Chapter 8 — Motion
Welcome to HSLC Guru! This study material is carefully prepared for ASSEB Class 9 Science learners, offering a complete English-medium guide to Chapter 8 — Motion. Here you will find a clear chapter summary, all textbook questions with model answers, derivations of equations of motion, solved numericals, additional practice questions, fill-in-the-blanks, true/false statements, an exhaustive glossary and a handy formula table to revise quickly before any exam.
Chapter Summary
An object is said to be in motion if its position changes with time relative to a reference point (origin); otherwise it is said to be at rest. Both rest and motion are relative concepts because they depend on the observer’s frame of reference. For example, a passenger sitting inside a moving bus is at rest with respect to the bus but in motion with respect to a person standing on the road. To describe motion, we need to choose a fixed reference point and a reference frame. The total path length covered by a body is called distance (a scalar quantity), while the shortest straight-line directed path from the initial to the final position is called displacement (a vector quantity). Distance can never be negative or zero for a moving body, but displacement can be zero if the body returns to its starting position.
Motion may be classified as uniform motion, in which a body covers equal distances in equal intervals of time, and non-uniform motion, in which it covers unequal distances in equal intervals of time. Speed is defined as the distance travelled per unit time (speed = distance/time) and is a scalar quantity, whereas velocity is the displacement per unit time (velocity = displacement/time) and is a vector quantity. The SI unit of both speed and velocity is metre per second (m/s). When the velocity of an object changes with time, the body is said to be accelerating. Acceleration is defined as the rate of change of velocity with respect to time, given by a = (v − u)/t, where u is the initial velocity, v the final velocity and t the time taken. Its SI unit is metre per second squared (m/s²). Acceleration is positive when velocity increases and negative (called retardation or deceleration) when velocity decreases.
For a body moving with uniform acceleration along a straight line, three important equations of motion are used: (i) v = u + at, (ii) s = ut + ½at², and (iii) v² = u² + 2as, where s is the displacement. These equations can also be derived graphically from a velocity–time graph. Graphical representation of motion is a powerful tool: in a distance–time graph, the slope (gradient) of the line gives the speed of the body — a straight inclined line indicates uniform speed, while a curved line indicates non-uniform speed. In a velocity–time graph, the slope gives the acceleration of the body, and the area enclosed between the velocity–time line and the time axis gives the distance (or displacement) travelled.
When a body moves along a circular path with constant speed, the motion is called uniform circular motion. Although the speed remains constant, the direction of motion changes continuously, so the velocity changes and the body is said to be accelerated. The force that keeps the body moving along the circular path and is directed towards the centre is called centripetal force. Examples include a stone tied to a string and whirled in a circle, the motion of planets around the Sun and an athlete running on a circular track. The speed of such motion is given by v = 2πr/T, where r is the radius of the circle and T is the time period. Numerical problems based on these concepts strengthen our understanding of motion.
Textbook Questions and Answers
1-Mark Questions
Q1. Define motion.
Answer: An object is said to be in motion when its position changes continuously with time with respect to a fixed reference point (observer).
Q2. What is a reference point?
Answer: A fixed point or object with respect to which the position, rest or motion of a body is described is called a reference point or origin.
Q3. Define displacement.
Answer: Displacement is the shortest distance measured from the initial to the final position of an object in a particular direction. It is a vector quantity and its SI unit is metre (m).
Q4. What is the SI unit of acceleration?
Answer: The SI unit of acceleration is metre per second squared, written as m/s² or m s⁻².
Q5. What is uniform motion?
Answer: A body is said to be in uniform motion when it covers equal distances in equal intervals of time, however small the time interval may be.
Q6. Give one example of a body whose displacement is zero but distance is not zero.
Answer: An athlete running along a circular track and returning to the starting point covers a distance equal to the circumference, but the displacement is zero.
Q7. What does the slope of a distance–time graph represent?
Answer: The slope of a distance–time graph represents the speed of the moving body.
Q8. What is centripetal force?
Answer: The force acting on a body moving in a circular path and directed towards the centre of the circle is called centripetal force.
Q9. Define retardation.
Answer: Retardation (or deceleration) is the negative acceleration that takes place when the velocity of a body decreases with time.
Q10. Write the SI unit of velocity.
Answer: The SI unit of velocity is metre per second (m/s or m s⁻¹).
Q11. What is meant by average speed?
Answer: Average speed is the ratio of the total distance travelled by a body to the total time taken to cover that distance.
Q12. Convert 54 km/h into m/s.
Answer: 54 km/h = 54 × (5/18) m/s = 15 m/s.
2–3 Mark Questions
Q1. Distinguish between distance and displacement.
Answer:
| Distance | Displacement |
|---|---|
| It is the total path length covered by a body. | It is the shortest straight-line distance from the initial to the final position. |
| Scalar quantity (only magnitude). | Vector quantity (magnitude and direction). |
| Always positive and never zero for a moving body. | Can be positive, negative or zero. |
| Distance ≥ Displacement. | Displacement ≤ Distance. |
Q2. Differentiate between speed and velocity.
Answer:
| Speed | Velocity |
|---|---|
| Distance travelled per unit time. | Displacement per unit time. |
| Scalar quantity. | Vector quantity. |
| Always positive. | Can be positive, negative or zero. |
| SI unit: m/s. | SI unit: m/s. |
Q3. Why is rest and motion said to be relative?
Answer: Rest and motion are relative concepts because the state of a body — whether at rest or in motion — depends entirely on the observer or reference point chosen. A passenger sitting inside a moving train is at rest with respect to other passengers and the seat, but the same passenger is in motion with respect to a person standing on the platform. Hence, no body is absolutely at rest or absolutely in motion.
Q4. A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Find the acceleration.
Answer: Initial velocity, u = 18 km/h = 18 × (5/18) = 5 m/s.
Final velocity, v = 36 km/h = 36 × (5/18) = 10 m/s.
Time, t = 5 s.
Acceleration, a = (v − u)/t = (10 − 5)/5 = 1 m/s². Therefore, the acceleration of the car is 1 m/s².
Q5. What information do we get from a velocity–time graph?
Answer: From a velocity–time graph we can obtain three pieces of information: (i) the velocity of the body at any instant of time, (ii) the slope of the graph at any point gives the acceleration of the body, and (iii) the area enclosed between the v–t line and the time axis gives the distance (or displacement) travelled by the body in that interval of time.
Q6. A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find the velocity acquired and the distance travelled.
Answer: Given u = 0, a = 0.1 m/s², t = 2 min = 120 s.
Velocity, v = u + at = 0 + 0.1 × 120 = 12 m/s.
Distance, s = ut + ½at² = 0 + ½ × 0.1 × (120)² = 0.05 × 14400 = 720 m.
Therefore, the bus acquires a velocity of 12 m/s and travels 720 m.
Q7. Explain uniform circular motion with an example.
Answer: When a body moves along a circular path with constant speed, the motion is called uniform circular motion. Although the speed remains constant, the direction of motion changes continuously at every point of the path, so the velocity is not constant and the motion is accelerated. Example: a stone tied to a string and whirled around in a horizontal circle with a uniform speed performs uniform circular motion; the motion of the moon around the earth is another example.
5–6 Mark Questions
Q1. Derive the equation v = u + at graphically.
Answer: Consider a body moving with initial velocity u and uniform acceleration a. After time t, let its final velocity be v. Plot a velocity–time graph: time t is taken along the X-axis and velocity along the Y-axis. The graph is a straight line AB inclined to the time axis. From the graph, OA = u, BD = v and OD = t.
Now BC = BD − CD = BD − OA = v − u.
The slope of the line AB gives the acceleration:
a = slope = BC/AC = (v − u)/t.
Rearranging, at = v − u, hence v = u + at. This is the first equation of motion.
Q2. Derive the equation s = ut + ½at² graphically.
Answer: The distance travelled by the body in time t is equal to the area enclosed under the velocity–time graph (the trapezium OABD).
Distance, s = area of trapezium OABD = area of rectangle OACD + area of triangle ABC.
Area of rectangle OACD = OA × OD = u × t = ut.
Area of triangle ABC = ½ × AC × BC = ½ × t × (v − u) = ½ × t × at = ½at².
Therefore, s = ut + ½at². This is the second equation of motion.
Q3. Derive the equation v² = u² + 2as graphically.
Answer: The distance s is given by the area of trapezium OABD:
s = ½ × (sum of parallel sides) × distance between them = ½ × (OA + BD) × OD = ½ × (u + v) × t.
From the first equation of motion, t = (v − u)/a. Substituting,
s = ½ × (u + v) × (v − u)/a = (v² − u²)/2a.
Therefore, 2as = v² − u², which gives v² = u² + 2as. This is the third equation of motion.
Q4. A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming uniform acceleration, find (i) the acceleration and (ii) the distance travelled by the train to attain this velocity.
Answer: Given u = 0, v = 72 km/h = 72 × (5/18) = 20 m/s, t = 5 min = 300 s.
(i) Acceleration, a = (v − u)/t = (20 − 0)/300 = 1/15 m/s² ≈ 0.067 m/s².
(ii) Distance, s = ut + ½at² = 0 + ½ × (1/15) × (300)² = ½ × (1/15) × 90000 = 45000/15 = 3000 m = 3 km.
Alternatively, using v² = u² + 2as: 20² = 0 + 2 × (1/15) × s, hence s = 400 × 15 / 2 = 3000 m. Therefore, the acceleration is 0.067 m/s² and distance travelled is 3 km.
Q5. A car moving with a velocity of 20 m/s applies brakes and comes to rest after travelling 50 m. Calculate the retardation and the time taken to stop.
Answer: Given u = 20 m/s, v = 0, s = 50 m.
Using v² = u² + 2as: 0 = (20)² + 2 × a × 50.
0 = 400 + 100a, hence a = −400/100 = −4 m/s².
So the retardation is 4 m/s².
Using v = u + at: 0 = 20 + (−4) × t, t = 20/4 = 5 s.
Therefore, the car stops in 5 seconds with a retardation of 4 m/s².
Q6. Explain with the help of suitable graphs (i) uniform speed, (ii) non-uniform speed, and (iii) uniformly accelerated motion.
Answer: (i) Uniform speed: For a body moving with uniform speed, equal distances are covered in equal intervals of time. The distance–time graph is a straight line inclined to the time axis. The slope of this line gives the constant speed.
(ii) Non-uniform speed: For a body moving with non-uniform (changing) speed, unequal distances are covered in equal intervals of time. The distance–time graph is a curved line whose slope continuously changes.
(iii) Uniformly accelerated motion: When a body moves with uniformly increasing velocity, the velocity–time graph is a straight line inclined to the time axis with positive slope. The slope of the line gives the constant acceleration, and the area enclosed under the line up to time t gives the distance travelled in time t.
Additional Practice Questions
Multiple Choice Questions
Q1. The SI unit of speed is:
(a) km/h (b) m/s (c) cm/s (d) m/s²
Answer: (b) m/s.
Q2. Which of the following is a vector quantity?
(a) Distance (b) Speed (c) Displacement (d) Time
Answer: (c) Displacement.
Q3. The slope of a velocity–time graph gives:
(a) Distance (b) Displacement (c) Acceleration (d) Speed
Answer: (c) Acceleration.
Q4. A body moving with uniform velocity has:
(a) Constant speed and direction (b) Variable speed (c) Variable direction (d) Both speed and direction variable
Answer: (a) Constant speed and direction.
Q5. If a body moves along a circular path of radius r with time period T, its speed is:
(a) πr/T (b) 2πr/T (c) πr²/T (d) 2πr²/T
Answer: (b) 2πr/T.
Q6. The area under a velocity–time graph gives:
(a) Acceleration (b) Distance (c) Speed (d) Time
Answer: (b) Distance.
Q7. A body starts from rest and acquires a velocity of 10 m/s in 5 s. Its acceleration is:
(a) 1 m/s² (b) 2 m/s² (c) 5 m/s² (d) 10 m/s²
Answer: (b) 2 m/s².
Q8. Which of the following equations of motion is dimensionally correct?
(a) v = u + at (b) s = ut + ½at² (c) v² = u² + 2as (d) All of these
Answer: (d) All of these.
Q9. The motion of the earth around the sun is an example of:
(a) Linear motion (b) Random motion (c) Uniform circular motion (d) Oscillatory motion
Answer: (c) Uniform circular motion.
Q10. 36 km/h is equal to:
(a) 6 m/s (b) 10 m/s (c) 12 m/s (d) 36 m/s
Answer: (b) 10 m/s.
Fill in the Blanks
Q1. The shortest distance from the initial to the final position is called __________.
Answer: displacement.
Q2. The rate of change of velocity is called __________.
Answer: acceleration.
Q3. The slope of a distance–time graph gives the __________ of the body.
Answer: speed.
Q4. The force directed towards the centre during circular motion is called __________ force.
Answer: centripetal.
Q5. The SI unit of acceleration is __________.
Answer: m/s² (metre per second squared).
True or False
Q1. Distance is a vector quantity.
Answer: False. Distance is a scalar quantity.
Q2. A body moving along a circular path with constant speed has zero acceleration.
Answer: False. The direction of velocity changes continuously, so the body is accelerated.
Q3. The slope of a velocity–time graph gives acceleration.
Answer: True.
Q4. Displacement can never be greater than distance.
Answer: True.
Q5. Negative acceleration is called retardation.
Answer: True.
Q6. A body moving with uniform velocity must have zero acceleration.
Answer: True. If the velocity does not change in magnitude or direction, then a = 0.
Glossary
| Term | Meaning |
|---|---|
| Rest | State of a body whose position does not change with time relative to a reference point. |
| Motion | State of a body whose position changes with time relative to a reference point. |
| Reference point | A fixed point or object used as the origin to describe the position or motion of a body. |
| Distance | Total path length covered by a body during its motion (scalar). |
| Displacement | Shortest directed distance from the initial to the final position (vector). |
| Uniform motion | Motion in which equal distances are covered in equal intervals of time. |
| Non-uniform motion | Motion in which unequal distances are covered in equal intervals of time. |
| Speed | Distance travelled per unit time (scalar). |
| Velocity | Displacement per unit time (vector). |
| Acceleration | Rate of change of velocity with respect to time. |
| Retardation | Negative acceleration; rate of decrease of velocity. |
| Uniform circular motion | Motion of a body along a circular path with constant speed. |
| Centripetal force | Force directed towards the centre of the circular path that keeps the body in circular motion. |
| Scalar quantity | Physical quantity having magnitude only. |
| Vector quantity | Physical quantity having both magnitude and direction. |
Formula Table
| Quantity | Formula | SI Unit |
|---|---|---|
| Speed | Speed = Total distance / Total time | m/s |
| Average speed | v(avg) = Total distance / Total time | m/s |
| Velocity | v = Displacement / Time | m/s |
| Acceleration | a = (v − u)/t | m/s² |
| First equation of motion | v = u + at | — |
| Second equation of motion | s = ut + ½at² | — |
| Third equation of motion | v² = u² + 2as | — |
| Speed in circular motion | v = 2πr/T | m/s |
| Conversion | 1 km/h = 5/18 m/s | — |
| Conversion | 1 m/s = 18/5 km/h | — |