1
To state the position of an object we need —
💡 Hint:
The school and the railway station.
Explanation:
An origin is needed. If the school is 2 km north of the station, the station is the origin.
Class 9 · Science · Chapter 8 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Motion
Motion
How to use this page:
1. Read — 8.1 walls · 8.2 train · 8.3 court · 8.4 odometer · 8.5 table · 8.6 walk · 8.7 thunder · 8.8 acceleration · 8.9 train · 8.10 cycles · 8.11 stone, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. Use the distance-time figure — the straight line shows uniform motion and the rising curve shows non-uniform motion.
In the Bihar book this is chapter 8. In the NCERT 2026-27 Exploration book, motion is chapter 4. Progress stays in this browser.
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- 1 Motion, rest and a reference point — Activities 8.1 and 8.2
- 2 Distance and displacement — Activities 8.3 and 8.4
- 3 Uniform motion and speed — Activities 8.5 and 8.6
- 4 Velocity and the distance of lightning — Activity 8.7
- 5 Acceleration — Activity 8.8
- 6 Graphs — Activities 8.9 and 8.10
- 7 The three equations of motion
- 8 Uniform circular motion — Activity 8.11
- Chapter winner — every lesson at mastery ★
🎬 Shorts
This chapter has no Short on the channel yet, so ten empty cards are not shown. The notes, diagrams and checks are complete.
📄 PDF download — notes + answer key
Motion, rest and a reference point — Activities 8.1 and 8.2
गति, विराम और निर्देश बिंदु — क्रियाकलाप 8.1 और 8.2 · Section 8.1 · Activities 8.1, 8.2
A position needs an originNotes
An object is in motion when its position changes with time. The motion of air is read from dust and from leaves. Sunrise and the seasons are tied to the motion of the Earth, but we do not see that motion directly.
From a moving bus the roadside trees seem to go backward. A person on the road sees the bus moving. Inside the bus the passengers see one another at rest. So motion and rest depend on the observer.
To state a position we choose a reference point, called the origin. A village school is 2 km north of the railway station. Here the station is the origin. Another origin can be chosen. This chapter first takes motion along a straight line, then equations and graphs, and at the end circular motion.
Activities 8.1 and 8.2 — the wall and the trainActivity
Activity 8.1: are the classroom walls at rest or in motion? Relative to the room they are at rest. Relative to a frame outside the Earth they move with the Earth. Without a reference point the answer is incomplete.
Activity 8.2: you sit in a stationary train and the train beside you moves, so your own train seems to move. That is relative motion. A look at the platform through your window breaks the illusion. Uncontrolled motion, as in a flood or a tsunami, is dangerous. Controlled motion, as in hydro-electric power, is useful.
Worked exampleExample
Question: Why does a bus passenger see a fellow passenger at rest and a tree in motion?
Answer: The fellow passenger’s position does not change relative to the bus, so that passenger is at rest. The tree’s position changes relative to the bus, so the tree appears in motion. For a person standing on the road the bus is in motion. All three statements are right from their own reference points.
10-second revision
- A position is incomplete without an origin
- A wall is at rest relative to the room
- A stationary train can seem to move because of the train beside it
Board tip · BSEBBoard tip
In an answer about rest or motion, name the reference point: the room, the road or the Earth.
Board tip · CBSEBoard tip
Give the train as the example of relative motion. Do not write only “everything moves”.
Check your understandingall correct = mastery ★
2
The classroom walls are at rest from every reference point.
Explanation:
False — they are at rest relative to the room, and they can also be described as moving with the Earth. Activity 8.1.
3
A stationary train seems to move when —
💡 Hint:
Activity 8.2.
Explanation:
The motion of the neighbouring train creates the illusion that your train is moving.
4
Define motion. Name one motion that is not seen directly.
Model answer (board format)
Motion is a change of position with time. The motion of air is inferred from dust or leaves. The motion of the Earth is also not seen directly, but sunrise and the seasons are tied to it.
Check your answer:
Distance and displacement — Activities 8.3 and 8.4
दूरी और विस्थापन — क्रियाकलाप 8.3 और 8.4 · Section 8.1.1 · Activities 8.3, 8.4
Path length and the shortest gapNotes
An object leaves O, passes C and B, reaches A, and then returns through B to C. OA = 60 km. When the path is OA + AC = 60 km + 35 km = 95 km, that is the distance. Distance is written as a magnitude only, with no direction.
Displacement is the shortest distance from the start to the end. From O to A the distance and the displacement are both 60 km. From O to A and then to B the distance is 60 km + 25 km = 85 km and the magnitude of the displacement is 35 km. If the object returns to O the displacement is 0 and the distance is 60 km + 60 km = 120 km. The magnitude of displacement cannot exceed the distance.
Activities 8.3 and 8.4 — the court and the odometerActivity
Activity 8.3: take a metre scale and a long rope. Walk from one corner of a basketball court to the opposite corner along the sides. Measure the path with the rope. The displacement is the diagonal, so it is shorter than the distance.
Activity 8.4: an odometer shows the distance travelled. From Bhubaneswar to New Delhi the odometer difference is 1850 km. That is the distance. The straight map distance is the magnitude of the displacement and it is less than 1850 km, because the road is not straight.
Worked exampleExample
Question: A farmer moves along the boundary of a square field of side 10 m in 40 s. What is the magnitude of his displacement after 2 min 20 s?
Formula: distance = speed × time. One round = 4 × side.
Substitution: perimeter = 4 × 10 m = 40 m. Speed = 40 m / 40 s = 1 m/s. Time = 2 × 60 s + 20 s = 140 s. Distance = 1 m/s × 140 s = 140 m. Rounds = 140 m / 40 m = 3.5. After three full rounds he is at the start. Half a round = 20 m = two sides, so the opposite corner.
Displacement = √(10² + 10²) m = √200 m = 10√2 m ≈ 14.1 m.
10-second revision
- Distance has only a magnitude; displacement has a direction
- Zero displacement can sit with a non-zero distance
- An odometer measures distance, not the straight map distance
Board tip · BSEBBoard tip
In the farmer answer write all three figures: 140 s, 3.5 rounds and 10√2 m. Do not stop at 14 m.
Board tip · CBSEBoard tip
“Displacement is greater than distance” is a wrong option. The magnitude is always smaller than or equal to the distance.
Check your understandingall correct = mastery ★
1
If an object moves and returns to the start —
💡 Hint:
From O to A and back to O.
Explanation:
The displacement is 0. The distance can be 120 km if OA = 60 km.
2
The magnitude of displacement can be greater than the distance travelled.
Explanation:
False — the shortest path cannot be longer than the whole path.
3
From Bhubaneswar to New Delhi the odometer shows a distance of ______ km.
💡 Hint:
Activity 8.4. The displacement is smaller.
Explanation:
1850 km. That is the distance, not the straight displacement.
4
An odometer measures —
💡 Hint:
The distance meter of a car.
Explanation:
The odometer measures distance. A speedometer shows speed.
5
Find the magnitude of displacement after 3.5 rounds of a square of side 10 m. Write the formula and the unit.
Model answer (board format)
After three rounds the farmer is at the start. Half a round is two sides, so he is at the opposite corner. Displacement = √(10² + 10²) = √200 = 10√2 m ≈ 14.1 m.
Check your answer:
Uniform motion and speed — Activities 8.5 and 8.6
एकसमान गति और चाल — क्रियाकलाप 8.5 और 8.6 · Sections 8.1.2 and 8.2 · Table 8.1
Equal distances in equal timesNotes
If an object covers 5 m in every second it is in uniform motion. Unequal distances in equal times are non-uniform motion, as with a car on a crowded street or a person jogging in a park. The time interval should be small.
Speed = distance / time. v = s / t. The SI unit is m/s. cm/s and km/h are also used. Average speed = total distance / total time. A car that covers 100 km in 2 h has an average speed of 50 km/h. It need not have held that speed the whole time.
The factor from km/h to m/s is 5/18. 50 km/h = 50 × (5/18) = 250/18 = 13.9 m/s.
Activities 8.5 and 8.6 — the table and a walking distanceActivity
In Table 8.1, every 15 min the distance of A grows as 10, 20, 30, 40, 50, 60, 70 m. Each step is 10 m, so A is uniform. The distances of B are 12, 19, 23, 35, 37, 41, 44 m. The steps are 7, 4, 12, 2, 4, 3 m, so B is non-uniform.
Activity 8.6: measure the time from your house to the bus stop or the school. Take the average walking speed as 4 km/h. Distance = speed × time. If the time is in hours the distance comes out in kilometres. 15 min = 0.25 h, so distance = 4 km/h × 0.25 h = 1 km.
Worked exampleExample
Question: An object travels 16 m in 4 s and then another 16 m in 2 s. Find the average speed.
Formula: average speed = total distance / total time.
Substitution: total distance = 16 m + 16 m = 32 m. Total time = 4 s + 2 s = 6 s. Average speed = 32 m / 6 s = 5.33 m/s.
Answer: 5.33 m/s. This is not a uniform speed, because the object first moved at 4 m/s and then at 8 m/s.
10-second revision
- Uniform motion: equal distances in equal times
- Average speed = total distance / total time, unit m/s
- 50 km/h = 13.9 m/s, factor 5/18
Board tip · BSEBBoard tip
In Table 8.1 write the steps. Only “A is straight” stays incomplete.
Board tip · CBSEBoard tip
Do not push a direction into average speed. Direction belongs to velocity. Multiply by 5/18 to go from km/h to m/s.
Check your understandingall correct = mastery ★
1
In Table 8.1 the motion of object A is —
💡 Hint:
The steps of B are not equal.
Explanation:
A is uniform. B is non-uniform because the steps are 7, 4, 12, 2, 4 and 3 m.
2
An average speed of 50 km/h means the car stayed at 50 km/h at every instant.
Explanation:
False — it is total distance divided by total time. The speed in between can be higher or lower.
3
The SI unit of speed is ______.
💡 Hint:
km/h is used, but it is not the SI unit.
Explanation:
m/s. 1 km/h = 5/18 m/s.
4
The average speed for 16 m in 4 s and then 16 m in 2 s is —
💡 Hint:
Total distance 32 m, total time 6 s.
Explanation:
32 m / 6 s = 5.33 m/s. 4 and 8 are the speeds of the separate pieces.
5
At 4 km/h, the distance walked in 30 min is —
💡 Hint:
30 min = 0.5 h. Distance = speed × time.
Explanation:
4 km/h × 0.5 h = 2 km. The method of Activity 8.6.
6
An odometer goes from 2000 km to 2400 km and the trip takes 8 h. Find the average speed in km/h and in m/s.
Model answer (board format)
Formula: average speed = distance / time. Distance = 2400 km − 2000 km = 400 km. Speed = 400 km / 8 h = 50 km/h. In m/s: 50 × 5/18 = 250/18 = 13.9 m/s.
Check your answer:
Velocity and the distance of lightning — Activity 8.7
वेग और बिजली की दूरी — क्रियाकलाप 8.7 · Section 8.2.1 · Activity 8.7
Add a direction and speed becomes velocityNotes
Velocity is speed with a stated direction. Change the speed, the direction, or both, and the velocity changes. Average velocity = displacement / time. When the velocity changes at a uniform rate, vav = (u + v) / 2. Speed and velocity share the unit m/s.
The magnitude of average velocity equals the average speed only when the motion is straight and the direction does not change, because then the distance and the magnitude of displacement are equal. In uniform motion the velocity stays constant and the change in velocity is zero.
Activity 8.7 — lightning and thunderActivity
In a cloud the lightning is seen first and the thunder is heard later, because light is much faster than sound. Measure the time gap with a watch. The speed of sound in air is 346 m/s.
Distance = speed × time. For a delay of 3 s the distance = 346 m/s × 3 s = 1038 m. The time of the light is left out of this sum because it is very small. If a spaceship signal arrives in 5 min and the speed of light is 3 × 108 m/s, the distance = 3 × 108 × 300 = 9 × 1010 m.
Worked exampleExample
Question: Usha swims 180 m in one minute in a 90 m pool, from one end to the other and back. Find her average speed and average velocity.
Formula: average speed = total distance / total time. Average velocity = displacement / total time.
Substitution: distance = 180 m. Time = 1 min = 60 s. Displacement = 0 m, because she returns. Average speed = 180 m / 60 s = 3 m/s. Average velocity = 0 m / 60 s = 0 m/s.
10-second revision
- Velocity = displacement / time, with a direction
- At a uniform rate, average velocity is (u + v) / 2
- Speed of sound is 346 m/s; distance = 346 × time
Board tip · BSEBBoard tip
In the Usha answer write both the speed 3 m/s and the velocity 0 m/s. Do not swap them.
Board tip · CBSEBoard tip
In the lightning figure keep the unit m. Do not turn 346 into km/h when the time is in seconds.
Check your understandingall correct = mastery ★
1
The magnitude of average velocity equals the average speed when —
💡 Hint:
Then distance = magnitude of displacement.
Explanation:
Along a straight line without a change of direction. On returning, the velocity can be zero while the speed is not.
2
To change the velocity, the speed must change.
Explanation:
False — a change of direction alone also changes the velocity. Circular motion is that case.
3
When velocity changes at a uniform rate, the average velocity is ______.
💡 Hint:
The arithmetic mean of the initial and the final.
Explanation:
(u + v) / 2.
4
If the thunder arrives 2 s late, the distance of the lightning is about —
💡 Hint:
The speed is 346 m/s.
Explanation:
Distance = 346 m/s × 2 s = 692 m.
5
Usha swims 180 m and returns in 60 s. Find the average speed and the average velocity with the formulas.
Model answer (board format)
Average speed = 180 m / 60 s = 3 m/s. The displacement is 0 m. Average velocity = 0 m / 60 s = 0 m/s.
Check your answer:
Acceleration — Activity 8.8
त्वरण — क्रियाकलाप 8.8 · Section 8.3 · Activity 8.8
Change in velocity divided by timeNotes
Acceleration is the rate of change of velocity. a = (v − u) / t. The SI unit is m/s². Acceleration along the velocity is positive and acceleration opposite the velocity is negative.
If the velocity increases or decreases by equal amounts in equal times, the acceleration is uniform. Free fall is uniformly accelerated motion. If a car in traffic changes its velocity by unequal amounts, the acceleration is non-uniform.
Rahul starts from rest and reaches 6 m/s in 30 s. a = (6 m/s − 0) / 30 s = 0.2 m/s². In the next 5 s the velocity becomes 4 m/s. a = (4 − 6) / 5 = −0.4 m/s². The negative sign shows the brakes.
Activity 8.8 — four faces of accelerationActivity
Pick one example each from everyday motion. (a) Acceleration along the motion: a bicycle gaining speed. (b) Acceleration against the motion: applying brakes. (c) Uniform acceleration: free fall. (d) Non-uniform acceleration: a car on a crowded road.
Uniform acceleration does not mean uniform speed. The speed is changing, but the rate of that change is steady.
Worked exampleExample
Question: A bus slows from 80 km/h to 60 km/h in 5 s. Find the acceleration.
Formula: a = (v − u) / t. First convert km/h to m/s with the factor 5/18.
Substitution: u = 80 × 5/18 = 200/9 m/s = 22.2 m/s. v = 60 × 5/18 = 50/3 m/s = 16.7 m/s. v − u = 150/9 − 200/9 = −50/9 m/s. a = (−50/9) / 5 = −10/9 m/s² ≈ −1.11 m/s².
Answer: −1.11 m/s². The sign says the acceleration is opposite the velocity.
10-second revision
- a = (v − u) / t, unit m/s²
- Brakes give a negative acceleration
- Free fall is uniform acceleration
Board tip · BSEBBoard tip
Do not subtract km/h with a time in seconds. First make m/s with 5/18, then (v − u) / t.
Board tip · CBSEBoard tip
Do not write negative acceleration as a different physical quantity. It is the same a, and the sign shows the opposite direction.
Check your understandingall correct = mastery ★
1
The SI unit of acceleration is —
💡 Hint:
Velocity divided by time.
Explanation:
m/s². The unit of speed is m/s.
2
Free fall is an example of non-uniform acceleration.
Explanation:
False — free fall is uniformly accelerated motion. A car in traffic is non-uniform acceleration.
3
From rest to 6 m/s in 30 s, the acceleration is —
💡 Hint:
u = 0.
Explanation:
a = (6 − 0) / 30 = 0.2 m/s².
4
A train starts from rest and reaches 40 km/h in 10 min. Find the acceleration in m/s². Write the formula, the substitution and the unit.
Model answer (board format)
a = (v − u) / t. u = 0. v = 40 × 5/18 = 100/9 m/s. t = 10 × 60 = 600 s. a = (100/9) / 600 = 1/54 m/s² ≈ 0.0185 m/s².
Check your answer:
Graphs — Activities 8.9 and 8.10
ग्राफ — क्रियाकलाप 8.9 और 8.10 · Section 8.4 · Tables 8.2 to 8.5
Distance-time and velocity-timeNotes
On a distance-time graph time is on the x-axis and distance on the y-axis. At uniform speed, distance is proportional to time, so the graph is a straight line. The slope is the speed: v = (s₂ − s₁) / (t₂ − t₁). A straight line parallel to the time axis means rest.
In Table 8.2 the car’s distance at 0, 2, 4, 6, 8, 10 and 12 s is 0, 1, 4, 9, 16, 25 and 36 m. That is not a straight line. A curve shows non-uniform speed. On a velocity-time graph, uniform velocity is a line parallel to the time axis, such as 40 km/h. For uniform acceleration the velocity-time graph is a straight slanted line. The area between this graph and the time axis is the magnitude of the displacement. For non-uniform acceleration the shape can be anything.
Activities 8.9 and 8.10 — a train and two bicyclesActivity
Activity 8.9: stations A, B and C. B is 120 km from A, departure 08:15, arrival 11:15, so 3 h and a speed of 40 km/h. C is 60 km from B, departure 11:30, arrival 13:00, so 1.5 h and again 40 km/h. A 15 min stop at each station is a horizontal line on the distance-time graph. Between stops the graph is a straight line because the motion is taken as uniform.
Activity 8.10: Feroz and Sania leave at the same time and take the same route. Feroz reaches 3.6 km by 8:20. Sania completes the same distance at 8:25. Feroz is faster. The five-minute pieces of both are not equal, so both are non-uniform. Draw both lines on one graph.
Worked exampleExample
Question: On a distance-time graph A is (1 s, 4 m) and B is (5 s, 20 m), and AB is a straight line. Find the speed.
Formula: v = (s₂ − s₁) / (t₂ − t₁).
Substitution: v = (20 m − 4 m) / (5 s − 1 s) = 16 m / 4 s = 4 m/s.
Answer: 4 m/s. Because the line is straight, this speed is uniform.
10-second revision
- A straight distance-time line is uniform speed
- A distance-time curve is non-uniform speed
- The area of a velocity-time graph is the magnitude of the displacement
Board tip · BSEBBoard tip
In a graph answer name the axes: time horizontal, distance or velocity vertical.
Board tip · CBSEBoard tip
The area is asked for the velocity-time graph, not the distance-time graph. Do not mix the two.
Check your understandingall correct = mastery ★
1
The distance-time graph of uniform speed is —
💡 Hint:
A curve is non-uniform speed. A horizontal line is rest.
Explanation:
A straight slanted line. A line parallel to the time axis is rest.
2
The area between a velocity-time graph and the time axis gives the magnitude of the displacement.
Explanation:
True — for uniform velocity this is the area of a rectangle.
3
If a distance-time graph is parallel to the time axis, the object is —
💡 Hint:
The distance is not increasing.
Explanation:
At rest. A horizontal velocity-time line means uniform velocity, not rest.
4
From station A to B the train covers 120 km in 3 h, so the speed is ______ km/h.
💡 Hint:
Activity 8.9. B to C has the same speed.
Explanation:
40 km/h. The next 60 km in 1.5 h is also 40 km/h.
5
Feroz finishes 3.6 km at 8:20 and Sania at 8:25. Who is faster?
💡 Hint:
Less time, the same distance.
Explanation:
Feroz. He arrives 5 min earlier. Activity 8.10.
6
Table 8.2 has distances 0, 1, 4, 9, 16, 25, 36 m at intervals of 2 s. What will the graph look like, and why?
Model answer (board format)
The distance-time graph will be a curve, not a straight line. In equal times the distances increase by 1, 3, 5, 7, 9 and 11 m, so the speed is non-uniform.
Check your answer:
The three equations of motion
गति के तीन समीकरण · Section 8.5 · uniform acceleration
Only uniform acceleration and a straight lineNotes
For an object moving in a straight line with uniform acceleration there are three equations.
v = u + at, s = ut + ½at², v² = u² + 2as
u is the initial velocity, v the final velocity, a the acceleration, t the time and s the distance. The first is the velocity-time relation, the second the position-time relation, and the third the position-velocity relation. The third comes by removing t from the first two. On the velocity-time graph, s is the sum of the areas of a rectangle and a triangle. For brakes, keep a negative. These equations do not apply directly to non-uniform acceleration.
Worked exampleExample
Question: A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Find the acceleration and the distance.
Formula: a = (v − u) / t and s = ut + ½at².
Substitution: u = 18 × 5/18 = 5 m/s. v = 36 × 5/18 = 10 m/s. t = 5 s. a = (10 m/s − 5 m/s) / 5 s = 1 m/s². s = (5 m/s)(5 s) + ½(1 m/s²)(5 s)² = 25 m + 12.5 m = 37.5 m.
Answer: The acceleration is 1 m/s² and the distance is 37.5 m.
A second pattern, the brakes
Brakes give a = −6 m/s², t = 2 s, and at the stop v = 0. From v = u + at, 0 = u + (−6 m/s²)(2 s), so u = 12 m/s. Then s = ut + ½at² = (12 m/s)(2 s) + ½(−6 m/s²)(2 s)² = 24 m − 12 m = 12 m. The car travels another 12 m before it stops. This is why a gap is kept between vehicles.
10-second revision
- v = u + at
- s = ut + ½at²
- v² = u² + 2as, and these are for uniform acceleration
Board tip · BSEBBoard tip
When you write the three equations, say that they are for uniform acceleration and a straight line.
Board tip · CBSEBoard tip
Dropping the unit costs marks. Write m/s, m/s² and m as different units.
Check your understandingall correct = mastery ★
1
v² = u² + 2as applies when —
💡 Hint:
All three equations share one condition.
Explanation:
Uniform acceleration on a straight line. Do not apply it directly to non-uniform acceleration.
2
In s = ut + ½at², if a is negative the second term of the distance is negative.
Explanation:
True — in the brake example, 24 m − 12 m = 12 m.
3
Starting from rest, the distance in time t at uniform acceleration a is ______.
💡 Hint:
With u = 0 the first term drops.
Explanation:
½at². The full equation is s = ut + ½at².
4
For u = 0, a = 0.1 m/s² and t = 120 s, the speed is —
💡 Hint:
v = u + at.
Explanation:
v = 0 + 0.1 × 120 = 12 m/s. The distance would be ½ × 0.1 × 14400 = 720 m.
5
A bus starts from rest at 0.1 m/s² for 2 min. Find the speed gained and the distance.
Model answer (board format)
v = u + at = 0 + (0.1 m/s²)(120 s) = 12 m/s. s = ut + ½at² = 0 + ½(0.1)(120)² = ½ × 0.1 × 14400 = 720 m.
Check your answer:
Uniform circular motion — Activity 8.11
एकसमान वृत्तीय गति — क्रियाकलाप 8.11 · Section 8.6 · Activity 8.11
The speed is steady, the direction is notNotes
On a rectangular path an athlete changes direction four times, at the corners. On a hexagon, six times, and on an octagon, eight times. Keep adding sides and the path approaches a circle, and the direction changes continuously. Even if the magnitude of the speed is constant, the velocity changes, so this is accelerated motion.
Motion on a circular path at uniform speed is called uniform circular motion. If the radius is r and the time for one round is t, the speed is v = 2πr / t. The Moon, the Earth, a satellite and a cyclist on a circular track at steady speed are examples.
Activity 8.11 — a stone on a threadActivity
Tie a small stone to one end of a thread. Hold the other end and whirl the stone in a circle at constant speed. Then let the thread go.
As soon as it is released the stone flies straight along the tangent to the circle. Release it at different places and the direction is not the same each time. On the circle the direction of the velocity is along the tangent at every point. A hammer or a discus, once released, also goes along the direction it had at the release.
Worked exampleExample
Question: An athlete completes one round of a circular track of diameter 200 m in 40 s. Find the distance and the displacement in 2 min 20 s.
Formula: distance of one round = 2πr. v = 2πr / t. Take π = 22/7.
Substitution: r = 100 m. One round = 2 × (22/7) × 100 = 4400/7 m. Time = 140 s. Rounds = 140/40 = 3.5. Distance = 3.5 × (4400/7) m = 2200 m. 3.5 rounds means the opposite point, so displacement = diameter = 200 m. Speed = (4400/7) / 40 = 15.7 m/s.
10-second revision
- Uniform circular motion is accelerated because the direction changes
- v = 2πr / t
- A released stone goes along the tangent
Board tip · BSEBBoard tip
Do not swap diameter and radius. If the diameter is 200 m then r = 100 m. At 3.5 rounds the displacement is the diameter.
Board tip · CBSEBoard tip
“The speed is constant so the acceleration is zero” is wrong. The direction is changing, so the velocity is changing.
Check your understandingall correct = mastery ★
1
In uniform circular motion, what stays constant is —
💡 Hint:
The direction changes at every point.
Explanation:
The magnitude of the speed is constant. The velocity is not, because the direction changes.
2
A stone released from the thread moves toward the centre of the circle.
Explanation:
False — it goes straight along the tangent at that instant. Activity 8.11.
3
If the radius is r and the time for one round is t, the speed is ______.
💡 Hint:
Circumference divided by time.
Explanation:
v = 2πr / t.
4
One round of a 200 m diameter takes 40 s. Find the distance and the displacement in 140 s. Take π = 22/7.
Model answer (board format)
r = 100 m. One round = 2πr = 4400/7 m. Rounds = 140/40 = 3.5. Distance = 3.5 × 4400/7 = 2200 m. Displacement = 200 m, because 3.5 rounds leave the athlete at the opposite point.
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❓ Full question bank — with answers and explanations — 67 questions
Pick a type. The 39 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ140 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA120 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/14Pick one option. A wrong try brings a hint.
1
The reference point is called —
💡 Hint:
The school is 2 km north of the station.
Explanation:
The origin.
2
From O to A and back to O, with OA = 60 km, the distance is —
💡 Hint:
The displacement is 0.
Explanation:
120 km. The displacement is zero.
3
Object B of Table 8.1 is —
💡 Hint:
The steps are 7, 4, 12, 2, 4, 3 m.
Explanation:
Non-uniform. A is uniform.
4
32 m in 6 s. The average speed is —
💡 Hint:
Total distance divided by total time.
Explanation:
32/6 = 5.33 m/s.
5
50 km/h is about —
💡 Hint:
Multiply by 5/18.
Explanation:
50 × 5/18 = 13.9 m/s.
6
After one minute, Usha’s average velocity is —
💡 Hint:
She returns.
Explanation:
0 m/s. The average speed is 3 m/s.
7
In a = (v − u) / t, a bus from 80 to 60 km/h in 5 s has acceleration about —
💡 Hint:
First make m/s.
Explanation:
−10/9 m/s² ≈ −1.11 m/s².
8
A curve on a distance-time graph shows —
💡 Hint:
Table 8.2.
Explanation:
Non-uniform speed. A straight line is uniform speed.
9
The area of a velocity-time graph gives —
💡 Hint:
A rectangle or a trapezium.
Explanation:
The magnitude of the displacement.
10
For u = 5 m/s, a = 1 m/s² and t = 5 s, the distance is —
💡 Hint:
s = ut + ½at².
Explanation:
25 + 12.5 = 37.5 m.
11
After 3.5 rounds of a 200 m diameter, the displacement is —
💡 Hint:
The opposite point. The distance is 2200 m.
Explanation:
200 m. The distance is 2200 m.
12
A stone released from the thread goes —
💡 Hint:
Activity 8.11.
Explanation:
Along the tangent.
13
A speedometer shows —
💡 Hint:
The odometer shows distance.
Explanation:
Speed. The book reads the speedometer for the velocity-time table.
14
20 km/h going and 30 km/h returning, equal distances. The average speed of the whole trip is —
💡 Hint:
Not the arithmetic mean.
Explanation:
24 km/h. The time is d/20 + d/30 and the total distance is 2d.
True or false
0/81
Displacement can never be zero.
Explanation:
False — return to the start and the displacement is zero while the distance is not.
2
A car on a crowded street is usually in non-uniform motion.
Explanation:
True — the distances in equal times are not equal.
3
Average velocity is always (u + v) / 2, whatever the change.
Explanation:
False — this form is only when the velocity changes at a uniform rate. The general definition is displacement divided by time.
4
The unit of acceleration is m/s².
Explanation:
True — a = (v − u) / t.
5
A horizontal distance-time line shows uniform speed.
Explanation:
False — it shows rest. Uniform speed is a slanted straight line.
6
v = u + at applies unchanged to non-uniform acceleration.
Explanation:
False — the three equations are for uniform acceleration and a straight line.
7
Uniform circular motion has acceleration.
Explanation:
True — even if the speed is constant, the direction changes.
8
An odometer measures displacement.
Explanation:
False — it measures the distance travelled. Bhubaneswar to Delhi, 1850 km, is a distance.
Fill in the blanks
0/81
Average speed = total distance / ______.
💡 Hint:
Total time.
Explanation:
v = s / t.
2
To convert km/h into m/s, multiply by ______.
💡 Hint:
5/18.
Explanation:
The reverse factor is 18/5.
3
The speed of sound in air is taken as ______ m/s.
💡 Hint:
346 m/s.
Explanation:
Activity 8.7.
4
Free fall is an example of ______ acceleration.
💡 Hint:
Uniform.
Explanation:
The non-uniform example is a car in traffic.
5
The area of a velocity-time graph gives the magnitude of the ______.
💡 Hint:
Displacement.
Explanation:
The slope of distance-time is the speed.
6
The second equation is s = ut + ______.
💡 Hint:
½at².
Explanation:
The first is v = u + at.
7
In uniform circular motion v = ______.
💡 Hint:
2πr / t.
Explanation:
Circumference divided by time.
8
Feroz finishes 3.6 km ______ min before Sania.
💡 Hint:
5 min.
Explanation:
8:20 and 8:25.
Match
0/21
Match the quantity with its formula or unit.
Column B: A. (u + v) / 2 · B. 2πr / t · C. m/s · D. m/s²
1. Speed
2. Acceleration
3. Average velocity, uniform rate
4. Circular speed
Explanation:
Speed is m/s, acceleration m/s², average velocity (u+v)/2, and on a circle 2πr/t.
2
Match the graph with its meaning.
Column B: A. Rest · B. Uniform velocity · C. Uniform speed · D. Non-uniform speed
1. Distance-time, straight slanted
2. Distance-time curve
3. Distance-time horizontal
4. Velocity-time horizontal
Explanation:
A slanted line is uniform speed, a curve is non-uniform, a horizontal distance-time line is rest, a horizontal velocity-time line is uniform velocity.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): An object that has moved can have zero displacement.
Reason (R): On returning to the initial position the shortest distance is zero.
Explanation:
Both are true and R explains A.
2
Assertion (A): The SI unit of acceleration is m/s².
Reason (R): An odometer measures the distance travelled.
Explanation:
Both are true, but R does not explain the unit of acceleration.
3
Assertion (A): Uniform circular motion is accelerated motion.
Reason (R): There is acceleration only when the magnitude of the speed changes.
Explanation:
A is true. R is false — a change of direction also changes the velocity.
4
Assertion (A): The magnitude of displacement can be greater than the distance.
Reason (R): Displacement is the shortest distance from the start to the end.
Explanation:
A is false. R is true — this is why the magnitude is not greater than the distance.
5
Assertion (A): The average speed of 20 km/h going and 30 km/h returning is not 25 km/h.
Reason (R): The times for equal distances are not equal, so the arithmetic mean is not used.
Explanation:
Both are true and R explains A. The average is 24 km/h.
Balance the equation
0/4This chapter has no chemical equation. Below is only coefficient practice on H2 + O2 → H2O.
1
This is coefficient practice, not a result of this chapter. The motion lesson has no chemical equation. Balance H2 + O2 → H2O (a blank means 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
Coefficient practice again. This equation is not related to v = u + at. Keep the coefficient of oxygen as 1. Balance H2 + O2 → H2O (a blank means 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
3
A student fills 1, 1, 1. This is not a result of the motion chapter, only an atom-counting exercise. Fill the correct coefficients of H2 + O2 → H2O (a blank means 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
4
Coefficient practice before the numerical questions. The distance-time graph does not give this equation. Balance H2 + O2 → H2O (a blank means 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
Classify
0/21
Mark each example as uniform motion or non-uniform motion.
5 m in every second
A car on a crowded street
Object A of Table 8.1
Object B of Table 8.1
Explanation:
A and 5 m every second are uniform. The crowded-street car and B are non-uniform.
2
Mark each statement as distance or displacement.
The odometer’s 1850 km
The straight length on the map
Becoming zero on return
The rope walked along the sides
Explanation:
The odometer and the rope are distance. The straight map length, and becoming zero on return, are displacement.
Very short answer
0/51
What is an origin?
Model answer (board format)
The reference point from which the position of an object is stated is the origin.
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2
Write the formula and the SI unit of average speed.
Model answer (board format)
Average speed = total distance / total time. The SI unit is m/s.
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3
Write the formula and the unit of acceleration.
Model answer (board format)
a = (v − u) / t. The unit is m/s².
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4
Write the formula for the speed in uniform circular motion. Also write what the letters mean.
Model answer (board format)
v = 2πr / t. r is the radius and t is the time for one round.
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5
What does an odometer measure?
Model answer (board format)
An odometer measures the distance travelled by the vehicle.
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Short answer
0/41
16 m in 4 s and then 16 m in 2 s. Find the average speed. Write the formula, the substitution and the unit.
Model answer (board format)
Average speed = total distance / total time = (16 m + 16 m) / (4 s + 2 s) = 32 m / 6 s = 5.33 m/s.
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2
A train at 90 km/h gets brakes of −0.5 m/s². Find the distance until it stops.
Model answer (board format)
u = 90 × 5/18 = 25 m/s. v = 0. From v² = u² + 2as, 0 = 625 + 2(−0.5)s. s = 625 m.
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3
On a distance-time graph, describe the shapes of uniform speed, non-uniform speed and rest.
Model answer (board format)
Uniform speed is a straight slanted line. Non-uniform speed is a curve. Rest is a straight line parallel to the time axis. The slope gives the speed.
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4
A boat starts from rest at 3.0 m/s² for 8.0 s. Find the distance.
Model answer (board format)
s = ut + ½at² = 0 + ½(3.0 m/s²)(8.0 s)² = ½ × 3 × 64 = 96 m.
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Long answer
0/31
Write four differences between distance and displacement. Find the farmer’s displacement for a 10 m square and 2 min 20 s.
Model answer (board format)
Distance is the path length, magnitude only. Displacement is the shortest gap, with a direction. Distance is not zero once he has moved. Displacement can be zero. The magnitude of displacement is not greater than the distance. Farmer: time 140 s, speed 1 m/s, distance 140 m, 3.5 rounds, displacement = 10√2 m ≈ 14.1 m.
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2
Write the three equations and say when they apply. Find the acceleration and distance of a car going from 18 km/h to 36 km/h in 5 s.
Model answer (board format)
v = u + at, s = ut + ½at², v² = u² + 2as. These are for uniform acceleration on a straight line. u = 5 m/s, v = 10 m/s, t = 5 s. a = (10 − 5)/5 = 1 m/s². s = 5×5 + ½×1×25 = 37.5 m.
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3
Why is uniform circular motion called accelerated? What does Activity 8.11 show? A satellite of radius 42250 km takes 24 h for one round. Find the speed in km/h. Take π = 3.14.
Model answer (board format)
The speed is constant but the direction changes at every point, so the velocity changes and there is acceleration. The released stone goes along the tangent, not toward the centre. v = 2πr / t = 2 × 3.14 × 42250 km / 24 h = 265330 / 24 = 11055 km/h.
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BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination. Annual questions will be added only when a source page is available.
1
Where is the farmer after 2 min 20 s?
💡 Hint:
3.5 rounds.
Explanation:
At the opposite corner. The displacement is 10√2 m.
2
If a velocity-time graph is parallel to the time axis —
💡 Hint:
A horizontal distance-time line is rest.
Explanation:
The velocity is uniform. The acceleration is zero.
3
80 km/h is written in m/s as ______, as a fraction.
💡 Hint:
200/9 m/s.
Explanation:
80 × 5/18.
4
Joseph jogs 300 m of a straight road in 150 s and then 100 m back in 60 s. Find the average speed and average velocity from A to C.
Model answer (board format)
Distance = 400 m. Time = 210 s. Average speed = 400/210 = 40/21 m/s ≈ 1.90 m/s. Displacement = 200 m. Average velocity = 200/210 = 20/21 m/s ≈ 0.95 m/s.
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5
The model set asks for coefficients separately. This is not an equation of motion. Balance H2 + O2 → H2O (a blank means 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
6
How is speed read from a graph? Why is Table 8.2 called a curve? What is the area of a velocity-time graph? Find one slope: distance from 0 to 20 m in 4 s.
Model answer (board format)
The slope of a distance-time graph is the speed, v = (s₂ − s₁)/(t₂ − t₁). In Table 8.2 the distance increases by 1, 3, 5, 7, 9 and 11 m in equal times, so the graph is a curve and the speed is non-uniform. The area of a velocity-time graph is the magnitude of the displacement. Slope = 20 m / 4 s = 5 m/s.
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CBSE-style questions
0/6These are competency-based practice questions. They are not copies of a CBSE paper.
1
A car goes 3 km east and then 3 km west, back to the start. The distance and the magnitude of displacement are —
💡 Hint:
Back at the start.
Explanation:
Distance 6 km, displacement 0.
2
A student writes 25 km/h as the average of 20 km/h and 30 km/h for equal distances going and returning. The correct average is —
💡 Hint:
Add the times, not the speeds.
Explanation:
24 km/h. 2d / (d/20 + d/30) = 24.
3
Assertion (A): When brakes are applied the acceleration is taken as negative.
Reason (R): The unit of negative acceleration is km.
Explanation:
A is true. R is false — the unit is still m/s².
4
Assertion (A): The distance-time graph of Table 8.2 is a curve.
Reason (R): The distance covered in equal time intervals is not equal.
Explanation:
Both are true and R explains A.
5
A driver says the car that brakes from 52 km/h and stops in 5 s will travel farther than the car that brakes from 3 km/h and stops in 10 s, because the speed is higher. Without drawing the full graph, what must be checked using area?
Model answer (board format)
The area of the velocity-time graph gives the distance. The first car’s triangle is tall and narrow, the second is short and wide. The comparison is of areas, not of the initial speed alone. Convert the speed to m/s and find ½ × speed × time.
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6
A stone is thrown upward at 5 m/s. The downward acceleration is 10 m/s². Find the greatest height and the time to reach it.
Model answer (board format)
Take upward as positive. In v = u + at, 0 = 5 + (−10)t, so t = 0.5 s. s = ut + ½at² = 5(0.5) + ½(−10)(0.5)² = 2.5 − 1.25 = 1.25 m.
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🧠 What you learned + equation sheet
What you learned
- Motion and rest depend on a reference point. A wall is also moving with the Earth.
- Distance is the path length. Displacement is the shortest gap and it can be zero.
- Average speed = total distance / total time. Velocity has a direction.
- a = (v − u) / t. The unit is m/s². Brakes give a negative acceleration.
- On a distance-time graph a straight line is uniform motion and a curve is non-uniform. The area under a v-t graph is the displacement.
- v = u + at, s = ut + ½at², v² = u² + 2as. On a circle, v = 2πr / t.
| What | Keep this |
|---|---|
| Average speed | total distance / total time |
| Average velocity | (u + v) / 2 when the change is uniform |
| Acceleration | a = (v − u) / t, unit m/s² |
| First equation | v = u + at |
| Distance equation | s = ut + ½at² |
| Circle | v = 2πr / t |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.