📘 Motion – Class 9 Science NCERT Solutions (In-text + Exercises)
Motion is one of the most fascinating topics in physics. From the movement of planets to the simple walk of a human being, everything involves motion. In this blog, we will cover all in-text questions and exercise questions from NCERT Class 9 Science Chapter 8 – Motion along with detailed answers, explanations, and extra insights.
📝 In-text Questions and Answers
❓ Question 1
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
✅ Answer:
Yes, an object can have zero displacement even if it has travelled a distance.
For example:
If a person walks 10 m east and then returns 10 m west to the starting point, the distance travelled = 20 m, but the displacement = 0, because the initial and final positions are the same.
❓ Question 2
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?
✅ Answer:
- Side of square = 10 m
- Perimeter = 4 × 10 = 40 m
- Time for 1 round = 40 s
- Time given = 2 min 20 s = 140 s
- Number of rounds = 140 ÷ 40 = 3 full rounds + 20 s
👉 In 3 full rounds, the farmer comes back to the starting point.
👉 In the next 20 seconds, he covers half the perimeter = 20 m (since speed = 1 m/s).
Thus, after 3.5 rounds, he will be at the point diagonally opposite to the starting point.
- Displacement = length of diagonal = √(10² + 10²) = √200 = 14.14 m
✅ Final Answer: 14.14 m
❓ Question 3
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.
✅ Answer:
Both statements are incorrect. The correct fact is:
- Displacement can be zero (e.g., if the object returns to its starting point).
- Displacement is always less than or equal to the distance, never greater.
✅ Correct Statement: Displacement can be equal to distance, but never greater than distance.
📝 In-text Questions and Answers (Speed & Velocity)
❓ Question 1
Distinguish between speed and velocity.
✅ Answer:
| Speed | Velocity |
|---|---|
| Scalar quantity (only magnitude) | Vector quantity (magnitude + direction) |
| Distance travelled per unit time | Displacement per unit time |
| Can never be negative | Can be negative, zero, or positive (since it depends on direction) |
| Example: A car moves at 60 km/h | Example: A car moves at 60 km/h towards the east |
❓ Question 2
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
✅ Answer:
The magnitude of average velocity = average speed when an object moves along a straight line without changing direction.
Example:
- A car travelling 100 km straight on a highway without turning back → distance = displacement, hence average speed = average velocity.
❓ Question 3
What does the odometer of an automobile measure?
✅ Answer:
An odometer is a device fitted in vehicles that measures the total distance travelled by the automobile.
👉 It does not measure displacement, only distance.
❓ Question 4
What does the path of an object look like when it is in uniform motion?
✅ Answer:
When an object is in uniform motion, it covers equal distances in equal intervals of time.
- The path looks like a straight line in a distance–time graph.
- In real life, this means the object is moving in a straight path at a constant speed.
❓ Question 5
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, i.e., 3 × 10⁸ m/s.
✅ Answer:
- Time taken = 5 minutes = 5 × 60 = 300 s
- Speed of signal = 3 × 10⁸ m/s
- Distance = Speed × Time = (3 × 10⁸) × 300 = 9 × 10¹⁰ m
✅ Final Answer: Distance of spaceship = 9 × 10¹⁰ m
📝 In-text Questions and Answers (Acceleration)
❓ Question 1
When will you say a body is in (i) uniform acceleration, (ii) non-uniform acceleration?
✅ Answer:
-
Uniform Acceleration:
When a body’s velocity changes by equal amounts in equal intervals of time.
👉 Example: A freely falling body near Earth’s surface (g = 9.8 m/s²). -
Non-uniform Acceleration:
When a body’s velocity changes by unequal amounts in equal intervals of time.
👉 Example: A car moving on a busy road, slowing down and speeding up irregularly.
❓ Question 2
A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.
✅ Answer:
- Initial velocity, u = 80 km/h = (80 × 1000 / 3600) = 22.22 m/s
- Final velocity, v = 60 km/h = (60 × 1000 / 3600) = 16.67 m/s
- Time, t = 5 s
a = `\frac{v - u}{t} = \frac{16.67 - 22.22}{5} = -1.11 \`, m/s^2
✅ Final Answer: Acceleration = –1.11 m/s² (negative sign shows retardation).
❓ Question 3
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.
✅ Answer:
- Initial velocity, u = 0 (starting from rest)
- Final velocity, v = 40 km/h = (40 × 1000 / 3600) = 11.11 m/s
- Time, t = 10 minutes = 600 s
a = `\frac{v - u}{t} = \frac{11.11 - 0}{600} = 0.0185 \`, m/s^2
✅ Final Answer: Acceleration = 0.0185 m/s²
📝 In-text Questions and Answers (Graphs of Motion)
❓ Question 1
What is the nature of the distance–time graphs for uniform and non-uniform motion of an object?
✅ Answer:
- Uniform Motion: The distance–time graph is a straight line because the object covers equal distances in equal intervals of time.
- Non-uniform Motion: The distance–time graph is a curved line because the object covers unequal distances in equal intervals of time.
👉 Straight line = constant speed.
👉 Curved line = speed is changing.
❓ Question 2
What can you say about the motion of an object whose distance–time graph is a straight line parallel to the time axis?
✅ Answer:
If the distance–time graph is a horizontal line (parallel to the time axis), it means the distance does not change with time.
👉 The object is at rest.
Example: A parked car.
❓ Question 3
What can you say about the motion of an object if its speed–time graph is a straight line parallel to the time axis?
✅ Answer:
If the speed–time graph is a horizontal straight line, it means the object is moving with constant speed (uniform motion).
Example: A train moving at 60 km/h steadily.
📝 In-text Questions and Answers (Equations of Motion)
❓ Question 1
A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
✅ Answer:
- Initial velocity, u = 0 (rest)
- Acceleration, a = 0.1 m/s²
- Time, t = 2 min = 120 s
(a) Speed acquired
`v = u + at = 0 + (0.1)(120) = 12 \`, m/s
(b) Distance travelled
`s = ut + \tfrac{1}{2} at^2
= 0 + \tfrac{1}{2} (0.1)(120^2)
= 0.05 \times 14400 = 720 \`, m
✅ Final Answer: Speed = 12 m/s, Distance = 720 m
❓ Question 2
A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of –0.5 m/s². Find how far the train will go before it is brought to rest.
✅ Answer:
- Initial velocity, u = 90 km/h = (90 × 1000 / 3600) = 25 m/s
- Final velocity, v = 0 (rest)
- Acceleration, a = –0.5 m/s²
Using the equation:
`v^2 = u^2 + 2as
0 = (25)^2 + 2(-0.5)(s)
0 = 625 – s
s = 625 \`, m
✅ Final Answer: The train will travel 625 m before stopping.
❓ Question 3
A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?
✅ Answer:
- Initial velocity, u = 0 (starts from rest)
- Acceleration, a = 2 cm/s² = 0.02 m/s²
- Time, t = 3 s
`v = u + at = 0 + (0.02)(3) = 0.06 \`, m/s
✅ Final Answer: Velocity after 3 s = 0.06 m/s
❓ Question 4
A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?
✅ Answer:
- Initial velocity, u = 0
- Acceleration, a = 4 m/s²
- Time, t = 10 s
`s = ut + \tfrac{1}{2} at^2
= 0 + \tfrac{1}{2} (4)(10^2)
= 2 \times 100 = 200 \`, m
✅ Final Answer: The racing car covers 200 m in 10 seconds.
❓ Question 5
A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
✅ Answer:
- Initial velocity, u = 5 m/s
- Final velocity at highest point, v = 0
- Acceleration, a = –10 m/s²
Using
`v^2 = u^2 + 2as
0 = (5)^2 + 2(-10)(s)
0 = 25 – 20s
s = 1.25 \`, m
Now,
`v = u + at
0 = 5 + (-10)t
t = 0.5 \`, s
✅ Final Answer: Maximum height = 1.25 m, Time to reach = 0.5 s
📝 In-text Questions and Answers (Uniform Circular Motion)
❓ Question
In the activity with a stone tied to a thread and whirled in a circular path, if you release the thread, can you tell the direction in which the stone moves after it is released?
✅ Answer:
When the stone is released, it moves along a straight line tangential to the circular path at the point of release.
👉 This is because of Newton’s First Law of Motion (Law of Inertia): a body continues to move in the direction of its velocity unless acted upon by an external force.
👉 While the stone is tied, the tension in the thread provides the centripetal force to keep it moving in a circle. Once released, no force acts toward the center, so it flies off tangentially.
🏁 Motion – Class 9 Science NCERT Exercise Solutions (Complete)
Based on NCERT Class 9 Science, Chapter: Motion — Exercises, pp. 85–86.
Q1) Athlete on a circular track
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?
Given: Diameter = 200 m ⇒ Circumference = π × D = 200π m
Time per round = 40 s; Total time = 140 s ⇒ Rounds = 140/40 = 3.5 rounds
- Distance = 3.5 × 200π = 700π m ≈ 2199.1 m
- Displacement after 3.5 rounds = diameter = 200 m (he’s at the point opposite the start)
✅ Answer: Distance ≈ 2199.1 m; Displacement = 200 m (along the line through the circle’s center).
Q2) Joseph jogs on a straight 300 m road
A→B in 2 min 30 s, then 100 m back to C in 1 min. Find average speeds and velocities for (a) A→B, (b) A→C.
(a) A→B
- Distance = 300 m; Time = 150 s
- Average speed = 300/150 = 2 m/s
- Average velocity = displacement/time = 300/150 = 2 m/s towards B
(b) A→C
- Distance = 300 + 100 = 400 m; Time = 150 + 60 = 210 s
- Average speed = 400/210 ≈ 1.90 m/s
- Average velocity = (displacement A→C)/time = 200/210 ≈ 0.95 m/s towards B
Q3) Average speed for a two-way trip
Abdul’s onward average speed is 20 km/h and return is 30 km/h along the same route. Find overall average speed.
For equal distances:
`\text{Average speed}=\frac{2uv}{u+v}=\frac{2\times20\times30}{20+30}=\mathbf{24\ km/h}`
Q4) Motorboat with constant acceleration
From rest, a motorboat accelerates at 3.0 m/s² for 8.0 s. How far does it travel?
s=`\tfrac12 at^2=\tfrac12\times3.0\times(8.0)^2=\mathbf{96\ m}`
Q5) Braking car / speed–time graph (method guide)
Your exercise page shows a speed–time graph for a car. You’re asked to (a) find distance in the first 4 s (and shade that area), and (b) identify the uniform motion segment. Here’s how to do it with your edition’s numbers:
- (a) Distance in first 4 s = Area under v–t curve from 0 to 4 s.
- If it’s a straight line up from 0, the area is a triangle: .
- If it’s partly flat, add rectangle + triangle as needed.
- (b) Uniform motion corresponds to a horizontal (flat) segment of the v–t graph (constant speed). Just quote the time interval from the graph’s x-axis.
📌 Plug in the exact speeds you read at 4 s (and the flat segment times) from your printed figure to get the numeric distance and the interval(s) of uniform motion.
Q6) Three objects A, B, C on a distance–time graph
The figure in the exercise shows three lines. Answer: (a) fastest, (b) ever same point?, (c) distance of C when B passes A, (d) distance of B when it passes C. Method:
- (a) Fastest ⇒ steepest slope (largest Δdistance/Δtime).
- (b) Same point ⇒ Do the lines intersect? If any two lines cross, those two objects are at the same position at the same time.
- (c) Read the time when B passes A (their lines intersect). Then read C’s distance at that same time from C’s line.
- (d) Similarly, find when B passes C (their intersection) and read B’s distance there.
📌 Use the exact coordinates from your book’s axes for final numerical answers.
Q7) Ball dropped from 20 m (g = 10 m/s²)
Find impact velocity and time.
- (downward)
Q8) Speed–time graph for a car (another figure)
(a) Distance in first 4 s = area under v–t from 0–4 s. (b) Uniform motion part?
- (a) Compute area under the curve (triangle/rectangle/trapezium as per the figure).
- (b) Look for flat (horizontal) segment(s)—that’s where speed is constant.
📌 Read heights (speeds) and widths (time intervals) directly from the printed axes to finalize the number.
Q9) Which situations are possible? Give an example each.
(a) Constant acceleration but zero velocity — Possible (instantaneously).
Example: A ball thrown straight up has constant acceleration (gravity) and at the topmost point its velocity is momentarily zero.
(b) Acceleration with uniform speed — Possible.
Example: Uniform circular motion at constant speed has acceleration towards the center (centripetal) while the speed stays constant.
(c) Motion in one direction with acceleration perpendicular to it — Possible.
Example: A charged particle moving straight into a uniform magnetic field (field perpendicular to velocity) experiences acceleration perpendicular to motion (and bends into a circle/arc).
Q10) Artificial satellite in circular orbit
Orbit radius = 42,250 km; period = 24 h. Find speed.
Convert to SI: ,
v=`\frac{2\pi r}{T}=\frac{2\pi(4.225\times10^7)}{86400}\approx \mathbf{3.08\times10^3\ m/s}\ (\approx \mathbf{3.08\` km/s
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