Class 9 Science Important Questions and Answers: Describing Motion Around Us, Kinematic Equations and Numericals
Premium NCERT, Board Exam, Exemplar and Olympiad-level question answers for Class 9 Science Motion. Learn distance, displacement, speed, velocity, acceleration, graphs, equations of motion and numerical problem-solving.
Short Introduction
Motion is one of the most important chapters in Class 9 Physics. It explains how objects change their position with time and how this change can be measured using distance, displacement, speed, velocity and acceleration.
This chapter also introduces graphical representation of motion and three important kinematic equations used for solving numerical problems based on uniformly accelerated motion.
Scoring area: Definitions, differences, graph-based questions and numerical problems from equations of motion are very important for CBSE exams.
Chapter Overview
1. Motion and Rest
An object is in motion if its position changes with time with respect to a reference point.
2. Distance and Displacement
Distance is the total path length, while displacement is the shortest distance between initial and final positions.
3. Speed and Velocity
Speed is distance travelled per unit time. Velocity is displacement per unit time in a given direction.
4. Acceleration
Acceleration is the rate of change of velocity with time.
5. Graphs of Motion
Distance-time and velocity-time graphs help describe motion visually.
6. Equations of Motion
Three kinematic equations are used for uniformly accelerated motion.
Important Keywords
Important Formulas
Speed
Speed = Distance ÷ Time
Average Speed
Average speed = Total distance ÷ Total time
Velocity
Velocity = Displacement ÷ Time
Acceleration
a = (v – u) ÷ t
First Equation of Motion
v = u + at
Second Equation of Motion
s = ut + 1/2 at2
Third Equation of Motion
v2 – u2 = 2as
Distance from Velocity-Time Graph
Distance = Area under velocity-time graph
Symbols: u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement or distance covered in a given direction.
Important Very Short Answer Questions
Q1. What is motion?
Answer: An object is said to be in motion if its position changes with time with respect to a reference point.
Q2. What is rest?
Answer: An object is said to be at rest if its position does not change with time with respect to a reference point.
Q3. Define distance.
Answer: Distance is the actual length of the path travelled by an object.
Q4. Define displacement.
Answer: Displacement is the shortest distance between the initial and final positions of an object in a given direction.
Q5. Is distance a scalar or vector quantity?
Answer: Distance is a scalar quantity because it has only magnitude and no direction.
Q6. Is displacement a scalar or vector quantity?
Answer: Displacement is a vector quantity because it has both magnitude and direction.
Q7. Define speed.
Answer: Speed is the distance travelled by an object per unit time.
Q8. Define velocity.
Answer: Velocity is the displacement of an object per unit time in a specified direction.
Q9. Define acceleration.
Answer: Acceleration is the rate of change of velocity with time.
Q10. What is retardation?
Answer: Retardation is negative acceleration. It occurs when the velocity of an object decreases with time.
Q11. What is the SI unit of speed?
Answer: The SI unit of speed is metre per second, written as m/s.
Q12. What is the SI unit of acceleration?
Answer: The SI unit of acceleration is metre per second square, written as m/s2.
Short Answer Questions
Q1. Differentiate between distance and displacement.
Answer:
| Distance | Displacement |
|---|---|
| It is the actual path length travelled by an object. | It is the shortest distance between initial and final positions. |
| It is a scalar quantity. | It is a vector quantity. |
| It can never be negative. | It can be positive, negative or zero. |
| Distance is always greater than or equal to displacement. | Displacement can be less than or equal to distance. |
Q2. Differentiate between speed and velocity.
Answer:
| Speed | Velocity |
|---|---|
| Speed is distance travelled per unit time. | Velocity is displacement per unit time. |
| It is a scalar quantity. | It is a vector quantity. |
| It has only magnitude. | It has both magnitude and direction. |
| Speed can never be negative. | Velocity can be positive, negative or zero. |
Q3. What is uniform motion? Give one example.
Answer: An object is said to be in uniform motion if it covers equal distances in equal intervals of time. Example: a car moving at a constant speed on a straight road.
Q4. What is non-uniform motion? Give one example.
Answer: An object is said to be in non-uniform motion if it covers unequal distances in equal intervals of time. Example: a bus moving in city traffic.
Q5. What does the slope of a distance-time graph represent?
Answer: The slope of a distance-time graph represents the speed of the object.
Q6. What does the slope of a velocity-time graph represent?
Answer: The slope of a velocity-time graph represents the acceleration of the object.
Q7. What does the area under a velocity-time graph represent?
Answer: The area under a velocity-time graph represents the distance or displacement covered by the object.
Q8. Can displacement be zero even when distance is not zero?
Answer: Yes. If an object returns to its starting point, its displacement is zero, but the distance travelled is not zero.
Q9. Why is velocity called a vector quantity?
Answer: Velocity is called a vector quantity because it has both magnitude and direction.
Q10. A body is moving with constant velocity. What is its acceleration?
Answer: If a body is moving with constant velocity, its acceleration is zero because there is no change in velocity.
Long Answer Questions
Q1. Explain the three equations of motion for uniformly accelerated motion.
Answer: The three equations of motion are used when an object moves with uniform acceleration.
- First equation: v = u + at
- Second equation: s = ut + 1/2 at2
- Third equation: v2 – u2 = 2as
Here, u is initial velocity, v is final velocity, a is acceleration, t is time and s is displacement.
Q2. Describe the nature of distance-time graphs for different types of motion.
Answer:
- A straight line parallel to the time axis shows that the object is at rest.
- A straight sloping line shows uniform motion.
- A curved line shows non-uniform motion.
- A steeper line shows greater speed.
- The slope of the distance-time graph gives speed.
Q3. Describe the nature of velocity-time graphs for different types of motion.
Answer:
- A straight line parallel to the time axis shows constant velocity.
- A straight line sloping upward shows uniform acceleration.
- A straight line sloping downward shows retardation or negative acceleration.
- The slope of a velocity-time graph gives acceleration.
- The area under the velocity-time graph gives distance or displacement.
Q4. Why are equations of motion not used for all types of motion?
Answer:
- The standard equations of motion are valid only for uniformly accelerated motion.
- Uniform acceleration means acceleration remains constant with time.
- If acceleration changes continuously, these equations cannot be directly applied.
- For non-uniform acceleration, more advanced methods are needed.
Q5. Explain how circular motion can be accelerated motion even when speed remains constant.
Answer:
- In uniform circular motion, the speed of the object remains constant.
- However, the direction of motion changes continuously.
- Velocity depends on both magnitude and direction.
- Since direction changes, velocity changes.
- Therefore, circular motion is accelerated motion even when speed is constant.
Important Numericals with Solutions
Numerical 1: A car travels 120 km in 3 hours. Find its average speed.
Solution:
Distance = 120 km
Time = 3 h
Average speed = Total distance ÷ Total time
Average speed = 120 ÷ 3 = 40 km/h
Numerical 2: A body starts from rest and reaches a velocity of 20 m/s in 5 s. Find acceleration.
Solution:
u = 0 m/s, v = 20 m/s, t = 5 s
a = (v – u) ÷ t
a = (20 – 0) ÷ 5 = 4 m/s2
Numerical 3: A train moving at 10 m/s accelerates uniformly at 2 m/s2 for 5 s. Find final velocity.
Solution:
u = 10 m/s, a = 2 m/s2, t = 5 s
v = u + at
v = 10 + 2 × 5
v = 10 + 10 = 20 m/s
Numerical 4: A body starts from rest and accelerates at 3 m/s2 for 4 s. Find the distance travelled.
Solution:
u = 0 m/s, a = 3 m/s2, t = 4 s
s = ut + 1/2 at2
s = 0 × 4 + 1/2 × 3 × 42
s = 1/2 × 3 × 16 = 24 m
Distance travelled = 24 m
Numerical 5: A car moving with velocity 15 m/s is brought to rest in 5 s. Find retardation.
Solution:
u = 15 m/s, v = 0 m/s, t = 5 s
a = (v – u) ÷ t
a = (0 – 15) ÷ 5 = -3 m/s2
Retardation = 3 m/s2
Numerical 6: A body moving with initial velocity 5 m/s accelerates at 2 m/s2 and covers 60 m. Find final velocity.
Solution:
u = 5 m/s, a = 2 m/s2, s = 60 m
v2 – u2 = 2as
v2 – 52 = 2 × 2 × 60
v2 – 25 = 240
v2 = 265
v = √265 = 16.28 m/s approximately
Numerical 7: A scooter covers 200 m in 20 s. Find its speed in m/s.
Solution:
Distance = 200 m, Time = 20 s
Speed = Distance ÷ Time
Speed = 200 ÷ 20 = 10 m/s
Numerical 8: A body has initial velocity 12 m/s and final velocity 4 m/s in 4 s. Find acceleration.
Solution:
u = 12 m/s, v = 4 m/s, t = 4 s
a = (v – u) ÷ t
a = (4 – 12) ÷ 4 = -2 m/s2
Acceleration = -2 m/s2
The negative sign shows retardation.
Case-Study Based Questions
Case Study 1: Journey of a Car
A car travels 40 km towards the east and then 30 km towards the west. The total time taken is 2 hours. The car does not return to its starting point completely.
Q1. What is the total distance travelled by the car?
Answer: Total distance = 40 km + 30 km = 70 km.
Q2. What is the displacement of the car?
Answer: Displacement = 40 km – 30 km = 10 km towards east.
Q3. Find the average speed of the car.
Answer: Average speed = Total distance ÷ Total time = 70 ÷ 2 = 35 km/h.
Q4. Is distance greater than displacement in this case?
Answer: Yes. Distance is 70 km, while displacement is only 10 km towards east.
Case Study 2: Motion of a Train
A train starts from rest and moves with uniform acceleration. After 10 seconds, its velocity becomes 30 m/s. The motion of the train can be studied using equations of motion.
Q1. What is the initial velocity of the train?
Answer: Since the train starts from rest, initial velocity u = 0 m/s.
Q2. Find the acceleration of the train.
Answer: a = (v – u) ÷ t = (30 – 0) ÷ 10 = 3 m/s2.
Q3. Which equation is used to find final velocity?
Answer: The equation v = u + at is used to find final velocity.
Q4. What type of acceleration is shown here?
Answer: The train shows uniform acceleration because acceleration remains constant.
Case Study 3: Graph-Based Motion
A velocity-time graph of an object is a straight line parallel to the time axis. The velocity of the object remains 12 m/s for 8 seconds.
Q1. What type of motion is represented by the graph?
Answer: The graph represents uniform motion with constant velocity.
Q2. What is the acceleration of the object?
Answer: Acceleration is zero because velocity is constant.
Q3. Find the distance covered in 8 seconds.
Answer: Distance = Velocity × Time = 12 × 8 = 96 m.
Q4. What does the area under a velocity-time graph represent?
Answer: It represents distance or displacement covered by the object.
Assertion-Reason Questions
Choose the correct option:
A. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
C. Assertion is true but Reason is false.
D. Assertion is false but Reason is true.
Q1. Assertion: Distance can never be negative.
Reason: Distance is the actual length of the path travelled by an object.
Answer: A. Both are true and Reason correctly explains Assertion.
Q2. Assertion: Displacement can be zero even if distance is not zero.
Reason: If an object returns to its starting point, final position and initial position are the same.
Answer: A. Both are true and Reason correctly explains Assertion.
Q3. Assertion: Speed is a vector quantity.
Reason: Speed has only magnitude and no direction.
Answer: D. Assertion is false but Reason is true.
Q4. Assertion: Slope of a velocity-time graph gives acceleration.
Reason: Acceleration is the rate of change of velocity with time.
Answer: A. Both are true and Reason correctly explains Assertion.
Q5. Assertion: In uniform circular motion, velocity remains constant.
Reason: Direction of motion changes continuously in circular motion.
Answer: D. Assertion is false but Reason is true.
Q6. Assertion: Equations of motion are used for uniformly accelerated motion.
Reason: In uniformly accelerated motion, acceleration remains constant.
Answer: A. Both are true and Reason correctly explains Assertion.
Exam Tips
- Always write the given values before solving a numerical.
- Convert km/h into m/s when required by multiplying by 5/18.
- Convert m/s into km/h by multiplying by 18/5.
- Use correct SI units: speed in m/s and acceleration in m/s2.
- Remember that negative acceleration means retardation.
- For graph questions, mention slope and area under the graph.
- Use equations of motion only when acceleration is uniform.
- In difference questions, write answers in table format for better marks.
Quick Revision Box
Motion
Change in position with time with respect to a reference point.
Distance
Total path length covered by an object.
Displacement
Shortest distance between initial and final positions in a direction.
Speed
Distance travelled per unit time.
Velocity
Displacement per unit time in a given direction.
Acceleration
Rate of change of velocity with time.
Distance-Time Graph
Slope gives speed.
Velocity-Time Graph
Slope gives acceleration and area gives displacement.
Equations of Motion
Used only for uniformly accelerated motion.
FAQ Section
What is the main focus of Class 9 Science Motion?
The main focus is to understand motion using distance, displacement, speed, velocity, acceleration, graphs and equations of motion.
What are the three equations of motion?
The three equations of motion are v = u + at, s = ut + 1/2 at² and v² – u² = 2as.
When can equations of motion be used?
Equations of motion can be used only when the object moves with uniform acceleration.
What is the difference between distance and displacement?
Distance is the actual path travelled, while displacement is the shortest distance between initial and final positions in a given direction.
What does the slope of a distance-time graph show?
The slope of a distance-time graph shows speed.
What does the slope of a velocity-time graph show?
The slope of a velocity-time graph shows acceleration.
What does the area under a velocity-time graph represent?
The area under a velocity-time graph represents distance or displacement.
What is retardation?
Retardation is negative acceleration. It occurs when velocity decreases with time.
How can km/h be converted into m/s?
To convert km/h into m/s, multiply the value by 5/18.
How can m/s be converted into km/h?
To convert m/s into km/h, multiply the value by 18/5.
Final Conclusion
The chapter Describing Motion Around Us builds the foundation of Physics for Class 9 students. To score well, students should master definitions, differences, formulas, graphs and numerical problems. The most important areas are distance-displacement, speed-velocity, acceleration, velocity-time graph and the three equations of motion.
Regular practice of numericals and graph-based questions will help students perform strongly in NCERT questions, CBSE board exams, Exemplar-level questions and Olympiad-level reasoning.
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