4
Question Question 4
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2 for 8.0 s. How far does the boat travel during this time?
Given,
Initial velocity (u) = 0
Acceleration (a) = 3 ms-2
Time period (t) = 8 s
Distance travelled (S) = ?
According to the second equation of motion,
S = ut + at2
Substituting we get,
S = 0 + x 3 x 8 x 8
= 96 m
∴ The motorboat travels a distance of 96 m
BRIGHT TUTORIALS
BRIGHT TUTORIALS
CBSE Class IX | Academic Year 2026-2027
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Science | Chapter 7: MotionWeb Content — Quick Reference
Chapter 7: Motion — Quick Reference
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