Class 9 Science Chapter 7 – Motion
📑 Table of Contents
- 🧪 Intext Questions – Page 100
- 🧮 Intext Questions – Page 102
- 📘 End-of-Chapter Numerical Questions
- 📈 Velocity-Time Graphs
🧪 Intext Questions – Page 100
Q1. An object has moved through a distance. Can it have zero displacement?
Yes. If an object moves in a circular path and returns to the starting point, displacement is zero though distance covered is not.
Q2. 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?
Side = 10 m, Perimeter = 40 m
Time = 140 s → Number of rounds = 140 / 40 = 3.5
After 3.5 rounds, the farmer is at the opposite corner → displacement = diagonal
Diagonal = √(10² + 10²) = √200 = 14.14 m
Time = 140 s → Number of rounds = 140 / 40 = 3.5
After 3.5 rounds, the farmer is at the opposite corner → displacement = diagonal
Diagonal = √(10² + 10²) = √200 = 14.14 m
Q3. Which of the following is true for displacement?
(a) False. Displacement can be zero.
(b) False. Displacement can be less than or equal to distance.
Correct: None
(b) False. Displacement can be less than or equal to distance.
Correct: None
🧮 Intext Questions – Page 102
Q5. Signal from a spaceship reached ground in 5 minutes. What is the distance of the spaceship? (Speed of signal = 3 × 108 m/s)
Time = 5 min = 300 s
Distance = Speed × Time = 3 × 108 × 300 = 9 × 1010 m
Distance = Speed × Time = 3 × 108 × 300 = 9 × 1010 m
📘 End-of-Chapter Numerical Questions
Q1. An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed?
Total distance = 16 + 16 = 32 m
Total time = 4 + 2 = 6 s
Average speed = 32 / 6 = 5.33 m/s
Total time = 4 + 2 = 6 s
Average speed = 32 / 6 = 5.33 m/s
Q2. A car moves half distance at 40 km/h and other half at 60 km/h. What is the average speed?
Average speed = 2ab / (a + b)
= (2 × 40 × 60) / (40 + 60) = 4800 / 100 = 48 km/h
= (2 × 40 × 60) / (40 + 60) = 4800 / 100 = 48 km/h
Q3. A car accelerates from 5 m/s to 15 m/s in 5 s. Find acceleration and distance.
u = 5 m/s, v = 15 m/s, t = 5 s
Acceleration a = (v – u) / t = (15 – 5) / 5 = 2 m/s²
Distance s = ut + (1/2)at² = 5×5 + 0.5×2×25 = 25 + 25 = 50 m
Acceleration a = (v – u) / t = (15 – 5) / 5 = 2 m/s²
Distance s = ut + (1/2)at² = 5×5 + 0.5×2×25 = 25 + 25 = 50 m
Q4. A body covers 80 m in 5 s and 120 m in next 5 s. Find initial velocity and acceleration.
Using equations: s = ut + 1/2 at²
For 1st 5s: 80 = 5u + (1/2)×a×25 → 80 = 5u + 12.5a …(1)
For total 10s: 200 = 10u + 0.5×a×100 → 200 = 10u + 50a …(2)
Solving (1)×2 = 160 = 10u + 25a
Subtracting: (2) – (1×2): 40 = 25a → a = 1.6 m/s²
Put in (1): 80 = 5u + 12.5×1.6 = 5u + 20 → u = (80 – 20)/5 = 12 m/s
For 1st 5s: 80 = 5u + (1/2)×a×25 → 80 = 5u + 12.5a …(1)
For total 10s: 200 = 10u + 0.5×a×100 → 200 = 10u + 50a …(2)
Solving (1)×2 = 160 = 10u + 25a
Subtracting: (2) – (1×2): 40 = 25a → a = 1.6 m/s²
Put in (1): 80 = 5u + 12.5×1.6 = 5u + 20 → u = (80 – 20)/5 = 12 m/s
Q5. A train moving at 72 km/h is stopped in 1 min. Find acceleration and distance.
u = 72 km/h = 20 m/s, v = 0, t = 60 s
a = (v – u)/t = –20 / 60 = –1/3 m/s²
s = ut + 0.5 at² = 20×60 + 0.5×(–1/3)×3600 = 1200 – 600 = 600 m
a = (v – u)/t = –20 / 60 = –1/3 m/s²
s = ut + 0.5 at² = 20×60 + 0.5×(–1/3)×3600 = 1200 – 600 = 600 m
📈 Velocity-Time Graphs
1. Uniform Acceleration
Velocity increases linearly with time. The graph is a straight line sloping upward.
2. Uniform Deceleration
Velocity decreases linearly with time. The graph is a straight line sloping downward.
3. Zero Acceleration
Velocity remains constant. The graph is a horizontal line.
Prepared for CBSE Class 9 Science Students – Chapter 7: Motion