CBSE Class 9 Science Question 9 of 10

Motion — Question 2

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Question 2

A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? NCERT Class 9 Science CBSE Solutions.

Given,

Time period of travel = 2 minutes 20 seconds

Convert min into sec

1 min = 60 sec

So 2 min 20 sec = (2 x 60) sec + 20 sec = 120 + 20 = 140 sec

Side of the given square field = 10 m

Hence, the perimeter of a square = 4 x side = 4 x 10 = 40 m

Time taken by the farmer to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of = 4040\dfrac{40}{40} x 1 = 1 m

Hence, distance covered by the farmer in 140 sec = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m

= total distanceperimeter\dfrac{\text{total distance}}{\text{perimeter}}

= 14040\dfrac{140}{40}

= 3.5

If the farmer starts from point A, after 3.5 rounds he will be at point C of the field.

∴ Displacement (from Pythagoras theorem)

AC=(AB)2+(BC)2=(10)2+(10)2=100+100=200=102=10×1.414=14.14 m\text{AC} = \sqrt{(\text{AB})^2 + (\text{BC})^2} \\[1em] = \sqrt{(10)^2 + (10)^2} \\[1em] = \sqrt{100 + 100} \\[1em] = \sqrt{200} \\[1em] = 10\sqrt{2} \\[1em] = 10 \times 1.414 \\[1em] = 14.14 \text{ m}

Hence, the magnitude of displacement = 14.14 m

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Science | Chapter 7: MotionWeb Content — Quick Reference

Chapter 7: Motion — Quick Reference

distance displacement speed velocity acceleration equations of motion graphs circular motion

Quick Revision Points

  • Distance (scalar, total path) vs Displacement (vector, shortest path)
  • Speed = distance/time (scalar); Velocity = displacement/time (vector)
  • Acceleration a = (v−u)/t; unit: m/s²
  • Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
  • d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
  • 1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
  • Uniform circular motion: constant speed, changing velocity (direction changes)
Exam Tips for Chapter 7
  • Read the detailed chapter notes for complete coverage of all NCERT topics.
  • Practice all NCERT in-text and back exercise questions — they are frequently asked in exams.
  • Focus on comparison tables, diagrams, and definitions — these are high-scoring areas.
  • For numericals (if applicable), practice at least 20 problems of varying difficulty.
  • Refer to the practice question bank (200+ questions) for thorough preparation.
Related Resources
  • Detailed Notes: ch07-motion.html
  • Practice Questions: 100+ questions with answers in 05-practice-questions/
  • Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
  • Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html