Class 9 · ScienceNCERT 2026 Revised Syllabus
Chapter 4: Describing Motion Around Us (0906CH04)
Exploration · Displacement, velocity, acceleration, equations of motion.
Chapter code: 0906ch04 · Book: Exploration · Class 9 Science
Find complete NCERT solutions for Class 9 Science, Exploration Chapter 4. Every question has a clear, step-by-step answer for chapter code 0906CH04.
Pause and Ponder(5 questions)
1
When will the displacement be zero for an athlete running back and forth on a straight track? What is the total distance in that case?
School Level Solution
Ans.
Displacement is zero when the athlete returns to the starting point after running to the end and back. If track length = d, total distance = 2d while displacement = 0.
2
Fuel used in a vehicle depends on which: (i) Total distance travelled; (ii) Displacement?
School Level Solution
Ans.
Fuel consumption depends on (i) Total distance travelled. Fuel is burned for every metre of actual movement, regardless of direction. Displacement measures only straight-line change in position.
3
A ball rolls down an inclined track O to D. Is its motion a straight line? Are total distance and displacement equal at A, B, C, D?
School Level Solution
Ans.
The ball follows the curved inclined path, not a true straight line. At intermediate positions A, B, C: total distance > displacement (path curves). Only near flat portion D might they be approximately equal.
4
Drive 200 km north in 3 h, then 200 km south in 2 h. Find average speed and average velocity for the trip.
School Level Solution
Ans.
Total distance = 200 + 200 = 400 km; Total time = 3 + 2 = 5 h
Average speed = 400 ÷ 5 = 80 km h⁻¹
Total displacement = 200N – 200S = 0; Average velocity = 0 km h⁻¹
Average speed = 400 ÷ 5 = 80 km h⁻¹
Total displacement = 200N – 200S = 0; Average velocity = 0 km h⁻¹
5
Under what conditions is (i) magnitude of average velocity equal to average speed; (ii) average velocity zero while average speed is non-zero?
School Level Solution
Ans.
(i) When object moves in a straight line without reversing direction, total distance = total displacement.
(ii) When object returns to starting point, displacement = 0 → average velocity = 0; but distance ≠ 0 → average speed ≠ 0.
(ii) When object returns to starting point, displacement = 0 → average velocity = 0; but distance ≠ 0 → average speed ≠ 0.
Activities — How to Do(4 questions)
1
Activity 4.1 (ball thrown upwards): a ball goes straight up from O to B and falls back to O. Is this motion in a straight line? Fill Table 4.1 (distance vs displacement at O, A, B, C, O) and choose the true statement about displacement.
School Level Solution
Ans.
Up-and-down along one line is still straight-line motion. Distance keeps accumulating (O→B→O = twice the height); displacement rises to max at B then falls back to zero at O. Hence displacement is never more than distance — option (iii): magnitude is less than or equal to total distance.
2
Activity 4.2 (car acceleration): look up the 0–100 km/h time for various cars and compute average acceleration. How is it calculated, and what are the units?
School Level Solution
Ans.
Note each car’s 0–100 km/h time T from the internet. Convert: 100 km/h = 27.8 m s⁻¹. Average acceleration a = 27.8 ÷ T m s⁻². Example: T = 10 s → a ≈ 2.78 m s⁻². Shorter T means sportier (larger) acceleration.
3
Activity 4.3 (plot a position–time graph): draw axes on graph paper, put time on X and position on Y, choose a scale (e.g., 5 divisions = 1 s; 5 divisions = 20 m), plot Table 4.3 and join the points. What shape do you get, and what does it mean?
School Level Solution
Ans.
Mark time (1 s, 2 s …) on X and position (20 m, 40 m …) on Y; plot each pair and join. The points form a straight line (Fig. 4.11c) — position grows uniformly with time, meaning constant (uniform) velocity.
4
Activity 4.4 (velocity from the graph): on segment AB draw lines parallel to the axes to form triangle ABC. What do sides BC and CA represent, and how do you get average velocity?
School Level Solution
Ans.
BC (vertical) = change in position (s₂ − s₁); CA (horizontal) = change in time (t₂ − t₁). Read s₁, s₂, t₁, t₂ off the graph; average velocity v = (s₂ − s₁) ÷ (t₂ − t₁) — the slope of AB. Steeper slope means faster motion.