What is necessary to describe the position of an object?

The distance and direction from a reference point

Only the distance from a reference point

The distance and direction from a reference point

Only the direction from a reference point

Which of the following pairs correctly classifies physical quantities as scalar and vector?

Speed – scalar, Velocity – vector

Distance – vector, Displacement – scalar

Speed – scalar, Velocity – vector

Displacement – scalar, Average speed – vector

Average velocity – scalar, Distance – vector

A train moving on a straight track has its position recorded as: at t=0 s, 0 m; t=5 s, 150 m; t=10 s, 300 m; t=15 s, 450 m. Which statement is correct?

The train is in uniform motion with constant speed 30 m/s.

The train is in non-uniform motion with constant speed 30 m/s.

The train is in uniform motion with increasing speed.

The train is in uniform motion with constant speed 30 m/s.

Assertion (A): When a stone whirled in a circular path has constant speed, its acceleration is non-zero. Reason (R): Acceleration is produced only if there is a change in the magnitude of velocity.

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

The positions of a moving object at different times are: at t=0 s, position = 0 m; t=2 s, position = 10 m; t=4 s, position = 30 m; t=6 s, position = 30 m; t=8 s, position = 10 m. Determine the time interval during which the object is at rest, and calculate the average velocity between t=2 s and t=4 s.

Object is at rest between t=4 s and t=6 s (position unchanged). Average velocity from t=2 s to t=4 s = 10 m/s in the positive direction.

Two runners, A and B, start a race on a straight track from the same point. Runner A maintains a steady speed of 8 m/s for 40 seconds. Runner B runs at 6 m/s for the first 25 seconds and then increases to 10 m/s for the next 15 seconds. (a) Who covers more distance in the 40-second period? (b) What is the difference in distances covered? (c) What can you say about the nature of their motions?

(a) Runner A covers more distance. (b) Difference = 20 m. (c) Runner A’s motion is uniform; Runner B’s motion is non‑uniform.

Explain the concept of displacement. A cyclist rides 500 m north, then turns around and rides 300 m south. Determine the total distance covered and the magnitude of her displacement. Provide the direction of displacement.

Total distance = 500 m + 300 m = 800 m. Magnitude of displacement = 500 m - 300 m = 200 m, direction north.

The distance covered by a car moving on a straight road at various instants of time is recorded as follows: at t=0 s, 0 m; t=2 s, 10 m; t=4 s, 20 m; t=6 s, 30 m; t=8 s, 40 m; t=10 s, 50 m. Plot a distance-time graph for this motion (use a sketch) and determine whether the motion is uniform or non-uniform. Also calculate the average speed of the car.

The distance-time graph is a straight line passing through the origin, indicating uniform motion (constant speed). Average speed = total distance / total time = 50 m / 10 s = 5 m/s.

Differentiate between distance and displacement. Provide two points of difference and give one example each where (i) both have the same magnitude, and (ii) they have different magnitudes.

Distance is the total length of the path covered by an object, while displacement is the shortest distance between the initial and final positions along with direction. Two differences: (1) Distance is a scalar quantity having only magnitude, whereas displacement is a vector having both magnitude and direction. (2) Distance can never be zero for a moving object, but displacement can be zero if the object returns to the starting point. Example (i): When a car moves 100 m north on a straight road without turning, distance and displacement both are 100 m north. Example (ii): A car moves 100 m north and then 50 m south, total distance is 150 m but displacement is 50 m north.

A particle moves along a straight line such that its position x (in metres) at time t (in seconds) is given by x = 4t - t². The particle moves for the time interval t = 0 s to t = 4 s. (1) Find the initial velocity of the particle. (2) At what time does the particle reverse its direction of motion? (3) Calculate the total distance travelled by the particle in the given interval.

Initial velocity = 4 m/s. Particle reverses at t = 2 s. Total distance travelled = 8 m.

Class 9 Science Chapter 4: Describing Motion Around Us — Important Questions & Sample Paper

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Chapter 4 'Describing Motion Around Us' lays the foundation of kinematics by introducing the basic concepts needed to describe the motion of objects. It begins by differentiating between rest and motion, emphasizing the need for a reference point. The chapter then clarifies two fundamental ideas: distance as the total path length (a scalar) and displacement as the shortest straight-line distance between start and end points (a vector). Students learn to calculate average speed as total distance divided by time and average velocity as displacement divided by time. The concept of acceleration is introduced for non-uniform motion, along with uniform acceleration. Graphical representation of motion is a key focus: plotting and interpreting distance-time and velocity-time graphs teaches students how to extract information like speed, displacement, and acceleration. The chapter also derives the three equations of motion for objects moving with constant acceleration along a straight line, and concludes with uniform circular motion as an example of accelerated motion despite constant speed. Typical exam questions range from straightforward calculations of average speed and velocity to application-based problems combining multiple directions, such as a person walking east then west, or a vehicle moving north then east. Conceptual questions often ask to distinguish distance and displacement with real-life examples, and graph-based questions test the ability to read slopes and areas under curves.

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Science — Describing Motion Around Us

Class 9Time: 3 hrsMax Marks: 80

SECTION A

1.
What is necessary to describe the position of an object?
(a) Only the distance from a reference point(b) The distance and direction from a reference point(c) Only the direction from a reference point(d) The speed of the object
1
2.
Which of the following pairs correctly classifies physical quantities as scalar and vector?
(a) Distance – vector, Displacement – scalar(b) Speed – scalar, Velocity – vector(c) Displacement – scalar, Average speed – vector(d) Average velocity – scalar, Distance – vector
1
3.
If a reference point is chosen and an object's position relative to it does not change with time, the object is said to be:
(a) in motion(b) accelerating(c) at rest(d) in uniform motion
1

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In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

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Q1. What is necessary to describe the position of an object?
1 mark
Multiple Choice (MCQ)
(A) Only the distance from a reference point(B) The distance and direction from a reference point(C) Only the direction from a reference point(D) The speed of the object
▸ Answer▾ Answer
The distance and direction from a reference point
Q2. Which of the following pairs correctly classifies physical quantities as scalar and vector?
1 mark
Multiple Choice (MCQ)
(A) Distance – vector, Displacement – scalar(B) Speed – scalar, Velocity – vector(C) Displacement – scalar, Average speed – vector(D) Average velocity – scalar, Distance – vector
▸ Answer▾ Answer
Speed – scalar, Velocity – vector
Q3. If a reference point is chosen and an object's position relative to it does not change with time, the object is said to be:
1 mark
Multiple Choice (MCQ)
(A) in motion(B) accelerating(C) at rest(D) in uniform motion
▸ Answer▾ Answer
at rest
Q4. A train moving on a straight track has its position recorded as: at t=0 s, 0 m; t=5 s, 150 m; t=10 s, 300 m; t=15 s, 450 m. Which statement is correct?
1 mark
Multiple Choice (MCQ)
(A) The train is in non-uniform motion with constant speed 30 m/s.(B) The train is in uniform motion with increasing speed.(C) The train is in uniform motion with constant speed 30 m/s.(D) The train is at rest.
▸ Answer▾ Answer
The train is in uniform motion with constant speed 30 m/s.
Q5. Assertion (A): When a stone whirled in a circular path has constant speed, its acceleration is non-zero. Reason (R): Acceleration is produced only if there is a change in the magnitude of velocity.
1 mark
Assertion–Reason
(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.
▸ Answer▾ Answer
A is true but R is false.
Q6. A particle travels from point A to point B in a straight line with a constant speed of 20 m/s. It immediately returns from B to A with a constant speed of 30 m/s. Calculate the average speed for the entire round trip. (Hint: The average speed is not simply the average of the two speeds.)
2 marks
Short Answer
▸ Answer▾ Answer
24 m/s
Q7. The positions of a moving object at different times are: at t=0 s, position = 0 m; t=2 s, position = 10 m; t=4 s, position = 30 m; t=6 s, position = 30 m; t=8 s, position = 10 m. Determine the time interval during which the object is at rest, and calculate the average velocity between t=2 s and t=4 s.
2 marks
Short Answer
▸ Answer▾ Answer
Object is at rest between t=4 s and t=6 s (position unchanged). Average velocity from t=2 s to t=4 s = 10 m/s in the positive direction.
Q8. Two runners, A and B, start a race on a straight track from the same point. Runner A maintains a steady speed of 8 m/s for 40 seconds. Runner B runs at 6 m/s for the first 25 seconds and then increases to 10 m/s for the next 15 seconds. (a) Who covers more distance in the 40-second period? (b) What is the difference in distances covered? (c) What can you say about the nature of their motions?
3 marks
Short Answer
▸ Answer▾ Answer
(a) Runner A covers more distance. (b) Difference = 20 m. (c) Runner A’s motion is uniform; Runner B’s motion is non‑uniform.
Q9. Explain the concept of displacement. A cyclist rides 500 m north, then turns around and rides 300 m south. Determine the total distance covered and the magnitude of her displacement. Provide the direction of displacement.
3 marks
Short Answer
▸ Answer▾ Answer
Total distance = 500 m + 300 m = 800 m. Magnitude of displacement = 500 m - 300 m = 200 m, direction north.
Q10. The distance covered by a car moving on a straight road at various instants of time is recorded as follows: at t=0 s, 0 m; t=2 s, 10 m; t=4 s, 20 m; t=6 s, 30 m; t=8 s, 40 m; t=10 s, 50 m. Plot a distance-time graph for this motion (use a sketch) and determine whether the motion is uniform or non-uniform. Also calculate the average speed of the car.
5 marks
Long Answer
▸ Answer▾ Answer
The distance-time graph is a straight line passing through the origin, indicating uniform motion (constant speed). Average speed = total distance / total time = 50 m / 10 s = 5 m/s.
Q11. Differentiate between distance and displacement. Provide two points of difference and give one example each where (i) both have the same magnitude, and (ii) they have different magnitudes.
5 marks
Long Answer
▸ Answer▾ Answer
Distance is the total length of the path covered by an object, while displacement is the shortest distance between the initial and final positions along with direction. Two differences: (1) Distance is a scalar quantity having only magnitude, whereas displacement is a vector having both magnitude and direction. (2) Distance can never be zero for a moving object, but displacement can be zero if the object returns to the starting point. Example (i): When a car moves 100 m north on a straight road without turning, distance and displacement both are 100 m north. Example (ii): A car moves 100 m north and then 50 m south, total distance is 150 m but displacement is 50 m north.
Q12. A particle moves along a straight line such that its position x (in metres) at time t (in seconds) is given by x = 4t - t². The particle moves for the time interval t = 0 s to t = 4 s.
4 marks
Case Study
1. (i) Find the initial velocity of the particle.1 mark
2. (ii) At what time does the particle reverse its direction of motion?1 mark
3. (iii) Calculate the total distance travelled by the particle in the given interval.2 marks
▸ Answer▾ Answer
Initial velocity = 4 m/s. Particle reverses at t = 2 s. Total distance travelled = 8 m.

Frequently asked questions

What is the difference between distance and displacement, and why is it important in motion problems?

Distance is the total path length covered, a scalar quantity with only magnitude. Displacement is the shortest straight-line distance from the initial to final position, a vector quantity with magnitude and direction. This distinction is crucial because many exam questions test whether you use distance or displacement to compute speed (scalar) or velocity (vector). For example, if an object moves in a closed loop, distance is non-zero but displacement is zero, leading to average velocity being zero while average speed is not.

How do I correctly calculate average speed and average velocity when the motion involves multiple legs with different directions?

Average speed = total distance traveled / total time taken. Average velocity = total displacement / total time. For multiple segments, first draw a diagram. Add all segment lengths for total distance. For displacement, consider the vector sum: if segments are along the same line but opposite directions, subtract; if perpendicular, use the Pythagorean theorem. Then divide displacement magnitude by total time, and specify direction for velocity.

What are the typical graph-based questions in this chapter, and what should I look for?

You'll see distance-time and velocity-time graphs. For distance-time, the slope gives speed; a straight line indicates uniform motion, a curved line indicates non-uniform motion. For velocity-time, the slope gives acceleration, and the area under the graph gives displacement. Common questions ask to calculate speed from a distance-time slope, or find acceleration and displacement from a velocity-time graph. Practice interpreting the shape: horizontal line on velocity-time means constant velocity (zero acceleration), sloping line means uniform acceleration.

How can I avoid losing marks in numerical problems from this chapter?

Always list given data with proper units. Convert all units to SI (m, s, m/s) before plugging into formulas. Clearly distinguish between distance and displacement, speed and velocity. For average velocity, you must compute net displacement, not total distance. Use vector addition correctly. Lastly, watch out for conversions: minutes to seconds, km/h to m/s by multiplying by 5/18.

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