the y-axis is not changing as time passes. Since its position remains constant, the object is at rest.

2
The position-time graphs of two cars, Car X and Car Y, are plotted on the same axes. The straight line representing Car X is much steeper than the straight line representing Car Y. What can be concluded from this observation?
a Car X has a higher uniform velocity than Car Y.
b Car Y has a higher uniform velocity than Car X.
c Car X is accelerating faster than Car Y.
d Car X started its journey earlier than Car Y.
Correct Answer (a) Car X has a higher uniform velocity than Car Y. ✅
Explanation

The slope of a position-time graph determines the magnitude of the average velocity. A steeper slope means a greater change in position over the same amount of time, indicating a higher velocity.

3
For an object moving with constant acceleration, what physical quantity does the mathematical area under its velocity-time graph represent?
a The average speed of the object
b The total acceleration of the object
c The total displacement of the object
d The time interval of the motion
Correct Answer (c) The total displacement of the object ✅
Explanation

The area enclosed by the velocity-time graph and the time axis for any given time interval is geometrically equal to the magnitude of the displacement of the object during that interval.

4
An object starts from rest and accelerates uniformly. Its velocity-time graph is a straight diagonal line passing through the origin. How is the displacement (s) of this object mathematically related to time (t)?
a Displacement is directly proportional to time (s ∝ t).
b Displacement is inversely proportional to time (s ∝ 1/t).
c Displacement is directly proportional to the square of time (s ∝ t²).
d Displacement remains constant as time passes.
Correct Answer (c) Displacement is directly proportional to the square of time (s ∝ t²). ✅
Explanation

For an object starting from rest (u = 0) and moving with uniform acceleration, the kinematic equation is s = ½at². Because ½ and a are constants, the displacement s is directly proportional to the square of the time t². Geometrically, the area of the triangle under the v-t graph grows as a square of the base (time).

Chapters covered in CBSE Class 9 Science Latest Book – Exploration

  • Chapter 1: Exploration: Entering the World of Secondary Science
  • Chapter 2: Cell: The Building Block of Life
  • Chapter 3: Tissues in Action
  • Chapter 4: Describing Motion Around Us
  • Chapter 5: Exploring Mixtures and their Separation
  • Chapter 6: How Forces Affect Motion
  • Chapter 7: Work, Energy, and Simple Machines
  • Chapter 8: Journey Inside the Atom
  • Chapter 9: Atomic Foundations of Matter
  • Chapter 10: Sound Waves: Characteristics and Applications
  • Chapter 11: Reproduction: How Life Continues
  • Chapter 12: Patterns in Life: Diversity and Classification
  • Chapter 13: Earth as a System: Energy, Matter, and Life

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