Class 9 Science Exploration Chapter 4 Solutions (NCERT 2026–27) – Describing Motion Around Us

These Class 9 Science Exploration Chapter 4 solutions cover Describing Motion Around Us from the new NCF-2023 textbook (2026–27).

Class: 9 Subject: Science Book: Exploration Chapter: 4 Exercise: Revise, Reflect, Refine (16 Qs) Session: 2026–27
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Class 9 Science Exploration Chapter 4 Solutions – Overview

Chapter 4 of Exploration, Describing Motion Around Us, teaches how to describe motion precisely. It explains that rest and motion are relative (they depend on the reference point), the difference between distance and displacement and between speed and velocity, the meaning of acceleration, the three equations of motion for uniform acceleration, how to read distance–time and velocity–time graphs, and uniform circular motion. These Class 9 Science Exploration Chapter 4 solutions answer every textbook question step by step.

Key Concepts & Definitions

Rest and motion are relative: an object’s state depends on the chosen reference point.

Distance: total path length (scalar, only magnitude). Displacement: shortest distance from start to end, with direction (vector, can be zero).

Speed = distance/time (scalar). Velocity = displacement/time (vector). Acceleration = change in velocity/time (SI unit m s-2).

Graphs: the slope of a distance–time graph gives speed; the slope of a velocity–time graph gives acceleration; the area under a velocity–time graph gives displacement.

Uniform circular motion: motion at constant speed along a circle — the direction (hence velocity) keeps changing, so it is accelerated motion.

Equations of Motion (uniform acceleration)

v = u + at

s = ut + ½at2

v2 = u2 + 2as

where u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement. Unit conversion: km h-1 × 5/18 = m s-1.

“Think It Over” — Answers

How much distance does a moving vehicle cover before it stops after the brakes are applied?

ANSWERThe vehicle keeps moving for a short reaction distance (before braking) and then a braking distance while slowing to rest. The braking distance can be found from v2 = u2 + 2as (with v = 0).

Does this stopping distance depend on the speed at which we are moving?

ANSWERYes — strongly. The braking distance is proportional to the square of the speed (s ∝ u2), so doubling the speed makes the stopping distance about four times longer. This is why speed limits matter.

Class 9 Science Exploration Chapter 4 Solutions — Revise, Reflect, Refine

1. My father went to a shop 250 m away on a straight road, found he forgot the cloth bag, came home to take it, went to the shop again, bought provisions and came back home. What was the total distance travelled and his displacement from home?

ANSWER The trip is home→shop→home→shop→home = 4 × 250 m. Total distance = 1000 m. He ends where he started (home), so displacement = 0.

2. A student runs from the ground floor to the fourth floor to collect a book and then comes down to the classroom on the second floor. If each floor is 3 m high, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.

ANSWER (i) Up: ground → 4th = 4 × 3 = 12 m; down: 4th → 2nd = 2 × 3 = 6 m. Total distance = 12 + 6 = 18 m. (ii) Final position is the 2nd floor = 2 × 3 = 6 m above the start (displacement = 6 m, upward).

3. A girl riding her scooter finds the speedometer reading constant. Is it possible for her scooter to be accelerating, and if so, how?

ANSWERYes. The speedometer shows only speed. If she is turning (moving along a curve), the direction of her velocity changes, so the velocity changes even at constant speed — that is acceleration (a change in direction).

4. A car starts from rest and its velocity reaches 24 m s-1 in 6 s. Find the average acceleration and the distance travelled in these 6 s.

ANSWER a = (v − u)/t = (24 − 0)/6 = 4 m s-2. s = ut + ½at2 = 0 + ½(4)(62) = 72 m.

5. A motorbike moving with initial velocity 28 m s-1 and constant acceleration stops after travelling 98 m. Find the acceleration and the time taken to stop.

ANSWER v2 = u2 + 2as ⇒ 0 = 282 + 2a(98) ⇒ a = −784/196 = −4 m s-2 (deceleration). v = u + at ⇒ 0 = 28 + (−4)t ⇒ t = 7 s.

6. Fig. 4.27 shows a position–time graph of two objects A and B moving along parallel tracks in the same direction. Do A and B ever have equal velocity? Justify.

ANSWEROn a position–time graph the velocity is the slope of the curve. A and B have equal velocity at the instant when the tangents (slopes) of the two curves are parallel. If the graph shows a point where the two curves are equally steep, that is where their velocities are equal. (Read this point from Fig. 4.27.)

7. A graph (Fig. 4.28) shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 s. Choose the correct option(s). (i) The average velocity of both over 10 s is equal since they have the same initial and final positions. (ii) The average speeds of both over 10 s are equal since both cover equal distance in equal time. (iii) The average speed of A over 10 s is lower than that of B since it covers a shorter distance than B in 10 s. (iv) The average speed of A over 10 s is greater than that of B since B’s speed is lower in some segments.

ANSWER (i) is correct — A and B have the same initial and final positions, so they have the same displacement and hence the same average velocity over 10 s. For average speed, compare the total path length: since A covers a shorter distance than B in Fig. 4.28, option iii is also correct — A’s average speed is lower. Option ii is wrong because the distances are not equal, and option iv contradicts option iii.

8. A truck driver going at 54 km h-1 sees a 40 km h-1 speed limit and slows to 36 km h-1 in 36 s. What distance did he travel during this time (constant acceleration)?

ANSWER Convert: 54 km h-1 = 15 m s-1; 36 km h-1 = 10 m s-1. Distance = average velocity × time = ((15 + 10)/2) × 36 = 12.5 × 36 = 450 m.

9. A car starts from rest and accelerates uniformly to 20 m s-1 in 5 s, travels at 20 m s-1 for 10 s, then brakes to stop in 6 s. Find the total distance travelled.

ANSWER Phase 1 (speeding up): ½(0 + 20)(5) = 50 m. Phase 2 (constant): 20 × 10 = 200 m. Phase 3 (braking): ½(20 + 0)(6) = 60 m. Total = 50 + 200 + 60 = 310 m.

10. A bus travelling at 36 km h-1 sees an obstacle 30 m ahead. The driver takes 0.5 s to react before braking; then the bus slows at 2.5 m s-2. Will it stop before the obstacle?

ANSWER 36 km h-1 = 10 m s-1. Reaction distance = 10 × 0.5 = 5 m. Braking distance = u2/(2a) = 102/(2 × 2.5) = 20 m. Total stopping distance = 5 + 20 = 25 m, which is less than 30 m — yes, the bus stops about 5 m before the obstacle.

11. A student says, “The Earth moves around the Sun.” Can an object kept on the Earth be considered to be at rest?

ANSWERIt depends on the reference point. Relative to the Earth’s surface the object is at rest; relative to the Sun it is moving along with the Earth. Rest and motion are relative.

12. The velocity–time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas representing the displacement (i) while the cyclist moves with constant velocity, and (ii) when the velocity is decreasing. Also calculate the displacement and average acceleration in the 120 s interval.

ANSWER (i) The constant-velocity part is the flat (horizontal) portion of the graph; shade the rectangle under it.(ii) The decreasing part is the downward-sloping portion; shade the triangle/trapezium under it. Displacement = total area under the velocity–time graph (add the areas of all the regions). Average acceleration = (final velocity − initial velocity) ÷ 120 s. (Substitute the exact values read from Fig. 4.30.)

13. The velocity–time graph (Fig. 4.31) shows a runner’s velocity. Estimate the distance she ran.

ANSWERThe distance run equals the area under the velocity–time graph. Split the area into rectangles and triangles, find each area from the values in Fig. 4.31, and add them to get the total distance.

14. A car moves at a constant 6 m s-1 for 2 minutes, then accelerates at 1 m s-2 for 6 s. Find its displacement in the 2 min 6 s interval (draw a velocity–time graph).

ANSWER Phase 1 (constant): 6 × 120 = 720 m. Phase 2 (accelerating): s = ut + ½at2 = 6(6) + ½(1)(62) = 36 + 18 = 54 m. Total displacement = 720 + 54 = 774 m (this is the total area under the v–t graph).

15. Two cars A and B start from rest with constant acceleration. A reaches 5 m s-1 in 5 s; B reaches 3 m s-1 in 10 s. Plot both velocity–time graphs and calculate the displacement in the two intervals.

ANSWER aA = 5/5 = 1 m s-2; aB = 3/10 = 0.3 m s-2. Each graph is a straight line from the origin. Displacement = area of the triangle: car A in 5 s = ½(5)(5) = 12.5 m; car B in 10 s = ½(10)(3) = 15 m.

16. Rohan studies from 6 PM to 7:30 PM. For the tip of the minute hand (length 7 cm) during this interval, find its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity.

ANSWER Time = 90 minutes = 1.5 revolutions of the minute hand; r = 7 cm, circumference = 2πr = 44 cm. (i) Distance = 1.5 × 44 = 66 cm. (ii) After 1.5 revolutions the tip is diametrically opposite the start, so displacement = diameter = 14 cm. (iii) Speed = distance/time = 66 cm / 5400 s ≈ 0.0122 cm s-1 (= 0.73 cm min-1). (iv) Velocity = displacement/time = 14 cm / 5400 s ≈ 0.0026 cm s-1, directed along the diameter from start to end.

Common Mistakes to Avoid

Watch out for these

  • Confusing distance (scalar, total path) with displacement (vector, can be zero).
  • Forgetting to convert km h-1 to m s-1 (multiply by 5/18) before using the equations.
  • Mixing up graph rules: slope of distance–time = speed; slope of velocity–time = acceleration; area under velocity–time = displacement.
  • Thinking constant speed means no acceleration — circular motion is accelerated (direction changes).
  • Using a positive acceleration when an object is slowing down — deceleration is negative.
  • Leaving out units, or forgetting that displacement has a direction.

Extra Practice Questions

Very Short Answer Type Questions

Q1. What does the area under a velocity–time graph represent?

ANSWERThe displacement of the object.

Q2. Convert 72 km h-1 into m s-1.

ANSWER72 × 5/18 = 20 m s-1.

Q3. Is velocity a scalar or a vector quantity?

ANSWERA vector (it has both magnitude and direction).

Short Answer Type Questions

Q1. Differentiate between speed and velocity.

ANSWERSpeed is the distance covered per unit time (a scalar, always positive); velocity is the displacement per unit time (a vector, with direction, and it can be zero or negative).

Q2. A body covers equal distances in equal intervals of time. What is its acceleration?

ANSWERIt is in uniform motion (constant velocity along a straight line), so its acceleration is zero.

Long Answer Type Question

Q1. Derive the relation s = ut + ½at2 using a velocity–time graph.

ANSWER For uniform acceleration the v–t graph is a straight line from u to v over time t. Displacement = area under the line = area of rectangle (u × t) + area of triangle (½ × t × (v − u)). Since v − u = at, the triangle area = ½ × t × at = ½at2. Hence s = ut + ½at2.

MCQs & Assertion–Reason

1. Which of these is a vector quantity?

(a) distance    (b) speed    (c) displacement    (d) time

2. The SI unit of acceleration is:

(a) m s-1    (b) m s-2    (c) m    (d) s

3. Which is an equation of motion?

(a) v = u + at    (b) F = ma    (c) p = mv    (d) W = Fs

4. The slope of a distance–time graph gives:

(a) acceleration    (b) speed    (c) displacement    (d) time

5. The area under a velocity–time graph gives:

(a) acceleration    (b) speed    (c) displacement    (d) force

6. A body moving with constant speed in a circle:

(a) has zero acceleration    (b) is accelerating    (c) has constant velocity    (d) is at rest

7. A car’s velocity rises from 0 to 24 m s-1 in 6 s. Its acceleration is:

(a) 2 m s-2    (b) 4 m s-2    (c) 6 m s-2    (d) 24 m s-2

8. Displacement can be zero when the distance is not zero if the body:

(a) moves in a straight line    (b) returns to its starting point    (c) speeds up    (d) stops

9. 54 km h-1 equals:

(a) 10 m s-1    (b) 15 m s-1    (c) 18 m s-1    (d) 20 m s-1

10. Uniform velocity means the body moves with:

(a) changing speed    (b) constant speed in a straight line    (c) constant acceleration    (d) circular motion

Answer key: 1-(c), 2-(b), 3-(a), 4-(b), 5-(c), 6-(b), 7-(b), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A body moving in a circle at constant speed is accelerating.

Reason: Its direction of velocity keeps changing.

A-R 2. Assertion: Displacement can be zero even when distance is not zero.

Reason: Displacement is the shortest distance from the start to the end point.

A-R 3. Assertion: Distance is a scalar quantity.

Reason: It has only magnitude and no direction.

A-R 4. Assertion: The slope of a distance–time graph gives acceleration.

Reason: Slope equals the change in distance divided by time.

A-R 5. Assertion: 36 km h-1 equals 10 m s-1.

Reason: To convert km h-1 to m s-1, multiply by 5/18.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • Rest and motion are relative to a reference point.
  • Distance (scalar, total path) vs displacement (vector, shortest, can be zero).
  • Speed = distance/time; velocity = displacement/time; acceleration = change in velocity/time.
  • Equations of motion: v = u + at, s = ut + ½at2, v2 = u2 + 2as.
  • Graphs: slope of d–t = speed; slope of v–t = acceleration; area under v–t = displacement.
  • Uniform circular motion is accelerated because the direction of velocity changes.

Real-life Applications

These ideas explain everyday motion: why a vehicle’s stopping distance grows sharply with speed (so we keep speed limits and safe gaps), how a speedometer shows speed but not direction, how athletes and vehicles are analysed using velocity–time graphs, and why a satellite or a stone whirled on a string is constantly accelerating even at steady speed.

How to score full marks in this chapter

Always write the known values (u, v, a, t, s) with units, convert km h-1 to m s-1, and pick the right equation of motion. For graph questions, remember slope vs area. Show each step and keep the direction in displacement and velocity answers.

Frequently Asked Questions

What is Class 9 Science Exploration Chapter 4 about?

Describing motion — rest and motion as relative ideas, distance and displacement, speed and velocity, acceleration, the equations of motion, motion graphs and uniform circular motion.

What is the difference between distance and displacement?

Distance is the total path length (a scalar); displacement is the shortest straight-line distance from start to end with direction (a vector that can be zero).

What are the three equations of motion?

v = u + at, s = ut + ½at2, and v2 = u2 + 2as (for uniform acceleration).

Are these Class 9 Science Exploration Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Exploration textbook for 2026–27.

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