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Kriti
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Asked: 2026-08-20T04:31:34+00:00 In:

The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.

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For constant velocity, the area from 20–100 s is a rectangle: 3 × 80 = 240 m. For decreasing velocity, area from 100–120 s = ½(3+2)×20 = 50 m. Total displacement = 320 m; average acceleration = (2−0)/120 = 0.0167 m s⁻².

calculate the displacement and average acceleration in the 120 s time interval.cbse class 9 science exploration chapter 4 describing motion around us solutionsclass 9 science exploration chapter 4 question answerclass 9 science exploration chapter 4 solutionsvelocity-time graph from 0 s to 120 s for a cyclist
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    1. From the velocity-time graph:

      • Constant velocity (20–100 s): displacement = area of rectangle = 3 × 80 = 240 m.
      • Decreasing velocity (100–120 s): displacement = area of trapezium = ½(3+2) × 20 = 50 m.
      • Initial part (0–20 s): displacement = area of triangle = ½ × 20 × 3 = 30 m.

      Therefore, total displacement = 30 + 240 + 50 = 320 m. Average acceleration = (2−0)/120 = 0.0167 m s⁻².

       

      For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)

      https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/

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