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Class 9 Science: Describing Motion Around Us, 8 practice questions with answers

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These are 8 practice questions on Chapter 4, Describing Motion Around Us, of the new NCERT Class 9 Science textbook, Exploration (First Edition, April 2026). Each has a short worked answer, the units, and the textbook page it comes from. They cover distance and displacement, average speed, average velocity, average acceleration, the kinematic equations v = u + at and s = ut + ½at², and why uniform circular motion is accelerated. Try each one on paper before you open the answer. Speed also appears in sound: the Class 9 Sound Waves questions cover frequency, v = λν and echoes.

These questions are our own, drafted with AI help. Every number was re-computed by a test script, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Exploration: Science, Textbook for Grade 9, Chapter 4, printed pages 48 to 71):

Distance, speed and velocity

Question 1. When an object moves along a straight line, when does the total distance it covers equal the magnitude of its displacement? (When it moves in one direction without turning back; when it returns to its starting point; or when it moves forward and then comes back a shorter distance.)

Answer: when it moves in one direction without turning back. Turning back adds to the distance but reduces the displacement. Only motion in one direction keeps the two equal. (Page 51.)

Question 2. A cyclist rides 400 m along a straight road in 80 s and then rides back to the starting point in 120 s. What is the cyclist's average speed for the whole trip, in m/s?

Answer: 4 m/s. Average speed = total distance ÷ time interval = 800 m ÷ 200 s = 4 m/s. His average velocity is 0, because he ends where he started and the displacement is zero. (Equations 4.1 and 4.2, pages 52 and 53.)

Question 3. A cyclist rides 12 km in 40 minutes and then 8 km in the next 20 minutes. What is her average speed for the whole ride, in km per hour?

Answer: 20 km/h. Average speed = total distance ÷ total time = (12 + 8) km ÷ (40 + 20) min = 20 km in 1 hour = 20 km per hour. (Eq. 4.1, page 52; speed may be given in km/h, page 53.)

Acceleration

Question 4. A car moving on a straight road speeds up from 10 m/s to 25 m/s in 5 s. What is its average acceleration, in m/s²?

Answer: 3 m/s². a = (v − u) ÷ t = (25 − 10) m/s ÷ 5 s = 3 m/s². The car is speeding up, so the acceleration is in the direction of velocity. (Eq. 4.3c and the direction rule, page 54.)

Question 5. A train speeds up from 36 km/h to 72 km/h in 20 s. What is its average acceleration, in m/s²?

Answer: 0.5 m/s². Change the speeds to m/s first: 36 km/h = 36 × 1000 ÷ 3600 = 10 m/s and 72 km/h = 20 m/s. Average acceleration = change in velocity ÷ time = (20 − 10) ÷ 20 = 0.5 m/s². (Eq. 4.3c, page 54; the textbook converts km/h to m/s the same way in Example 4.3, page 55, with different speeds.)

Question 6. A merry-go-round turns at a constant speed. Why is a child sitting on it said to be accelerating? (The child's speed keeps increasing; the child covers no distance in one round; or the direction of the child's velocity keeps changing.)

Answer: the direction of the child's velocity keeps changing. In uniform circular motion the speed stays the same, but the direction of velocity changes continuously, so the motion is accelerated. Acceleration is non-zero whenever the velocity changes, in magnitude or in direction. (Pages 55, 66 and 67.)

Kinematic equations

Question 7. A scooter starts from rest and moves in a straight line with a constant acceleration of 2 m/s². How far does it travel in the first 6 s, in m?

Answer: 36 m. Use s = ut + ½at². With u = 0: s = ½ × 2 × 6² = 36 m. (Eq. 4.4b, page 64; the equation holds because the acceleration is constant, page 65.)

Question 8. A car moving at 5 m/s speeds up with a uniform acceleration of 2 m/s² for 6 s. (a) What is its velocity at the end (use v = u + at)? (b) How far does it go in the 6 s (use s = ut + ½at²)? (c) What is its average velocity over the 6 s, in m/s?

Answer: (a) v = 5 + 2 × 6 = 17 m/s. (b) s = 5 × 6 + ½ × 2 × 36 = 30 + 36 = 66 m. (c) Average velocity = displacement ÷ time = 66 ÷ 6 = 11 m/s; for uniform acceleration this equals (5 + 17) ÷ 2. (Equations 4.4a and 4.4b, pages 63 and 64; average velocity, Eq. 4.2, page 53.)

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Sources

All opened and read on 21 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 47.

The definitions and equations are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.