Given,

Mass of the car (m) = 1200 kg

Initial velocity (u) = 90 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

90 x $\dfrac{5}{18}$ = 25 ms^-1

Final velocity (v) = 18 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

18 x $\dfrac{5}{18}$ = 5 ms^-1

Time period (t) = 4 sec

As per the first equation of motion,

a = $\dfrac{\text{v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{5-25}{4}$ = -5ms^-2

Hence, acceleration of the car is -5 ms^-2

Negative sign shows retardation.

Initial momentum of the car = m × u = 1200 × 25 = 30,000 kgms^-1

Final momentum of the car = m × v = 1200 x 5 = 6,000 kgms^-1

∴ Change in momentum = mv - mu = 6,000 - 30,000 = -24,000 kgms^-1

Hence, change in momentum = 24,000 kgms^-1

We know, Force = mass x acceleration

Substituting we get,

F = 1200 × -5 = -6000 N

Hence, the magnitude of force required to slow down the vehicle = 6000 N

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

9403781999

Excellence in Education

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

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

9403781999

Excellence in Education

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