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These are complete exercise solutions for Samacheer Kalvi 9th Science Chapter 2 — Motion. Solve each question yourself first, then verify your method and answer below. Always show formula, substitution, calculation, and units in numerical problems. Pay special attention to the difference between scalar and vector quantities — it is the most frequently tested concept in this chapter.
Key Concepts (Quick Recap)
| Term | Definition |
|---|
| Distance | Total path length between two points; a scalar quantity (always positive, SI unit: m) |
| Displacement | Shortest straight-line path from initial to final position; a vector quantity (can be positive, negative, or zero, SI unit: m) |
| Speed | Rate of change of distance; scalar; formula: speed = distance / time; SI unit: m/s |
| Velocity | Rate of change of displacement; vector; formula: velocity = displacement / time; SI unit: m/s |
| Acceleration | Rate of change of velocity; formula: a = (v − u) / t; SI unit: m/s² or ms⁻² |
| Retardation (Deceleration) | Negative acceleration when velocity decreases with time; value is written with a minus sign |
| Uniform Motion | Object covers equal distances in equal intervals of time; velocity remains constant; acceleration = 0 |
| Non-uniform Motion | Object covers unequal distances in equal intervals of time; velocity changes; acceleration ≠ 0 |
| Uniform Circular Motion | Motion at constant speed along a circular path; direction changes continuously; centripetal acceleration = v²/r |
| Centripetal Force | Real force directed towards the centre of a circular path; formula: F = mv²/r |
| Centrifugal Force | Pseudo (virtual) force that appears in a rotating frame of reference; directed away from centre |
| Equation 1 | v = u + at — relates final velocity, initial velocity, acceleration, and time |
| Equation 2 | s = ut + ½at² — relates displacement, initial velocity, acceleration, and time |
| Equation 3 | v² = u² + 2as — relates final velocity, initial velocity, acceleration, and displacement (no time) |
| v-t Graph Area | Area under velocity-time graph = displacement; slope = acceleration |
I. Choose the Best Answer (MCQs)
Q1. The area under a velocity-time graph represents: (a) velocity (b) displacement ✓ (c) speed (d) acceleration
Answer(b) ✓ — Area under v-t graph = displacement (since area = v × t = displacement).
Q2. Which is most likely NOT a case of uniform circular motion? (a) Motion of Earth around the Sun (b) Motion of a toy train on a circular track (c) Motion of a racing car on a circular track ✓ (d) Motion of hour hand on a clock
Answer(c) ✓ — A racing car accelerates and decelerates; its speed is not constant, so it is not uniform circular motion.
Q3. Which graph represents uniform motion? A straight line through the origin on a distance-time graph (constant slope = constant speed).
Q4. The centrifugal force is: (a) a real force (b) reaction of centripetal force (c) virtual (pseudo) force ✓ (d) directed towards centre
Answer(c) ✓ — Centrifugal force is a pseudo force that appears in a rotating frame of reference; it is not a real force.
Q5. A body moves in a straight line from A to B with speed 20 m/s and returns from B to A with speed 30 m/s. The average speed for the entire trip is: (a) 25 m/s (b) 24 m/s ✓ (c) 50 m/s (d) 12 m/s
Answer(b) ✓ — Average speed = 2 × v₁ × v₂ / (v₁ + v₂) = 2 × 20 × 30 / (20 + 30) = 1200/50 = 24 m/s. This is the harmonic mean formula for equal distances.
Q6. If a body starts from rest and moves with uniform acceleration, the displacement is proportional to: (a) velocity (b) time (c) time² ✓ (d) time³
Answer(c) ✓ — From s = ut + ½at², with u = 0: s = ½at², so displacement is proportional to t².
Q7. The SI unit of acceleration is: (a) m/s (b) ms⁻² ✓ (c) m²/s (d) N·m
Answer(b) ✓ — Acceleration = change in velocity / time, so its SI unit is m/s² = ms⁻².
Q8. A particle completes one full revolution on a circular track of radius r. Its displacement is: (a) 2πr (b) πr (c) 0 ✓ (d) r
Answer(c) ✓ — After one full revolution, the initial and final positions coincide, so displacement = 0.
II. Fill in the Blanks
1. Speed is a scalar quantity whereas velocity is a vector quantity. 2. The slope of the distance-time graph at any point gives speed. 3. Negative acceleration is called retardation (deceleration). 4. Area under velocity-time graph shows displacement. 5. In uniform circular motion, the quantity that changes continuously is direction of velocity. 6. The slope of a velocity-time graph gives acceleration. 7. A body at rest has initial velocity zero. 8. For a body moving with uniform velocity, acceleration is zero. 9. The centripetal acceleration of a body moving with speed v on a circular path of radius r is given by v²/r.
III. True or False
1. The motion of a city bus in heavy traffic is uniform motion.
AnswerFalse (speed keeps changing — it is non-uniform motion).
2. Acceleration can get a negative value.
AnswerTrue (when velocity decreases, acceleration is negative = deceleration).
3. Distance covered never becomes zero but displacement can.
AnswerTrue (distance is always positive; displacement can be zero when you return to the start).
4. Velocity-time graph of a freely falling body is a straight line parallel to the x-axis.
AnswerFalse (a freely falling body accelerates, so the v-t graph is a straight line inclined to the x-axis, not parallel).
5. If the velocity-time graph is a straight line inclined to the x-axis, the displacement-time graph will be a straight line.
AnswerFalse (constant acceleration gives a parabolic s-t graph, not a straight line).
6. The displacement of a body can be greater than the distance travelled.
AnswerFalse (displacement is always less than or equal to distance; it can never exceed distance because displacement is the shortest path).
7. A body moving with constant speed on a circular path has no acceleration.
AnswerFalse (even though speed is constant, the direction of velocity changes continuously, producing centripetal acceleration directed towards the centre).
8. The distance-time graph of a body at rest is a straight line parallel to the time axis.
AnswerTrue (when a body is at rest, distance does not change with time, so the graph is a horizontal line).
IV. Assertion and Reason
Q1. A: Accelerated motion may be due to change in magnitude of velocity or direction or both. R: Acceleration is produced only by change in magnitude of velocity, not direction. → (c) A is true but R is false. Acceleration can result from change in speed, direction, or both.
Q2. A: Speedometer measures average speed. R: Average velocity = total displacement / total time. → (d) A is false (speedometer measures instantaneous speed) but R is true.
Q3. A: Displacement may be zero when distance is not zero. R: Displacement is the shortest distance between initial and final position. → (a) Both A and R are true, and R is the correct explanation. When you return to the starting point, distance > 0 but displacement = 0.
V. Match the Following
| Column I | Column II |
|---|
| 1. Uniform motion | (D) Constant speed on straight path |
| 2. Non-uniform motion | (C) Varying speed |
| 3. Circular motion | (A) Constant speed, changing direction |
| 4. Oscillatory motion | (B) To-and-fro about a mean position |
| 5. Retardation | (G) Negative acceleration |
| 6. Slope of v-t graph | (E) Acceleration |
| 7. Area under v-t graph | (F) Displacement |
| 8. Freely falling body | (H) Constant acceleration (g) directed downward |
VI. Answer Briefly (2-mark)
Q1. Define velocity.
Answer
Velocity is the rate of change of displacement with respect to time. It is a vector quantity, meaning it has both magnitude and direction. The SI unit of velocity is m/s (or ms⁻¹). Mathematically, velocity = displacement / time. Unlike speed, velocity can be positive, negative, or zero depending on the direction of motion.
Q2. Distinguish distance and displacement.
Answer
| Distance | Displacement |
|---|
| Total path length | Shortest path from start to end |
| Scalar quantity | Vector quantity |
| Always positive | Can be positive, negative, or zero |
| Depends on path | Independent of path |
| Can never be less than displacement | Can be equal to or less than distance |
For example, if a student walks 5 m forward and 5 m backward, the distance is 10 m but displacement is 0 m.
Q3. What is uniform motion?
Answer
An object is in uniform motion when it covers equal distances in equal intervals of time, regardless of how small those intervals are. In uniform motion, the velocity remains constant throughout, meaning there is no acceleration. The distance-time graph for uniform motion is a straight line passing through the origin, with its slope equal to the constant speed.
Q4. Compare speed and velocity.
Answer
| Speed | Velocity |
|---|
| Rate of change of distance | Rate of change of displacement |
| Scalar (magnitude only) | Vector (magnitude + direction) |
| Always positive or zero | Can be positive, negative, or zero |
speed = distance / time | velocity = displacement / time |
Speed tells us how fast an object moves; velocity tells us how fast and in which direction.
Q5. What is negative acceleration?
Answer
When the velocity of an object decreases with time, its acceleration is negative. Negative acceleration is also called retardation or deceleration. For example, when a car applies brakes and slows down from 20 m/s to 10 m/s in 2 seconds, the acceleration is a = (10 − 20) / 2 = −5 ms⁻². The negative sign indicates the body is decelerating.
Q6. Is uniform circular motion accelerated?
Answer
Yes, uniform circular motion is accelerated motion. Even though the speed remains constant, the direction of velocity changes continuously as the object moves along the circular path. This change in direction produces centripetal acceleration given by a = v²/r, directed towards the centre of the circular path. Since acceleration involves change in velocity (a vector), and direction is changing, the motion is accelerated.
Q7. What is uniform circular motion? Give two examples.
Answer
Uniform circular motion is the motion of an object moving at constant speed along a circular path. Although the speed is constant, the velocity changes because the direction of motion changes at every point. The acceleration responsible for this change in direction is called centripetal acceleration, directed towards the centre. Examples: (1) Earth revolving around the Sun; (2) Hands of a clock moving at constant speed.
VII. Answer in Detail (5-mark)
Q1. Derive the equations of motion by graphical method.
Answer
Consider a velocity-time graph for an object moving with initial velocity u, final velocity v, in time t, with uniform acceleration a. The graph is a straight line inclined to the time axis.
First equation: v = u + at
The slope of the v-t graph gives acceleration. From the graph, slope = (v − u) / t. Since slope = acceleration:
a = (v − u) / t
v = u + at
Second equation: s = ut + ½at²
The displacement equals the area under the v-t graph. This area consists of a rectangle (area = u × t) and a triangle (area = ½ × base × height = ½ × t × (v − u) = ½ × t × at = ½at²).
s = area of rectangle + area of triangle
s = ut + ½at²
Third equation: v² = u² + 2as
The area under the v-t graph can also be written as the area of a trapezium:
s = ½(u + v) × t
From the first equation, t = (v − u) / a. Substituting:
s = ½(u + v) × (v − u) / a
s = (v² − u²) / 2a
v² = u² + 2as
Key points to remember:
- These three equations are valid only for uniformly accelerated motion (constant a).
- All quantities must be in consistent SI units before substitution.
- Choose the equation that contains the known and unknown quantities you need.
Q2. Explain different types of motion with examples.
Answer
1. Linear (Rectilinear) motion: Movement along a straight line in a single direction. The path is a straight line and the object does not change direction. Example: A car moving on a straight highway, a stone falling vertically downward.
2. Circular motion: Movement along a circular path where the object continuously changes direction. When the speed is constant, it is called uniform circular motion. The acceleration is directed towards the centre (centripetal). Example: Earth orbiting the Sun, an athlete running on a circular track.
3. Oscillatory (Vibratory) motion: Repetitive to-and-fro motion about a fixed mean position. The object moves on either side of the equilibrium position in a periodic manner. Example: Pendulum of a clock swinging, a vibrating tuning fork, a child on a swing.
4. Random motion: Irregular, unpredictable motion with no fixed pattern or direction. The object may change its speed and direction abruptly and without pattern. Example: A fish swimming in water, dust particles in air, a flying mosquito.
5. Rotational motion: Motion of a body spinning about a fixed axis, where different points on the body trace circles of different radii. Example: Rotation of the Earth about its axis, a spinning top, blades of a ceiling fan.
Understanding the type of motion helps in selecting the right equations and approach to solve problems.
VIII. Numerical Problems
Q1. A ball is dropped from 20 m height. Find the time to strike the ground and the velocity at impact. (g = 10 ms⁻²)
Answer
u = 0, s = 20 m, a = g = 10 ms⁻²
Using s = ut + ½at²: 20 = 0 + ½ × 10 × t² → t² = 4 → t = 2 s
Using v = u + at: v = 0 + 10 × 2 = 20 ms⁻¹
Q2. An athlete runs on a circular track of diameter 200 m. He completes one round in 40 s. Find the distance and displacement at the end of 2 min 20 s (140 s).
Answer
Number of rounds = 140/40 = 3.5 rounds
Circumference = π × d = π × 200 ≈ 628.57 m
Distance = 3.5 × 628.57 ≈ 2199 m ≈ 2200 m
After 3.5 rounds, the athlete is at the diametrically opposite point from the start. Displacement = diameter = 200 m
Q3. A racing car has uniform acceleration of 4 ms⁻², starting from rest for 10 s.
Answer
u = 0, a = 4 ms⁻², t = 10 s
s = ut + ½at² = 0 + ½ × 4 × 100 = 200 m
Additional Numerical Problems
Q4. An object travels 16 m in 4 s and then another 16 m in 2 s. Average speed?
Answer
Total distance = 32 m, Total time = 6 s Average speed = 32/6 = 5.33 ms⁻¹
Q5. Sound heard 5 s after lightning. Speed of sound = 346 ms⁻¹. Distance?
Answer
d = v × t = 346 × 5 = 1730 m
Q6. A car brakes with deceleration 6 ms⁻², stopping in 2 s. Find initial velocity and distance.
Answer
v = 0, a = –6 ms⁻², t = 2 s u = v – at = 0 – (–6) × 2 = 12 ms⁻¹ s = ut + ½at² = 12 × 2 + ½ × (–6) × 4 = 24 – 12 = 12 m
Q7. A 900 kg car at 10 ms⁻¹ turns around a circle of radius 25 m. Find centripetal acceleration and force.
Answer
a = v²/r = 100/25 = 4 ms⁻² F = ma = 900 × 4 = 3600 N
Q8. A bus travels 20 km in 45 minutes. Average speed?
Answer
d = 20 km = 20,000 m, t = 45 min = 2700 s Speed = 20,000/2700 ≈ 7.4 ms⁻¹ (or ≈ 26.67 km/h)
Q9. A train goes from A to B at 100 km/h and returns at 80 km/h. Average velocity and average speed?
Answer
Average velocity = 0 (displacement = 0 for round trip) Average speed = 2 × 100 × 80 / (100 + 80) = 16000/180 = 88.89 km/h ≈ 90 km/h
Common Mistakes to Avoid
1. ❌ Confusing distance and displacement → ✅ Distance is total path (always positive, scalar); displacement is shortest path (can be zero or negative, vector).
2. ❌ Forgetting to use negative sign for deceleration → ✅ Deceleration means acceleration is negative. If a car decelerates at 5 ms⁻², use a = −5 ms⁻² in equations.
3. ❌ Substituting distance instead of displacement in velocity formula → ✅ Use displacement for velocity (v = displacement / time) and distance for speed (speed = distance / time).
4. ❌ Reading v-t graph slope as velocity → ✅ Slope of v-t graph = acceleration; area under v-t graph = displacement.
5. ❌ Confusing centripetal and centrifugal force → ✅ Centripetal force is real (acts towards centre); centrifugal force is pseudo (felt in rotating frame only).
6. ❌ Assuming uniform motion means the same as uniform circular motion → ✅ Uniform motion = constant velocity (a = 0); uniform circular motion = constant speed but changing direction (a ≠ 0).
7. ❌ Skipping units in numerical answers → ✅ Always write the formula first, substitute values with correct units, and include the final unit in your answer.
Exam Checklist
- [ ] Definitions of distance, displacement, speed, velocity, acceleration memorized with units
- [ ] Three equations of motion derived and memorized (
v = u + at, s = ut + ½at², v² = u² + 2as) - [ ] Distance-time and velocity-time graph interpretations practiced (slope meaning, area meaning)
- [ ] Circular motion: centripetal acceleration (
a = v²/r) and centripetal force (F = mv²/r) formulae known - [ ] Numerical problems: always write formula, substitute, calculate, include units
- [ ] Assertion-Reason question technique practiced (check A true, R true, R explains A)
- [ ] Difference tables (distance vs displacement, speed vs velocity) memorized
- [ ] Scalar vs vector distinction clear — direction matters for displacement, velocity, acceleration
- [ ] Average speed formula for round trips:
2v₁v₂/(v₁ + v2) when equal distances are covered - [ ] Sign conventions: deceleration is negative acceleration; downward motion uses
a = +g; upward motion uses a = −g - [ ] Special cases practiced: body at rest (u = 0), freely falling body (a = g), round trip (displacement = 0)
- [ ] Know the types of motion (linear, circular, oscillatory, random) with two examples each
Internal Links
- [Chapter 2 Guide — Motion](/school/tamil-nadu-state-board/class/9/science/motion)
- [All Class 9 Science Solutions](/school/tamil-nadu-state-board/class/9/science/solutions)
- [Class 9 Science Subject Guide](/school/tamil-nadu-state-board/class/9/science)
