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Describing Motion Around Us – Class 9 Science notes, important questions and MCQs

NCERT Class 9 Science, chapter 4. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Motion is a change of position with time. We describe it with distance, displacement, speed, velocity and acceleration, and with graphs and equations.

Basic quantities

Distance (scalar) is the total path length; displacement (vector) is the shortest distance with direction.
Speed = distance/time; velocity = displacement/time.
Average speed = total distance / total time.

Acceleration and graphs

Acceleration a = (v − u)/t; unit m/s²; negative acceleration is retardation.
Distance–time graph: slope = speed; straight line = uniform motion.
Velocity–time graph: slope = acceleration; area under the graph = displacement.

Equations of uniformly accelerated motion

v = u + at.
s = ut + ½at².
v² = u² + 2as.
Uniform circular motion has constant speed but changing velocity (direction changes).

Key terms

Scalar: quantity with magnitude only (distance, speed)
Vector: quantity with magnitude and direction (displacement, velocity)
Uniform motion: equal distances in equal intervals of time

Formulas

v = u + at
s = ut + ½at²
v² = u² + 2as
Circular speed v = 2πr/t

Important questions with answers

A runner completes one round of a circular track of radius 50 m. What are the distance and displacement?2 marks

Distance = 2πr = 2 × 3.14 × 50 = 314 m. Displacement = 0, because the runner returns to the starting point.

A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Find its acceleration and the distance covered.3 marks

u = 5 m/s, v = 10 m/s, t = 5 s. a = (10 − 5)/5 = 1 m/s². s = ut + ½at² = 25 + 12.5 = 37.5 m.

Explain what the slope and the area of a velocity–time graph tell us. A bus starts from rest, reaches 20 m/s in 10 s, moves steadily for 20 s, and then stops in 5 s. Find the total distance.5 marks

The slope of a v–t graph gives acceleration; the area under it gives displacement. Distance = area: accelerating part ½ × 10 × 20 = 100 m; steady part 20 × 20 = 400 m; braking part ½ × 5 × 20 = 50 m. Total = 550 m.

Quick revision

Speed is scalar; velocity is vector.
Slope of v–t graph = acceleration.
Area under v–t graph = displacement.

Exam tips

Convert km/h to m/s by multiplying with 5/18.
Displacement can be zero even when distance is not (round trip).

Practice MCQs

SI unit of acceleration is
1. m/s²
2. m/s
3. m²/s
4. km/h
Show answer
Answer: (a) m/s². m/s².
Which is a vector quantity?
1. distance
2. speed
3. time
4. displacement
Show answer
Answer: (d) displacement. It has direction.
72 km/h in m/s is
1. 7.2
2. 72
3. 20
4. 36
Show answer
Answer: (c) 20. 72 × 5/18 = 20.
A car goes from 10 m/s to 30 m/s in 5 s. Its acceleration is
1. 2 m/s²
2. 4 m/s²
3. 6 m/s²
4. 8 m/s²
Show answer
Answer: (b) 4 m/s². (30 − 10)/5 = 4.
An athlete runs one full round of a 400 m track. Displacement is
1. 400 m
2. 0 m
3. 200 m
4. 800 m
Show answer
Answer: (b) 0 m. Back to start.
Slope of a distance–time graph gives
1. acceleration
2. speed
3. displacement
4. force
Show answer
Answer: (b) speed. Distance per time.
A body starts from rest with a = 2 m/s². Distance covered in 5 s is
1. 10 m
2. 50 m
3. 25 m
4. 20 m
Show answer
Answer: (c) 25 m. s = ½ × 2 × 25 = 25 m.
A train moving at 20 m/s stops in 10 s with uniform retardation. Distance travelled while stopping is
1. 50 m
2. 200 m
3. 100 m
4. 400 m
Show answer
Answer: (c) 100 m. a = −2; v² = u² + 2as ⇒ 0 = 400 − 4s ⇒ s = 100 m.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.