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Motion – CBSE NCERT Study Resources

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Motion

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9th

9th - Science

Motion

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Motion

Chapter Overview

This chapter introduces the fundamental concepts of motion, including types of motion, distance and displacement, speed and velocity, acceleration, and the graphical representation of motion. Students will learn to differentiate between uniform and non-uniform motion and understand the equations governing uniformly accelerated motion.

Motion: A body is said to be in motion if its position changes with respect to a reference point (origin) over time.

Types of Motion

Motion can be classified into different types based on the path followed by the object:

  • Rectilinear Motion: Motion along a straight line.
  • Circular Motion: Motion along a circular path.
  • Periodic Motion: Motion that repeats itself after a fixed interval of time.

Distance and Displacement

Distance: The total path length covered by an object in motion. It is a scalar quantity.

Displacement: The shortest distance between the initial and final positions of an object. It is a vector quantity.

Speed and Velocity

Speed: The distance traveled by an object per unit time. It is a scalar quantity.

Velocity: The displacement of an object per unit time. It is a vector quantity.

Acceleration

Acceleration: The rate of change of velocity with respect to time. It is a vector quantity.

Graphical Representation of Motion

Motion can be represented graphically using distance-time and velocity-time graphs:

  • Distance-Time Graph: Shows how distance changes with time.
  • Velocity-Time Graph: Shows how velocity changes with time.

Equations of Uniformly Accelerated Motion

The three equations of motion for uniformly accelerated motion are:

  1. v = u + at
  2. s = ut + ½ at²
  3. v² = u² + 2as

Where:
u = initial velocity,
v = final velocity,
a = acceleration,
t = time,
s = displacement.

Uniform and Non-Uniform Motion

Uniform Motion: Motion in which an object covers equal distances in equal intervals of time.

Non-Uniform Motion: Motion in which an object covers unequal distances in equal intervals of time.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define displacement.
Answer:

Shortest distance between initial and final positions.

Question 2:
What is the SI unit of speed?
Answer:

Meter per second (m/s).

Question 3:
Give an example of non-uniform motion from NCERT.
Answer:

A car moving on a crowded road.

Question 4:
What does the slope of a distance-time graph represent?
Answer:

Speed of the object.

Question 5:
Name the device used to measure speed in vehicles.
Answer:

Speedometer.

Question 6:
What is acceleration?
Answer:

Rate of change of velocity.

Question 7:
Give a real-world example of uniform motion.
Answer:

Movement of Earth around the Sun.

Question 8:
What is the formula for average speed?
Answer:

Total distance / Total time.

Question 9:
What does a straight line on a distance-time graph indicate?
Answer:

Uniform motion.

Question 10:
Name the type of motion in a swinging pendulum.
Answer:

Oscillatory motion.

Question 11:
What is the acceleration due to gravity on Earth?
Answer:

9.8 m/s².

Question 12:
Give an NCERT example of rectilinear motion.
Answer:

A car moving on a straight road.

Question 13:
What is the difference between speed and velocity?
Answer:

Velocity includes direction, speed does not.

Question 14:
What does a negative slope in a velocity-time graph indicate?
Answer:

Deceleration.

Question 15:
Define uniform motion with an example.
Answer:

An object is said to be in uniform motion when it covers equal distances in equal intervals of time, regardless of the duration of the time interval.
Example: A car moving at a constant speed of 60 km/h on a straight highway.

Question 16:
What is the SI unit of acceleration?
Answer:

The SI unit of acceleration is meter per second squared (m/s²).

Question 17:
Differentiate between distance and displacement.
Answer:

Distance is the total path length covered by an object, while displacement is the shortest distance between the initial and final positions of the object.
Key difference: Distance is a scalar quantity (magnitude only), whereas displacement is a vector quantity (magnitude and direction).

Question 18:
A car accelerates from rest to 20 m/s in 5 seconds. Calculate its acceleration.
Answer:

Given: Initial velocity (u) = 0 m/s, Final velocity (v) = 20 m/s, Time (t) = 5 s.
Using the formula: a = (v - u)/t
a = (20 - 0)/5
a = 4 m/s².

Question 19:
What does the slope of a distance-time graph represent?
Answer:

The slope of a distance-time graph represents the speed of the object.
A steeper slope indicates higher speed, while a flatter slope indicates lower speed.

Question 20:
Give an example where an object has zero displacement but non-zero distance.
Answer:

Example: A person running around a circular track and returning to the starting point.
Here, displacement is zero (same start and end point), but distance is the circumference of the track.

Question 21:
What is the nature of motion when the acceleration of an object is zero?
Answer:

When acceleration is zero, the object is either at rest or moving with constant velocity (no change in speed or direction).

Question 22:
Convert 72 km/h into m/s.
Answer:

Given: Speed = 72 km/h.
To convert km/h to m/s, multiply by 5/18.
72 × (5/18) = 20 m/s.

Question 23:
What is the direction of acceleration for a freely falling object?
Answer:

The direction of acceleration for a freely falling object is downward, toward the center of the Earth, due to gravity.

Question 24:
Define non-uniform motion with an example.
Answer:

An object is said to be in non-uniform motion when it covers unequal distances in equal intervals of time.
Example: A car speeding up or slowing down in traffic.

Question 25:
What is the shape of the distance-time graph for an object at rest?
Answer:

The shape of the distance-time graph for an object at rest is a straight horizontal line, indicating no change in position over time.

Question 26:
A train covers 300 km in 5 hours. Calculate its average speed.
Answer:

Given: Distance = 300 km, Time = 5 hours.
Average speed = Total distance / Total time
= 300 km / 5 h
= 60 km/h.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
What is the SI unit of acceleration?
Answer:

The SI unit of acceleration is meter per second squared (m/s²).

It measures how much the velocity of an object changes per unit time.

Question 2:
A car accelerates from 10 m/s to 30 m/s in 5 seconds. Calculate its acceleration.
Answer:

Given:
Initial velocity (u) = 10 m/s
Final velocity (v) = 30 m/s
Time (t) = 5 s

Acceleration (a) = (v - u) / t
= (30 - 10) / 5
= 20 / 5
= 4 m/s².

Question 3:
What does the slope of a distance-time graph represent?
Answer:

The slope of a distance-time graph represents the speed of the object.

Steeper slope = Higher speed
Zero slope = Object at rest.

Question 4:
Give an example of non-uniform motion.
Answer:

Non-uniform motion occurs when an object covers unequal distances in equal intervals of time.

Example: A car slowing down as it approaches a traffic signal.

Question 5:
A train moves at a constant speed of 90 km/h for 2 hours. Calculate the distance covered.
Answer:

Given:
Speed = 90 km/h
Time = 2 hours

Distance = Speed × Time
= 90 × 2
= 180 km.

Question 6:
Define average speed.
Answer:

Average speed is the total distance traveled by an object divided by the total time taken.

Formula: Average speed = Total distance / Total time.

Question 7:
What is the direction of motion in a circular path at any instant?
Answer:

In a circular path, the direction of motion at any instant is tangential to the circle at that point.

This means the object moves along the tangent drawn at that point on the circle.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define uniform motion and give an example.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the duration of the time interval.

Example: A car moving at a constant speed of 60 km/h on a straight highway exhibits uniform motion because it covers 60 km every hour.

Question 2:
A cyclist covers 30 km in 2 hours. Calculate the average speed in m/s.
Answer:

Given: Distance = 30 km = 30,000 m, Time = 2 hours = 7200 s.

Formula: Average speed = Total distance / Total time.

Calculation:
Average speed = 30,000 m / 7200 s
= 4.17 m/s (approx).

Question 3:
What does the slope of a distance-time graph represent? Explain with a diagram reference.
Answer:

The slope of a distance-time graph represents the speed of the object.

  • A steeper slope indicates higher speed.
  • A zero slope (horizontal line) means the object is at rest.

Diagram reference: Imagine a graph where the x-axis is time and the y-axis is distance. A straight line rising upwards shows constant speed.

Question 4:
A car accelerates from rest to 20 m/s in 5 seconds. Calculate its acceleration.
Answer:

Given: Initial velocity (u) = 0 m/s, Final velocity (v) = 20 m/s, Time (t) = 5 s.

Formula: Acceleration (a) = (v - u) / t.

Calculation:
a = (20 m/s - 0 m/s) / 5 s
= 4 m/s².

Question 5:
Why is the motion of a pendulum considered oscillatory?
Answer:

The motion of a pendulum is oscillatory because it moves back and forth repeatedly about a fixed central position (mean position).

  • It follows a periodic path (same motion repeats at regular intervals).
  • The time taken for one complete oscillation is called time period.
Example: A swinging pendulum clock demonstrates oscillatory motion.

Question 6:
A cyclist covers 30 km in 2 hours. Calculate the average speed.
Answer:

Given: Distance = 30 km, Time = 2 hours


Formula: Average Speed = Total Distance / Total Time


Calculation:
Average Speed = 30 km / 2 h
Average Speed = 15 km/h

Question 7:
Explain why the distance-time graph for uniform motion is a straight line.
Answer:

In uniform motion, the object covers equal distances in equal intervals of time. This means the ratio of distance to time (speed) remains constant.


Since speed is constant, the distance-time graph is a straight line with a constant slope. The slope represents the speed of the object.


Example: If a car moves at 20 km/h, every hour it covers 20 km, resulting in a straight-line graph.

Question 8:
What is acceleration? Give its SI unit.
Answer:

Acceleration is the rate of change of velocity of an object with respect to time. It indicates how quickly an object speeds up, slows down, or changes direction.


Formula: Acceleration (a) = Change in Velocity (Δv) / Time Taken (Δt)


SI Unit: meter per second squared (m/s²)

Question 9:
A train accelerates from rest to 72 km/h in 5 minutes. Calculate its acceleration in m/s².
Answer:

Given:
Initial Velocity (u) = 0 m/s (at rest)
Final Velocity (v) = 72 km/h = (72 × 1000) / 3600 = 20 m/s
Time (t) = 5 minutes = 5 × 60 = 300 s


Formula: a = (v - u) / t


Calculation:
a = (20 m/s - 0 m/s) / 300 s
a = 20 / 300 ≈ 0.067 m/s²

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Define uniform motion and non-uniform motion with examples. How can we represent these motions graphically?
Answer:
Concept Overview

Uniform motion is when an object covers equal distances in equal intervals of time (e.g., a car moving at 60 km/h). Non-uniform motion involves unequal distances in equal time (e.g., a bus in traffic).

Process Explanation

We studied that a distance-time graph for uniform motion is a straight line, while for non-uniform motion, it is curved or zigzag.

Real-world Application

Our textbook shows a train moving at constant speed (uniform), while a bicycle slowing down near a signal (non-uniform).

Question 2:
Explain acceleration and derive its formula. Give an example where acceleration is negative.
Answer:
Concept Overview

Acceleration is the rate of change of velocity. Its formula is a = (v - u)/t, where v is final velocity, u is initial velocity, and t is time.

Process Explanation

We studied that when a car speeds up, acceleration is positive. If it slows down (e.g., braking), acceleration is negative (retardation).

Real-world Application

Our textbook shows a ball thrown upwards slowing down due to gravity, exhibiting negative acceleration.

Question 3:
What is distance and displacement? Illustrate with an example from NCERT.
Answer:
Concept Overview

Distance is the total path length (scalar), while displacement is the shortest path from start to end (vector).

Process Explanation

We studied that if a person walks 10m east and 5m west, distance is 15m, but displacement is 5m east.

Real-world Application

Our textbook shows a farmer moving around a field—distance depends on path, but displacement is straight-line from start to end.

Question 4:
Describe the three equations of motion and solve: A car accelerates from 10 m/s to 30 m/s in 5s. Find acceleration and distance.
Answer:
Concept Overview

The three equations are v = u + at, s = ut + ½at², and v² = u² + 2as.

Process Explanation

Given u = 10 m/s, v = 30 m/s, t = 5s. Using v = u + at, acceleration (a) = 4 m/s². Using s = ut + ½at², distance (s) = 100m.

Real-world Application

Our textbook shows similar problems, like a train accelerating uniformly.

Question 5:
What is circular motion? Is it accelerated? Explain with an NCERT example.
Answer:
Concept Overview

Circular motion is movement along a circular path. Even at constant speed, direction changes, so it is accelerated.

Process Explanation

We studied that a merry-go-round has constant speed but changing direction, causing centripetal acceleration.

Real-world Application

Our textbook shows the moon orbiting Earth—speed is constant, but acceleration exists due to direction change.

Question 6:
Explain acceleration and derive its formula. How is it different from deceleration?
Answer:
Concept Overview

Acceleration is the rate of change of velocity. Its formula is a = (v-u)/t, where v=final velocity, u=initial velocity, and t=time.

Process Explanation

We studied that acceleration occurs when speed increases, while deceleration (negative acceleration) happens when speed decreases.

Real-world Application

Our textbook shows a car speeding up (acceleration) and braking (deceleration) as examples.

Question 7:
Describe the three equations of motion and their applications. Give one NCERT example.
Answer:
Concept Overview

The three equations are: (1) v = u + at, (2) s = ut + ½at², (3) v² = u² + 2as. They relate displacement, velocity, acceleration, and time.

Process Explanation

We studied these equations for uniformly accelerated motion. For example, calculating the distance a car covers while accelerating.

Real-world Application

Our textbook shows a ball dropped from a height as an example of these equations.

Question 8:
What is circular motion? Is it accelerated motion? Justify with an example.
Answer:
Concept Overview

Circular motion is movement along a circular path. Even at constant speed, it is accelerated due to continuous change in direction.

Process Explanation

We studied that centripetal force causes this acceleration, as seen in planets orbiting the Sun.

Real-world Application

Our textbook shows a merry-go-round as an example of circular motion with acceleration.

Question 9:
Differentiate between distance and displacement. Illustrate with an NCERT example.
Answer:
Concept Overview

Distance is the total path length (scalar), while displacement is the shortest path from initial to final position (vector).

Process Explanation

We studied that displacement can be zero if the object returns to start, but distance is never zero.

Real-world Application

Our textbook shows a person walking around a circular track and returning to the start: distance is circumference, displacement is zero.

Question 10:
Define uniform motion and non-uniform motion with examples. How is speed calculated for each?
Answer:
Concept Overview

Uniform motion is when an object covers equal distances in equal intervals of time (e.g., a car moving at 60 km/h). Non-uniform motion involves unequal distances in equal time (e.g., a car accelerating).

Process Explanation

Speed is calculated as distance/time. For uniform motion, speed is constant. For non-uniform motion, average speed = total distance/total time.

Real-world Application

Our textbook shows a train moving uniformly, while a car in traffic shows non-uniform motion.

Question 11:
Explain acceleration with an example. How is it different from deceleration?
Answer:
Concept Overview

Acceleration is the rate of change of velocity (e.g., a car speeding up). Deceleration is negative acceleration (e.g., a car slowing down).

Process Explanation

Acceleration = (final velocity - initial velocity)/time. Deceleration is when acceleration is negative.

Real-world Application

Our textbook shows a bicycle accelerating downhill, while deceleration occurs when brakes are applied.

Question 12:
Describe the distance-time graph for uniform and non-uniform motion. What does the slope represent?
Answer:
Concept Overview

A straight line on a distance-time graph indicates uniform motion, while a curved line shows non-uniform motion.

Process Explanation

The slope represents speed. Steeper slope = higher speed. For non-uniform motion, the slope changes.

Real-world Application

Our textbook shows a cyclist moving uniformly, while a runner speeding up has a curved graph.

Question 13:
Differentiate between scalar and vector quantities in motion. Give one example of each.
Answer:
Concept Overview

Scalar quantities have magnitude only (e.g., distance). Vector quantities have magnitude and direction (e.g., displacement).

Process Explanation

Distance is how much ground is covered, while displacement is the shortest path with direction.

Real-world Application

Our textbook shows a car traveling 5 km (scalar) vs. moving 5 km east (vector).

Question 14:
Define uniform motion and non-uniform motion with examples. Also, explain how a distance-time graph differs for both types of motion.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the duration of those intervals. An example is a car moving at a constant speed of 60 km/h on a straight highway.


Non-uniform motion occurs when an object covers unequal distances in equal intervals of time. For example, a car accelerating from rest or decelerating due to traffic.


Distance-time graph for uniform motion: It is a straight line with a constant slope, indicating steady speed.
Distance-time graph for non-uniform motion: It is a curved line or a zig-zag pattern, showing changing speed.


Understanding these concepts helps in analyzing real-world scenarios like vehicle movement, sports, and celestial motions.

Question 15:
Derive the first equation of motion (v = u + at) using a velocity-time graph. Explain each step clearly.
Answer:

To derive v = u + at using a velocity-time graph, follow these steps:

1. Draw a straight line on the graph representing constant acceleration (a), starting from initial velocity (u) at time t = 0.
2. The slope of the line gives acceleration: a = (v - u)/t.
3. Rearrange the equation to solve for final velocity (v): v = u + at.

This equation shows how an object's velocity changes under uniform acceleration over time.

Question 16:
A cyclist covers 30 km in 2 hours. Calculate the average speed and explain whether the motion is uniform or non-uniform with reasoning.
Answer:

Average speed is calculated as:
Total distance / Total time = 30 km / 2 h = 15 km/h.

Since the question does not specify whether the cyclist maintained constant speed throughout, we cannot confirm uniform motion. Non-uniform motion is likely if the cyclist changed speed or stopped during the journey. Average speed alone does not indicate uniformity; a distance-time graph or additional data is needed for confirmation.

Question 17:
Define uniform motion and non-uniform motion with examples. Also, explain how distance-time graphs differ for both types of motion.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the duration of the time intervals. An example is a car moving at a constant speed of 60 km/h on a straight highway.

Non-uniform motion occurs when an object covers unequal distances in equal intervals of time. An example is a car accelerating from rest or decelerating to stop at a traffic signal.

The distance-time graph for uniform motion is a straight line with a constant slope, indicating steady speed.
For non-uniform motion, the graph is a curved line, showing varying speed (acceleration or deceleration).

Question 18:
Derive the three equations of motion (v = u + at, s = ut + ½ at², and v² = u² + 2as) using graphical method for uniformly accelerated motion. Explain each step clearly.
Answer:

First Equation (v = u + at):
Consider a velocity-time graph for an object with initial velocity u and uniform acceleration a.
The final velocity v after time t is given by the area under the graph (a straight line).
Slope of the graph = acceleration (a) = (v - u)/t
Rearranging gives: v = u + at

Second Equation (s = ut + ½ at²):
Distance s is the total area under the velocity-time graph.
It consists of a rectangle (area = ut) and a triangle (area = ½ × base × height = ½ × t × (v - u)).
Substituting v - u = at from the first equation, the triangle area becomes ½ at².
Thus, s = ut + ½ at²

Third Equation (v² = u² + 2as):
From the first equation, t = (v - u)/a.
Substitute this into the second equation: s = u(v - u)/a + ½ a(v - u)²/a².
Simplify to get v² = u² + 2as.

Question 19:
Define uniform motion and non-uniform motion with examples. Also, derive the equation for average speed and explain its significance in real-life scenarios.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the time duration. For example, a car moving at a constant speed of 60 km/h on a straight highway exhibits uniform motion.

Non-uniform motion occurs when an object covers unequal distances in equal intervals of time. For instance, a bicycle slowing down as it approaches a traffic signal shows non-uniform motion.

The equation for average speed is derived as follows:
Average Speed = Total Distance Travelled / Total Time Taken
Mathematically, it can be written as: vavg = Δs / Δt
Where:
- vavg = average speed
- Δs = total distance
- Δt = total time

In real-life scenarios, average speed helps in:

  • Planning travel time for long journeys.
  • Comparing the efficiency of different vehicles.
  • Analyzing traffic flow and road safety measures.

Question 20:
Define uniform motion and non-uniform motion with examples. Also, explain how a distance-time graph differs for both types of motion.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the duration of the time intervals. An example of uniform motion is a car moving at a constant speed of 60 km/h on a straight highway.

Non-uniform motion occurs when an object covers unequal distances in equal intervals of time. For example, a car accelerating from rest or decelerating to stop exhibits non-uniform motion.

The distance-time graph for uniform motion is a straight line with a constant slope, indicating steady speed.
The distance-time graph for non-uniform motion is a curved line or a zigzag line, showing changes in speed over time.

Understanding these concepts helps in analyzing real-world scenarios like traffic movement, sports performance, and even planetary motion.

Question 21:
Define uniform motion and non-uniform motion with examples. Explain how a distance-time graph helps distinguish between them.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the duration. An example is a car moving at a constant speed of 60 km/h on a straight highway.

Non-uniform motion occurs when an object covers unequal distances in equal intervals of time. For instance, a car accelerating from rest or slowing down due to traffic.

A distance-time graph helps distinguish between the two:

  • For uniform motion, the graph is a straight line with a constant slope, indicating steady speed.
  • For non-uniform motion, the graph is curved or irregular, showing changes in speed over time.
Question 22:
A train accelerates uniformly from rest to a speed of 72 km/h in 5 minutes. Calculate its acceleration and the distance covered during this period.
Answer:

Given:
Initial velocity (u) = 0 m/s (since the train starts from rest)
Final velocity (v) = 72 km/h = 72 × (5/18) = 20 m/s
Time (t) = 5 minutes = 300 seconds

Step 1: Calculate acceleration (a)
Using the formula: a = (v - u) / t
a = (20 m/s - 0 m/s) / 300 s
a = 20/300 = 0.0667 m/s²

Step 2: Calculate distance (s)
Using the formula: s = ut + ½at²
s = (0 × 300) + (½ × 0.0667 × 300²)
s = 0 + (0.0333 × 90000)
s = 3000 meters (or 3 km)

Conclusion: The train's acceleration is 0.0667 m/s², and it covers 3 km in 5 minutes.

Question 23:
Define uniform motion and non-uniform motion with examples. Derive the equation for average speed and explain its significance in real-life scenarios.
Answer:

Uniform motion refers to the movement of an object where it covers equal distances in equal intervals of time, regardless of the duration of the time intervals. An example is a car moving at a constant speed of 60 km/h on a straight highway.

Non-uniform motion occurs when an object covers unequal distances in equal intervals of time. For example, a car accelerating from rest or decelerating to stop exhibits non-uniform motion.

The equation for average speed is derived as follows:
Average Speed = Total Distance Travelled / Total Time Taken
For example, if a car travels 300 km in 5 hours, its average speed is 300 km / 5 h = 60 km/h.

Significance in real-life scenarios:

  • Helps in planning travel time for long-distance journeys.
  • Used in sports to analyze athletes' performance over a race.
  • Essential in logistics to estimate delivery times for goods.

Question 24:
A train accelerates uniformly from rest to a velocity of 72 km/h in 5 minutes. Calculate its acceleration and the distance covered during this period. Explain the concept of acceleration with a real-world application.
Answer:

Given:
Initial velocity (u) = 0 m/s (since the train starts from rest)
Final velocity (v) = 72 km/h = 72 × (1000/3600) = 20 m/s
Time (t) = 5 minutes = 5 × 60 = 300 s

Step 1: Calculate acceleration (a)
Using the equation v = u + at:
20 = 0 + a × 300
a = 20 / 300 = 0.0667 m/s²

Step 2: Calculate distance (s)
Using the equation s = ut + ½at²:
s = 0 × 300 + ½ × 0.0667 × (300)²
s = 0 + 0.0333 × 90000
s = 3000 m (or 3 km)

Concept of acceleration:
Acceleration is the rate of change of velocity with respect to time. It indicates how quickly an object speeds up, slows down, or changes direction.

Real-world application:
In elevators, controlled acceleration ensures smooth starts and stops for passenger comfort and safety. Excessive acceleration can cause discomfort, while too little may make the ride inefficient.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A car moves from point A to B at constant speed of 40 km/h and returns at 60 km/h. Calculate the average speed for the entire trip. Refer to NCERT Example 8.2.
Answer:
Case Summary
A car travels equal distances at two different speeds.
Scientific Principle
Average speed = Total distance / Total time (NCERT Eq. 8.6).
Solution Approach
  • Assume distance AB = 120 km (LCM of 40 & 60)
  • Time taken: 120/40 = 3h (going), 120/60 = 2h (returning)
  • Total distance = 240 km, Total time = 5h
  • Average speed = 240/5 = 48 km/h
Question 2:
A bicycle accelerates from rest to 10 m/s in 5 seconds. Determine the acceleration and distance covered using NCERT Activity 8.5 method.
Answer:
Case Summary
Bicycle starts from rest and gains speed uniformly.
Scientific Principle
v = u + at (Eq. 8.5) and s = ut + ½at² (Eq. 8.6) from NCERT.
Solution Approach
  • Initial velocity (u) = 0, Final velocity (v) = 10 m/s
  • Acceleration (a) = (10-0)/5 = 2 m/s²
  • Distance (s) = 0×5 + ½×2×5² = 25m
Question 3:
A train decelerates uniformly from 72 km/h to 36 km/h in 10s. Find the retardation and distance traveled, similar to NCERT Example 8.7.
Answer:
Case Summary
Train slows down from higher to lower speed.
Scientific Principle
Retardation is negative acceleration (NCERT Page 102).
Solution Approach
  • Convert speeds: 72 km/h = 20 m/s, 36 km/h = 10 m/s
  • Retardation = (10-20)/10 = -1 m/s² (negative sign shows deceleration)
  • Distance = (20²-10²)/(2×1) = 150m (using v²-u²=2as)
Question 4:
A ball is thrown vertically upwards with 20 m/s velocity. Calculate maximum height and time to return, referencing NCERT Figure 8.7.
Answer:
Case Summary
Vertical projectile motion under gravity.
Scientific Principle
At maximum height, final velocity becomes zero (NCERT Page 103).
Solution Approach
  • Using v²=u²+2as: 0=20²-2×10×s → s=20m
  • Time to reach top: t=(v-u)/a = (0-20)/(-10) = 2s
  • Total time = 2×2 = 4s (up + down time equal)
Question 5:
A car moves from point A to B at constant speed of 40 km/h and returns at 60 km/h. Case Summary: Calculate the average speed for the entire trip. Refer to NCERT Example 8.2.
Answer:
Case Summary: A car travels equal distances at two different speeds.
Scientific Principle: Average speed = Total distance / Total time (NCERT Eq. 8.6).
Solution Approach:
  • Assume distance AB = 120 km (LCM of 40 & 60 for easy calculation).
  • Time taken: 3h (A→B) + 2h (B→A) = 5h.
  • Total distance = 240 km.
  • Average speed = 240/5 = 48 km/h.

Our textbook shows similar problems with uniform motion.

Question 6:
A bicycle accelerates from rest to 6 m/s in 10s. Case Summary: Find its acceleration and distance covered. Use NCERT Activity 8.5 method.
Answer:
Case Summary: Bicycle starts from rest and gains speed.
Scientific Principle: Acceleration = (v-u)/t; Distance = ut + ½at² (NCERT Eq. 8.7, 8.8).
Solution Approach:
  • u = 0, v = 6 m/s, t = 10s.
  • Acceleration = (6-0)/10 = 0.6 m/s².
  • Distance = 0 + ½(0.6)(10)² = 30m.

Like Activity 8.5, we used kinematic equations for uniformly accelerated motion.

Question 7:
A train decelerates at 2 m/s² until stopping. Case Summary: If initial speed was 20 m/s, calculate stopping time and distance. Relate to NCERT Example 8.7.
Answer:
Case Summary: Train slows down to halt with constant deceleration.
Scientific Principle: v = u + at; v² = u² + 2as (NCERT Eq. 8.9, 8.10).
Solution Approach:
  • Final velocity v = 0, u = 20 m/s, a = -2 m/s².
  • 0 = 20 + (-2)t → t = 10s.
  • 0² = 20² + 2(-2)s → s = 100m.

This mirrors Example 8.7 where negative acceleration indicates deceleration.

Question 8:
An object is thrown vertically upwards at 20 m/s. Case Summary: Determine maximum height and time to return. Compare with NCERT Fig. 8.7.
Answer:
Case Summary: Object moves against gravity with initial velocity.
Scientific Principle: At max height, v = 0; Time of flight = 2u/g (NCERT Eq. 8.12).
Solution Approach:
  • Using v² = u² - 2gh → 0 = 20² - 2(10)h → h = 20m.
  • Time up = u/g = 2s → Total time = 4s.

Fig. 8.7 shows similar vertical motion under gravity's deceleration.

Question 9:
A car moves from rest and attains a velocity of 20 m/s in 5 seconds. Using the concept of acceleration, analyze its motion and compare it with NCERT's example of a train accelerating uniformly.
Answer:
Case Summary

A car starts from rest and reaches 20 m/s in 5 seconds. We need to find its acceleration.

Scientific Principle
  • Acceleration = (Final Velocity - Initial Velocity) / Time
  • Our textbook shows a similar example of a train accelerating uniformly.
Solution Approach

Using the formula, acceleration = (20 - 0)/5 = 4 m/s². Like the NCERT train example, this shows uniform acceleration.

Question 10:
A cyclist covers 30 meters in 10 seconds while moving at a constant speed. Explain how this demonstrates uniform motion and relate it to NCERT's example of a moving pendulum bob at constant speed.
Answer:
Case Summary

A cyclist moves 30 meters in 10 seconds at constant speed.

Scientific Principle
  • Uniform motion means equal distances in equal intervals of time.
  • NCERT shows a pendulum bob moving uniformly in a horizontal plane.
Solution Approach

Here, speed = 30/10 = 3 m/s. Since speed is constant, it's uniform motion, similar to the pendulum example.

Question 11:
A ball is thrown vertically upwards with a velocity of 15 m/s. Calculate the time taken to return to the thrower's hand using equations of motion and compare it with NCERT's free-fall example.
Answer:
Case Summary

A ball is thrown upwards at 15 m/s. We need to find its total time in air.

Scientific Principle
  • Time to reach maximum height: t = u/g (u = initial velocity, g = 9.8 m/s²).
  • NCERT's free-fall example uses similar principles.
Solution Approach

Time upwards = 15/9.8 ≈ 1.53 s. Total time = 2 × 1.53 ≈ 3.06 s, matching NCERT's concept of symmetrical motion.

Question 12:
A bus decelerates from 25 m/s to 5 m/s in 4 seconds. Determine its retardation and relate this to NCERT's example of a car applying brakes.
Answer:
Case Summary

A bus slows from 25 m/s to 5 m/s in 4 seconds.

Scientific Principle
  • Retardation = (Initial Velocity - Final Velocity) / Time.
  • NCERT shows a similar car braking example.
Solution Approach

Retardation = (25 - 5)/4 = 5 m/s². Like the NCERT example, this shows uniform deceleration due to braking.

Question 13:
A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. After this, it moves with a constant velocity for the next 20 seconds. The brakes are then applied, and the car comes to rest uniformly in 5 seconds.

Calculate:

  • The maximum velocity attained by the car.
  • The total distance traveled by the car.
Answer:

To solve this problem, we break it into three phases: acceleration, constant velocity, and deceleration.


Phase 1: Acceleration
Initial velocity (u) = 0 m/s
Acceleration (a) = 2 m/s²
Time (t) = 10 s
Final velocity (v) = u + at = 0 + (2 × 10) = 20 m/s
Distance (s₁) = ut + ½at² = 0 + ½ × 2 × (10)² = 100 m

Phase 2: Constant Velocity
Velocity (v) = 20 m/s
Time (t) = 20 s
Distance (s₂) = v × t = 20 × 20 = 400 m

Phase 3: Deceleration
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s
Time (t) = 5 s
Deceleration (a) = (v - u)/t = (0 - 20)/5 = -4 m/s²
Distance (s₃) = ut + ½at² = 20 × 5 + ½ × (-4) × (5)² = 100 - 50 = 50 m

Total Distance = s₁ + s₂ + s₃ = 100 + 400 + 50 = 550 m

The maximum velocity attained is 20 m/s, and the total distance traveled is 550 m.

Question 14:
A train travels a distance of 300 km at a uniform speed. If the speed had been 5 km/h less, it would have taken 2 hours more to cover the same distance.

Find the original speed of the train.

Answer:

Let the original speed of the train be v km/h.


Time taken at original speed (t₁) = Distance/Speed = 300/v hours
Reduced speed = (v - 5) km/h
Time taken at reduced speed (t₂) = 300/(v - 5) hours

According to the problem:
t₂ - t₁ = 2 hours
300/(v - 5) - 300/v = 2

Simplify the equation:
300v - 300(v - 5) = 2v(v - 5)
300v - 300v + 1500 = 2v² - 10v
1500 = 2v² - 10v
Divide by 2:
750 = v² - 5v
Rearrange:
v² - 5v - 750 = 0

Solve the quadratic equation:
v = [5 ± √(25 + 3000)] / 2
v = [5 ± √3025] / 2
v = [5 ± 55] / 2

Discarding the negative value (speed cannot be negative):
v = (5 + 55)/2 = 30 km/h

The original speed of the train was 30 km/h.

Question 15:
A car travels from point A to point B, covering a distance of 150 km in 3 hours. On the return trip, it takes 2 hours to cover the same distance.

Based on this case:

  • Calculate the average speed for the entire round trip.
  • Explain why average velocity for the entire journey is zero.
Answer:

Average speed is calculated as the total distance traveled divided by the total time taken.


Total distance = 150 km (A to B) + 150 km (B to A) = 300 km
Total time = 3 hours + 2 hours = 5 hours
Average speed = Total distance / Total time = 300 km / 5 h = 60 km/h

Average velocity is zero because the car returns to its starting point (A), resulting in zero displacement.


Displacement = 0 km (since initial and final positions are the same)
Average velocity = Displacement / Total time = 0 km / 5 h = 0 km/h
Question 16:
A cyclist accelerates uniformly from rest to a speed of 10 m/s in 5 seconds.

Based on this case:

  • Calculate the acceleration of the cyclist.
  • Plot a speed-time graph for this motion and determine the distance covered.
Answer:

Acceleration is the rate of change of velocity.


Initial velocity (u) = 0 m/s (since starting from rest)
Final velocity (v) = 10 m/s
Time (t) = 5 s
Acceleration (a) = (v - u) / t = (10 m/s - 0 m/s) / 5 s = 2 m/s²

For the speed-time graph:

  • X-axis: Time (s)
  • Y-axis: Speed (m/s)

The graph is a straight line starting from (0,0) to (5,10).
Distance covered = Area under the graph = ½ × base × height = ½ × 5 s × 10 m/s = 25 m
Question 17:
A car travels a distance of 150 km in 3 hours. Calculate its average speed. Later, it travels another 100 km in 2 hours. Find the total average speed for the entire journey.
Answer:

To calculate the average speed, we use the formula: Average Speed = Total Distance / Total Time.


Step 1: Calculate the average speed for the first part of the journey.


Distance = 150 km
Time = 3 hours
Average Speed = 150 km / 3 h = 50 km/h

Step 2: Calculate the average speed for the second part of the journey.


Distance = 100 km
Time = 2 hours
Average Speed = 100 km / 2 h = 50 km/h

Step 3: Calculate the total distance and total time for the entire journey.


Total Distance = 150 km + 100 km = 250 km
Total Time = 3 h + 2 h = 5 h

Step 4: Calculate the total average speed.


Average Speed = 250 km / 5 h = 50 km/h

The total average speed for the entire journey is 50 km/h. Interestingly, the average speed remains the same for both parts of the journey, which is a special case where the speeds are equal.

Question 18:
A cyclist starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. Calculate the final velocity and the distance covered during this time.
Answer:

This problem involves uniformly accelerated motion. We use the equations of motion:


Step 1: Calculate the final velocity using v = u + at.


Initial velocity (u) = 0 m/s (since the cyclist starts from rest)
Acceleration (a) = 2 m/s²
Time (t) = 10 s
Final velocity (v) = 0 + (2 × 10) = 20 m/s

Step 2: Calculate the distance covered using s = ut + ½at².


Distance (s) = (0 × 10) + (½ × 2 × 10²)
s = 0 + (1 × 100) = 100 m

The cyclist achieves a final velocity of 20 m/s and covers a distance of 100 m in 10 seconds. This demonstrates how acceleration affects both velocity and displacement over time.

Question 19:
A car travels a distance of 300 km in 5 hours. Calculate its average speed. If the car stops for 30 minutes during the journey, how does this affect the average speed? Explain with calculations.
Answer:

To calculate the average speed, we use the formula:
Average Speed = Total Distance / Total Time.
Given: Total Distance = 300 km, Total Time = 5 hours.

Step 1: Calculate initial average speed without considering the stop.
Average Speed = 300 km / 5 h = 60 km/h.

Step 2: Account for the 30-minute (0.5 hours) stop.
Total Time = 5 h + 0.5 h = 5.5 hours.

Step 3: Recalculate average speed with the stop included.
Average Speed = 300 km / 5.5 h ≈ 54.55 km/h.

Conclusion: The stop reduces the average speed because the total time increases while the distance remains unchanged.

Question 20:
A cyclist covers the first half of a distance at 12 km/h and the second half at 16 km/h. Find the average speed for the entire journey. Justify your answer with proper reasoning.
Answer:

To find the average speed for the entire journey, we use the harmonic mean formula since equal distances are covered at different speeds:
Average Speed = 2v₁v₂ / (v₁ + v₂).
Given: v₁ = 12 km/h, v₂ = 16 km/h.

Step 1: Plug the values into the formula.
Average Speed = (2 × 12 × 16) / (12 + 16) = 384 / 28 ≈ 13.71 km/h.

Explanation: The average speed is not the arithmetic mean (14 km/h) because the cyclist spends more time at the slower speed (12 km/h). The harmonic mean accounts for this time-weighted effect.

Question 21:
A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. It then moves at a constant velocity for the next 20 seconds.

(i) Calculate the distance covered by the car in the first 10 seconds.

(ii) Find the total distance traveled by the car in 30 seconds.

Answer:

Given: Initial velocity (u) = 0 m/s, Acceleration (a) = 2 m/s², Time (t) = 10 s.

(i) Distance covered in first 10 seconds:


Using the equation of motion: s = ut + ½ at²
Substitute the values:
s = 0 × 10 + ½ × 2 × (10)²
s = 0 + 1 × 100
s = 100 meters.

(ii) Total distance in 30 seconds:


After 10 seconds, the car attains a constant velocity (v).


Using v = u + at:
v = 0 + 2 × 10 = 20 m/s.
Distance covered at constant velocity = v × time = 20 × 20 = 400 meters.
Total distance = 100 m (accelerated) + 400 m (uniform) = 500 meters.
Question 22:
A cyclist covers the first half of a distance at 10 km/h and the second half at 15 km/h.

(i) Calculate the average speed for the entire journey.

(ii) Explain why the average speed is not simply the arithmetic mean of the two speeds.

Answer:

(i) Average speed calculation:


Let total distance = 2d.
Time for first half (t₁) = d / 10.
Time for second half (t₂) = d / 15.
Total time = t₁ + t₂ = (d/10 + d/15) = d(1/10 + 1/15) = d(3/30 + 2/30) = 5d/30 = d/6.
Average speed = Total distance / Total time = 2d / (d/6) = 12 km/h.

(ii) Explanation:


Average speed is a harmonic mean (not arithmetic mean) because it depends on time taken for each segment. Since the cyclist spends more time at the lower speed (10 km/h), the average speed is weighted toward the slower value. Arithmetic mean (12.5 km/h) would ignore this time bias.

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