Question Bank — Describing Motion Around Us

NCERT Class 9 Science · Chapter 4 · CBSE Board Pattern

Class 9 Science Chapter 4 Question Bank CBSE Board Pattern

QUESTION BANK

CBSE Class 09 Science – Chapter 4: Describing Motion Around Us
(Strictly based on NCERT text)

Section A — MCQs (10 questions, 1 mark each)

Q1. Which of the following physical quantities requires both magnitude and direction for complete description?
(a) Distance travelled
(b) Average speed
(c) Displacement
(d) Time interval

Q2. An athlete runs from O to A (100 m) and then back to B (40 m from O) in 16 s. The magnitude of displacement is
(a) 160 m
(b) 100 m
(c) 60 m
(d) 40 m

Q3. For an object moving in a straight line in one direction,
(a) magnitude of displacement is always greater than distance travelled
(b) distance travelled and magnitude of displacement are equal
(c) average speed is always equal to magnitude of average velocity
(d) both (b) and (c)

Q4. The SI unit of average acceleration is
(a) m s⁻¹
(b) m s⁻²
(c) m
(d) s

Q5. Assertion (A): In uniform circular motion, the speed of the object remains constant.
Reason (R): In uniform circular motion, the direction of velocity changes continuously.
(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.

Q6. A position-time graph that is a straight line parallel to the time axis indicates
(a) uniform motion
(b) accelerated motion
(c) object at rest
(d) non-uniform motion

Q7. The slope of a velocity-time graph gives
(a) displacement
(b) acceleration
(c) average speed
(d) distance travelled

Q8. Which of the following is true for the kinematic equations?
(a) They are valid only for motion with constant acceleration.
(b) They can be used when acceleration is changing.
(c) They relate only distance and time.
(d) They are valid only for circular motion.

Q9. Assertion (A): Average velocity of an object can be zero while its average speed is not zero.
Reason (R): Displacement can be zero while distance travelled is not zero.
(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.

Q10. In the velocity-time graph of an object moving with constant acceleration, the area enclosed between the line and the time axis represents
(a) acceleration
(b) velocity
(c) displacement
(d) speed

Section B — Very Short Answer (6 questions, 2 marks each)

Q11. Define displacement. How is it different from distance travelled? Give one example from the chapter where both are different.

Q12. An object is thrown vertically upwards from O, reaches B and falls back to O. What is its displacement when it returns to O? What is the total distance travelled?

Q13. State the condition under which the magnitude of average velocity equals average speed for motion in a straight line.

Q14. What does a curved position-time graph indicate about the motion of an object?

Q15. Write the three kinematic equations for motion in a straight line with constant acceleration.

Q16. Why is the motion of an object in uniform circular motion said to be accelerated even though its speed is constant?

Section C — Short Answer (5 questions, 3 marks each)

Q17. A ball rolls down an inclined track. Is its motion a straight-line motion? At different positions, compare the total distance travelled and the magnitude of displacement from the starting point O.

Q18. Sarang swims from one end of a 25 m pool to the other end and back in 50 s. Calculate his average speed and average velocity. What does the difference between the two values indicate?

Q19. A bus moving at 36 km h⁻¹ accelerates uniformly for 10 s and reaches 54 km h⁻¹. Calculate the magnitude of its average acceleration. In which direction does the acceleration act?

Q20. Draw a velocity-time graph for an object moving with constant velocity. From the graph, show how displacement can be calculated.

Q21. An object is dropped from a height. Its velocity increases by 9.8 m s⁻¹ in every successive second. Show that the average acceleration is constant and state its direction.

Section D — Long Answer (3 questions, 5 marks each)

Q22. (a) Define average speed and average velocity. Write their mathematical expressions.
(b) A car travels 200 km north in 3 h and then 200 km south in 2 h. Calculate average speed and average velocity for the entire trip.
(c) Under what condition is the magnitude of average velocity equal to average speed? (5 marks)

Q23. (a) What information can be obtained from a position-time graph?
(b) Draw a labelled position-time graph for an object moving with (i) constant velocity and (ii) changing velocity.
(c) Explain how average velocity is calculated from the slope of the position-time graph. (5 marks)

Q24. A car starts from rest with uniform acceleration and attains a velocity of 20 m s⁻¹ in 5 s. It then moves with this constant velocity for 10 s. Finally, brakes are applied and it stops in 6 s with uniform retardation.
(a) Draw a velocity-time graph for the entire motion.
(b) Calculate the total distance travelled using the graph (show working).
(c) Calculate the magnitude of acceleration and retardation. (5 marks)

Section E — Case/Source-Based (2 questions, 4 marks each)

Q25. Read the passage and answer the questions that follow:
“An athlete starts from O at t = 0 s, reaches B (40 m) at t = 4 s, then A (100 m) at t = 10 s and returns to B at t = 16 s. The total distance travelled is 160 m while displacement is 40 m in the positive direction.”

(i) Define displacement and state its magnitude and direction in the above case. (1 mark)
(ii) Calculate the average speed of the athlete between t = 0 s and t = 16 s. (1 mark)
(iii) Calculate the average velocity of the athlete between t = 0 s and t = 16 s. (1 mark)
(iv) Why are distance travelled and displacement not equal in this case? (1 mark)

Q26. Read the passage and answer the questions that follow:
“A vehicle is moving on a straight road. Its positions at different instants are: 0 m at 0 s, 20 m at 1 s, 40 m at 2 s, … up to 120 m at 6 s. The position-time graph is a straight line.”

(i) What does the straight-line position-time graph indicate about the motion? (1 mark)
(ii) Calculate the magnitude of velocity of the vehicle from the graph. (1 mark)
(iii) If the vehicle had been at rest at 40 m, how would the graph appear? (1 mark)
(iv) State one advantage of representing motion graphically. (1 mark)

▶ Answer Key Attempt all questions first,
then tap to reveal

Section A

  1. (c) — definition of vector quantity — 1 mark
  2. (d) — magnitude of displacement = OB = 40 m — 1 mark
  3. (d) — Note in NCERT — 1 mark
  4. (b) — SI unit of acceleration — 1 mark
  5. (a) — Both true; R explains constant speed with changing direction — 1 mark
  6. (c) — position not changing with time — 1 mark
  7. (b) — slope of v-t graph = acceleration — 1 mark
  8. (a) — valid only for constant acceleration — 1 mark
  9. (a) — Both true; R explains zero displacement with non-zero distance — 1 mark
  10. (c) — area under v-t graph = displacement — 1 mark

Section B

  1. Net change in position (magnitude + direction) — 1 mark; different from total path length (scalar) — 1 mark (example: athlete returning to B)
  2. Displacement = 0 — 1 mark; distance = 2 × OB — 1 mark
  3. When object moves in one direction only — 2 marks
  4. Velocity is changing (accelerated motion) — 2 marks
  5. v = u + at; s = ut + ½at²; v² = u² + 2as — 2 marks
  6. Direction of velocity changes continuously — 2 marks

Section C

  1. Yes, straight-line motion — 1 mark; distance > |displacement| at intermediate points — 2 marks
  2. Average speed = 1 m s⁻¹ — 1 mark; average velocity = 0 — 1 mark; difference shows motion in both directions — 1 mark
  3. a = 0.5 m s⁻² in direction of velocity — 3 marks (full working)
  4. Horizontal line; displacement = area under graph — 3 marks
  5. a = 9.8 m s⁻² constant in direction of motion — 3 marks

Section D

  1. Definitions + expressions (2 marks); calculation: speed = 80 km h⁻¹, velocity = 0 (2 marks); condition (1 mark)
  2. Information from graph (1 mark); two labelled graphs (2 marks); slope explanation (2 marks)
  3. Labelled v-t graph (2 marks); total distance = 260 m (working shown) (2 marks); a = 4 m s⁻², retardation = 10/3 m s⁻² (1 mark)

Section E

  1. (i) 40 m positive direction (1); (ii) 10 m s⁻¹ (1); (iii) 2.5 m s⁻¹ (1); (iv) object turned back (1)
  2. (i) Constant velocity (1); (ii) 20 m s⁻¹ (1); (iii) horizontal line at 40 m (1); (iv) visual comparison of motions (1)
← Revision Summary All Chapters

All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.