CBSE Class 9 Science Question 7 of 17

Force and Laws of Motion — Question 8

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Question 8

An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?

Answer

Given,

Mass (m) = 1500 kg

Acceleration (a) = -1.7 ms-2

Now, as Force = mass x acceleration

F = 1500 × (-1.7) = -2550 N

Hence, force required = -2550 N. (Negative sign indicates opposite direction.)

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CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
Science | Chapter 8: Force and Laws of MotionWeb Content — Quick Reference

Chapter 8: Force and Laws of Motion — Quick Reference

Newton laws inertia momentum F=ma conservation of momentum action reaction

Quick Revision Points

  • Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
  • First Law (Inertia): object stays at rest/uniform motion unless external force acts
  • Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
  • 1 N = force giving 1 kg an acceleration of 1 m/s²
  • Third Law: every action has equal and opposite reaction (on different bodies)
  • Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)
Exam Tips for Chapter 8
  • 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: ch08-force-and-laws-of-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
Web Content: Motion | Bright Tutorials
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BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
9403781999
Excellence in Education
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