Looking for accurate SEBA Class 9 Science Chapter 8 (Motion) textual solutions for the 2026–27 academic year? Access complete step-by-step textbook exercise answers covering distance and displacement, uniform and non-uniform motion, speed, velocity, acceleration, distance-time graphs, equations of motion, and uniform circular motion to score top marks in your exams.
Blue questions
Q1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans. Yes, an object can have zero displacement even if it has travelled some distance.
Example: If a person walks around a circular track and returns to the starting point, the distance travelled is non-zero, but the displacement is zero because the initial and final positions are the same.
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 from his initial position?
Ans.
Side of field = 10 m
Perimeter = 4 × 10 = 40 m
Time for 1 round = 40 s
Given time = 2 min 20 s = 140 s
Number of rounds = 140 ÷ 40 = 3.5 rounds
After 3 full rounds the farmer returns to the starting point (displacement = 0).
Remaining 0.5 round = 20 m
Half perimeter means he reaches the opposite corner of the square.
Displacement = diagonal of square
d = √(10² + 10²) = √200 = 10√2 m ≈ 14.14 m
Q3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Ans. Both statements are incorrect.
Displacement can be zero and is always less than or equal to the distance.
Q4. Distinguish between speed and velocity.
Ans.
Speed is the rate of distance covered by an object. It is a scalar quantity.
Velocity is the rate of displacement of an object. It is a vector quantity and has direction.
Q5. Under what condition is the magnitude of average velocity equal to the average speed?
Ans. When an object moves in a straight line in the same direction without turning back, distance = displacement. Therefore, average speed = average velocity.
Q6. What does the odometer of an automobile measure?
Ans. The odometer measures the total distance travelled by the vehicle.
Q7. What does the path of an object look like in uniform motion?
Ans. It appears as a straight line when plotted on a distance–time graph.
SEBA Class 9 Science Chapter 8 Solutions
Q8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance? (Speed of light = 3 × 10⁸ m/s)
Ans.
Time = 5 min = 300 s
Distance = speed × time
= (3 × 10⁸) × 300
= 9 × 10¹⁰ m
Q9. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Ans.
(i) If the velocity changes by equal amounts in equal intervals of time, the body has uniform acceleration.
(ii) If the velocity changes by unequal amounts in equal intervals of time, the body has non-uniform acceleration.
Q10. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration.
Ans.
Convert to m/s:
80 km/h = 22.22 m/s
60 km/h = 16.67 m/s
a = (v – u) / t
a = (16.67 – 22.22) / 5
a = -1.11 m/s²
Acceleration = -1.11 m/s² (retardation)
Q11. A train starting from rest attains 40 km/h in 10 min. Find its acceleration.
Ans.
Convert:
40 km/h = 11.11 m/s
10 min = 600 s
a = (11.11 – 0) / 600
a = 0.0185 m/s²
Acceleration = 0.0185 m/s²
Q12. What is the nature of distance–time graphs for uniform and non-uniform motion?
Ans.
Uniform motion -> straight line
Non-uniform motion -> curved line
Q13. What can you say about the motion of an object whose distance–time graph is a straight line parallel to the time axis?
Ans. The distance does not change with time -> the object is at rest.
Q14. What can you say about the motion of an object if its speed–time graph is a straight line parallel to the time axis?
Ans. The speed is constant -> the object is in uniform motion.
Class 9 Science Chapter 8 Question Answer SEBA 2026-27
Q15. What quantity is measured by the area under a velocity–time graph?
Ans. Displacement (or distance, depending on direction).
Q16. A bus starts from rest and accelerates at 0.1 m/s² for 2 minutes. Find (a) final speed (b) distance travelled.
Ans.
t = 2 min = 120 s
u = 0
(a)
v = u + at
v = 0 + 0.1 × 120 = 12 m/s
(b)
s = ut + (1/2)at²
s = 0 + (1/2)(0.1)(120²)
s = 0.05 × 14400 = 720 m
Q17. A train moving at 90 km/h stops with a uniform acceleration –0.5 m/s². Find the stopping distance.
Ans.
Convert: 90 km/h = 25 m/s
Using v² = u² + 2as
0 = 25² + 2(-0.5)s
0 = 625 – 1s
s = 625 m
Q18. A trolley going down an inclined plane has an acceleration of 2 cm/s². What will be its velocity after 3 s?
Ans.
Convert: 2 cm/s² = 0.02 m/s²
v = u + at
v = 0 + 0.02 × 3 = 0.06 m/s
Q19. A racing car accelerates at 4 m/s². What distance will it cover in 10 s after the start?
Ans.
s = ut + (1/2)at²
s = 0 + (1/2)(4)(10²)
s = 2 × 100 = 200 m
Motion Class 9 SEBA Textual Exercise Solutions
Q20. A stone is thrown upward with velocity 5 m/s. Acceleration = –10 m/s². Find (a) height reached (b) time taken to reach top.
Ans.
(a) Using v² = u² + 2as
0 = 5² + 2(-10)s
0 = 25 – 20s
s = 25 / 20 = 1.25 m
(b)
v = u + at
0 = 5 – 10t
10t = 5
t = 0.5 s
Textbook question
1. Circular Track Problem
Question: An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer:
Diameter of track, d = 200 m -> Radius, r = 100 m
Circumference of the track C = 2 × π × r = 2 × π × 100 ≈ 628 m
Time for one round = 40 s
Total time = 2 min 20 s = 140 s
Distance covered:
Number of rounds = 140 / 40 = 3.5 rounds
Distance = 3.5 × 628 ≈ 2198 m
Displacement:
After 3.5 rounds, the athlete is opposite to the starting point.
Displacement = diameter of track = 200 m
2. Joseph Jogging Problem
Question: Joseph jogs from A to B along a straight 300 m road in 2 min 30 s, then turns and jogs 100 m back to C in 1 min. Find average speed and velocity:
Answer:
(a) From A to B:
Distance s = 300 m, Time t = 2.5 min = 150 s
Average speed = s / t = 300 / 150 = 2 m/s
Displacement = 300 m (in same direction)
Average velocity = 300 / 150 = 2 m/s
(b) From A to C:
Distance covered = 300 + 100 = 400 m
Displacement = 300 – 100 = 200 m
Time = 150 + 60 = 210 s
Average speed = 400 / 210 ≈ 1.90 m/s
Average velocity = 200 / 210 ≈ 0.95 m/s
3. Abdul’s Trip Problem
Question: Abdul drives to school at 20 km/h and returns at 30 km/h. Find average speed for the round trip.
Answer:
Let distance one way = d km.
t1 = d / 20, t2 = d / 30
Total distance = 2d
Total time = t1 + t2 = (d / 20) + (d / 30)
LCM of 20 & 30 = 60
Total time = (3d / 60) + (2d / 60) = 5d / 60 = d / 12
v_avg = Total distance / Total time
v_avg = 2d / (d / 12) = 2d × (12 / d) = 24 km/h
4. Motorboat Acceleration Problem
Question: A motorboat starting from rest accelerates at 3.0 m/s² for 8 s. How far does it travel?
Answer:
Initial velocity, u = 0
Acceleration, a = 3 m/s²
Time, t = 8 s
s = ut + (1/2)at²
s = 0 + (1/2) × 3 × 8²
s = 1.5 × 64 = 96 m
SEBA Class 9 Science Chapter 8 Textbook Answers
5. Car Braking Problem
Question: A car moving at 52 km/h applies brakes.
Answer:
Convert 52 km/h = (52 × 1000) / 3600 ≈ 14.44 m/s
(a) Distance travelled while slowing down = area under speed-time graph (triangle if deceleration is uniform).
(b) Uniform motion part = horizontal part of speed-time graph before braking.
6. Distance-Time Graph Problem
Question: Fig. 7.10 shows distance-time graph of A, B, C.
Answer:
(a) Object with steepest slope = fastest -> identify slope.
(b) All at same point? -> check intersection.
(c) Distance travelled by C when B passes A -> read graph.
(d) Distance travelled by B when passing C -> read graph.
Note: Requires graph for exact numerical answers.
7. Free Fall Problem
Question: A ball is dropped from 20 m with g = 10 m/s². Find velocity on impact and time.
Answer:
Initial velocity u = 0
Distance s = 20 m, a = g = 10 m/s²
v² = u² + 2as
v² = 0 + 2 × 10 × 20 = 400
v = 20 m/s
v = u + at
20 = 0 + 10t
t = 2 s
8. Speed-Time Graph Problem
Question: Fig. 7.11 shows speed-time graph.
Answer:
(a) Distance travelled = area under graph (calculate triangle/rectangle area).
(b) Uniform motion = horizontal section of graph.
9. Acceleration Situations
Question: State which situations are possible with examples:
Answer:
(a) Constant acceleration but zero velocity -> Example: Ball at highest point in free fall.
(b) Acceleration but uniform speed -> Impossible (acceleration changes speed).
(c) Moving in one direction with acceleration perpendicular -> Example: Circular motion, velocity tangent, acceleration towards center.
Class 9 Science Chapter 8 Motion Question Answer
10. Satellite Problem
Question: An artificial satellite moves in circular orbit of radius 42250 km. Find speed if it takes 24 hours to revolve.
Answer:
Orbit radius r = 42250 × 10³ m
Time T = 24 h = 86400 s
Speed v = (2 × π × r) / T
v = (2 × π × 42250 × 10³) / 86400 ≈ 3074 m/s
🔬 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 9 Science Chapter 8 (Motion) Textual Exercise Solutions updated for the 2026–27 academic session. All textbook numericals, graphical derivations, and physics concepts strictly follow the latest revised Assam Board (SEBA) curriculum.
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