Class 9 Science: Describing Motion Around Us — Chapter Question Set

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net

displacement

time interval

displacement

m/s² (or m s⁻²)

9.8

velocity

acceleration

displacement

constant

uniform

direction

tangent

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stationary (or at rest)

False (An instant is a single reading; a time interval is the duration between two instants.)

True

False (Average speed cannot be zero if the object moves; average velocity can be zero if displacement is zero.)

True

False (Acceleration depends on how quickly velocity changes, not on how fast the object is moving.)

True

False (A curved graph indicates changing velocity, i.e., accelerated motion.)

True

False (Acceleration is non-zero because the direction of velocity changes continuously.)

True

False (A graph shows how position changes with time, not the route.)

False (They are valid only when acceleration is constant.)

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True

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A–2, B–1, C–4, D–5, E–3

A–3, B–4, C–2, D–1, E–5

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Distance is the total length of the path actually travelled by an object and is a scalar quantity. Displacement is the net change in position of the object between its initial and final positions; it is a vector quantity and has direction.

The magnitude of average velocity equals the average speed when the object moves along a straight line in one direction without turning back. In this case, the total distance travelled and the magnitude of the displacement are the same.

The average velocity is zero while the average speed is non-zero when the object returns to its starting point. Here the displacement is zero, but the distance travelled is not, so the average speed remains non-zero.

Total distance travelled = 400 km and total time = 5 h, so average speed = 400/5 = 80 km/h. Since she returns to her starting point, the displacement is zero, so the average velocity is 0 km/h.

Average acceleration is the change in velocity of an object divided by the time interval over which that change occurs. Its formula is a = (v − u)/(t₂ − t₁), where v is the final velocity and u is the initial velocity.

Converting the velocities, u = 10 m/s and v = 15 m/s, with t = 10 s. The average acceleration is a = (15 − 10)/10 = 0.5 m/s².

The average acceleration is in the direction of velocity when the speed of the object is increasing. It is opposite to the direction of velocity when the speed is decreasing, that is, when the object is slowing down.

The slope of a position-time graph represents the velocity of the object. It is calculated as the change in position divided by the change in time, that is, slope = (s₂ − s₁)/(t₂ − t₁).

The slope of a velocity-time graph represents the acceleration of the object. It is calculated as the change in velocity divided by the change in time.

The area enclosed between the velocity-time graph line and the time axis, for a given time interval, gives the displacement of the object during that interval.

The three equations are: v = u + at; s = ut + ½at²; and v² = u² + 2as, where u is initial velocity, v is final velocity, a is acceleration, t is time and s is displacement.

Since the car starts from rest, u = 0, v = 24 m/s and t = 6 s. The acceleration is a = 24/6 = 4 m/s², and the distance travelled is s = ½ × 4 × (6)² = 72 m.

Converting the velocities, u = 15 m/s and v = 10 m/s, with t = 36 s. The distance travelled is s = (u + v)/2 × t = (15 + 10)/2 × 36 = 12.5 × 36 = 450 m.

Uniform circular motion is the motion of an object along a circular path at a constant speed. An example is a child riding on a merry-go-round.

In uniform circular motion, the magnitude of velocity (that is, the speed) remains constant. However, the direction of velocity changes continuously as the object moves along the circle, so the velocity itself is not constant.

The displacement is zero, because the object returns to its starting position, so there is no net change in position.

Converting the speed, u = 20 m/s, and the car comes to rest so v = 0, with a = −4 m/s². Using v² = u² + 2as, the stopping distance is s = u²/(2a) = 400/8 = 50 m.

An instant of time is a single reading of the clock at a particular point in time. A time interval is the duration between two such instants, that is, the difference between two clock readings.

A graph shows how the position of an object changes with time relative to a chosen origin, not the actual physical path or route taken. It plots position against time, so it does not represent the geographical shape of the journey.

Scalars are quantities specified by their numerical value only, such as distance and speed. Vectors require both magnitude and direction for their complete description, such as displacement and velocity.

Introduction: To describe motion, we need to specify a reference point and track how an object's position changes with time.
Key Points:
1. Position and reference point — The distance and direction of an object from a reference point (origin O) describes its position at any instant.
2. Distance travelled — The total path length covered; for an athlete going from O to A (100 m) and back to B (40 m), distance = 160 m.
3. Displacement — The net change in position; for the same athlete, displacement = 40 m in the positive direction, which differs from distance.
Conclusion: Distance is the total path length while displacement is the net change in position, and they are equal only when the object moves in one direction without turning back.

Introduction: Average speed tells us how fast an object moves, while average velocity tells us how fast and in which direction its position changes.
Key Points:
1. Average speed — Total distance divided by time; Sarang swims 25 m + 25 m = 50 m in 50 s, giving average speed = 1 m/s.
2. Average velocity — Displacement divided by time; Sarang returns to his starting point, so displacement = 0 and average velocity = 0 m/s.
3. Key insight — Average speed and average velocity can be very different; speed uses distance while velocity uses displacement.
Conclusion: Average speed can be non-zero even when average velocity is zero, showing the importance of direction in describing motion.

Introduction: Average acceleration describes how quickly the velocity of an object changes over a time interval.
Key Points:
1. Accelerator pressed — Bus velocity increases from 10 m/s to 15 m/s in 10 s; a = 0.5 m/s² in the direction of velocity.
2. Brakes applied — Bus velocity decreases from 15 m/s to 0 m/s in 5 s; a = −3 m/s², opposite to the direction of velocity.
3. Sign convention — Positive acceleration means velocity is increasing; negative acceleration means velocity is decreasing.
Conclusion: The sign of acceleration tells us whether the object is speeding up or slowing down, and the direction of acceleration relative to velocity.

Introduction: A position-time graph provides a visual representation of how an object's position changes with time.
Key Points:
1. Plotting the graph — Time is shown on the X-axis and position on the Y-axis; points are plotted and connected to form the graph.
2. Straight line graph — Indicates constant velocity, as equal distances are covered in equal intervals of time.
3. Curved graph — Indicates that velocity is not constant and the object is in accelerated motion.
Conclusion: The shape of the position-time graph reveals whether the motion is uniform or accelerated, and its slope gives the velocity.

Introduction: A velocity-time graph shows how the velocity of an object changes with time and reveals key information about its motion.
Key Points:
1. Velocity at each instant — The height of the line above the time axis gives the velocity at any instant.
2. Slope gives acceleration — The slope of the line on a velocity-time graph gives the acceleration of the object.
3. Area gives displacement — The area enclosed by the line and the time axis equals the displacement in that time interval.
Conclusion: From a velocity-time graph, we can determine velocity, acceleration and displacement, making it a powerful tool for analysing motion.

Introduction: Kinematic equations provide a mathematical description of how the motion of an object changes with time when acceleration is constant.
Key Points:
1. Velocity equation — v = u + at, relating final velocity, initial velocity, acceleration and time.
2. Displacement equation — s = ut + ½at², relating displacement to initial velocity, acceleration and time.
3. Velocity-displacement equation — v² = u² + 2as, relating velocity, acceleration and displacement without time.
Conclusion: These equations allow us to predict the position or velocity of an object at a future time, provided the acceleration is constant.

Introduction: Uniform circular motion occurs when an object moves in a circular path with constant speed.
Key Points:
1. Constant speed, changing direction — The speed remains the same at every point, but the direction of velocity changes continuously.
2. Velocity is not constant — Since velocity depends on both magnitude and direction, the continuously changing direction means velocity is changing.
3. Acceleration is non-zero — Because the velocity is changing (in direction), the motion is accelerated even though the speed is constant.
Conclusion: In uniform circular motion, the acceleration arises solely from the continuous change in the direction of velocity.

Introduction: When brakes are applied, a vehicle travels some distance before stopping, and this depends on its initial velocity and deceleration.
Key Points:
1. Using v² = u² + 2as — With final velocity v = 0, the stopping distance s = u²/(2a), where a is the deceleration.
2. Velocity matters greatly — Doubling the initial velocity quadruples the stopping distance (s ∝ u²).
3. Real-world factors — Braking distance also depends on road surface, braking capacity and driver reaction time, requiring a safe following distance.
Conclusion: Understanding how stopping distance depends on velocity helps drivers maintain safe distances and avoid collisions.

Introduction: When an object is dropped from a height, it accelerates downward due to the Earth's gravitational force.
Key Points:
1. Constant acceleration — The velocity increases by equal amounts (9.8 m/s) in equal intervals of time, giving a constant acceleration of g = 9.8 m/s².
2. Increasing velocity — At t = 0, 1, 2, 3, 4 s, the velocities are 0, 9.8, 19.6, 29.4, 39.2 m/s respectively.
3. Direction of acceleration — The acceleration is in the direction of motion (downward), so it is positive when downward is taken as positive.
Conclusion: Free fall is a classic example of motion with constant acceleration, where velocity increases uniformly under gravity.

Introduction: The displacement of an object can be determined graphically from its velocity-time graph by finding the area under the curve.
Key Points:
1. Constant velocity — For a straight line parallel to the time axis, the area is a rectangle: displacement = velocity × time.
2. Changing velocity — For a slanted line, the area is a combination of a rectangle and a triangle: displacement = (u × t) + (½ × change in velocity × t).
3. Numerical example — For velocity increasing from 5 to 10 m/s over 10 s: displacement = (5 × 10) + (½ × 5 × 10) = 50 + 25 = 75 m.
Conclusion: The area under the velocity-time graph provides a convenient graphical method to calculate displacement for motion with constant acceleration.

Introduction: When two objects move along parallel tracks, their position-time graphs can be used to compare their velocities.
Key Points:
1. Slope comparison — The object with the steeper slope on the position-time graph has the higher velocity.
2. Same time interval — For the same time interval, the object that covers more displacement has a higher velocity.
3. Intersection point — If the graphs cross, the objects have the same position at that instant; if slopes are equal there, velocities are equal.
Conclusion: By comparing slopes and displacements on position-time graphs, we can determine which object is faster and when their velocities are equal.

Introduction: A velocity-time graph provides two important quantities — acceleration from its slope and displacement from its area.
Key Points:
1. Acceleration from slope — Slope = change in velocity ÷ change in time; for a straight line, this gives the constant acceleration.
2. Displacement from area — The area enclosed by the line and the time axis equals the displacement in that time interval.
3. Example calculation — For velocity changing from 5 to 10 m/s over 10 s: a = 5/10 = 0.5 m/s²; displacement = 75 m (area of trapezium).
Conclusion: Both acceleration and displacement can be determined graphically from a velocity-time graph, making it a versatile analytical tool.

Introduction: Fuel used by a vehicle depends on the total distance travelled, not on the displacement.
Key Points:
1. Distance determines fuel — Fuel consumption is proportional to the total path length (distance), regardless of direction changes.
2. Displacement can be zero — A vehicle returning to its start has zero displacement but still consumed fuel for the full distance.
3. Practical implication — Two trips with the same displacement may use very different amounts of fuel if the distances differ.
Conclusion: Distance, not displacement, determines fuel consumption, highlighting the practical difference between these two quantities.

Introduction: When solving motion problems, we can choose the origin and the positive direction as per our convenience.
Key Points:
1. Flexible choice — The origin can be at the ground level or at the point where the object is dropped, depending on what is convenient.
2. Positive direction — We can choose any direction as positive; e.g., downward can be positive for a falling object.
3. Consistency — Once chosen, the origin and positive direction should not be changed while solving the problem.
Conclusion: The freedom to choose origin and direction simplifies problem-solving, but consistency throughout the solution is essential.

Introduction: Uniform circular motion is an idealised model that serves as the foundation for understanding many real-world motions.
Key Points:
1. Idealised model — Constant speed and a perfectly circular path are often not met exactly in real situations.
2. Real-world examples — Motion of planets around the Sun, a vehicle making a circular turn, or a child on a merry-go-round.
3. Foundation for complex motion — It serves as the starting point for analysing more complex real-world circular motions.
Conclusion: Although idealised, uniform circular motion is a useful model that helps us understand and approximate many real-world circular motions.

(a) The speed is 36 km/h = 10 m/s, so the distance during reaction time is 10 × 0.5 = 5 m.
(b) Using s = u²/(2a), the braking distance is 100/5 = 20 m.
(c) Total stopping distance = 5 + 20 = 25 m, which is less than 30 m, so the bus stops in time.
(d) At higher speed, both the reaction distance and braking distance increase, so the bus might not stop before the obstacle.

(a) The acceleration is a = 20/5 = 4 m/s².
(b) Phase 1: ½ × 4 × 25 = 50 m; Phase 2: 20 × 10 = 200 m; Phase 3: ½ × 20 × 6 = 60 m.
(c) The total distance is 50 + 200 + 60 = 310 m.
(d) The graph is a line rising from the origin, then a horizontal line, then a line descending to zero — forming a trapezium.

(a) Using v² = u² + 2as, 0 = 784 + 2a(98), so a = −4 m/s².
(b) Using v = u + at, 0 = 28 + (−4)t, so t = 7 s.
(c) The acceleration is negative because the bike is slowing down.
(d) The negative sign indicates that the acceleration acts in the direction opposite to the velocity.

(a) 54 km/h = 15 m/s and 36 km/h = 10 m/s.
(b) The average acceleration is a = (10 − 15)/36 ≈ −0.14 m/s².
(c) The distance travelled is s = (15 + 10)/2 × 36 = 12.5 × 36 = 450 m.
(d) To stop completely, the driver would need greater deceleration and a longer distance.

(a) The data represents uniform motion, since equal distances are covered in equal time intervals.
(b) The velocity is v = 20 m/s.
(c) The position-time graph would be a straight line passing through the origin.
(d) The velocity-time graph would be a straight line parallel to the time axis at 20 m/s.

(a) No — the displacements in equal time intervals (1, 3, 5, 7, 9, 11 m) are increasing, so the velocity is not constant.
(b) The position-time graph is a curve.
(c) The velocity is increasing.
(d) This is accelerated (non-uniform) motion.

(a) No — although the speed is constant, the direction of motion is changing continuously, so the velocity is not constant.
(b) Yes, she is accelerating because the direction of her velocity is changing, and acceleration is the rate of change of velocity.
(c) On a straight road at constant speed, her acceleration becomes zero.
(d) Speed is only the magnitude (constant here), while velocity includes direction, which keeps changing on the circular track.

(a) The marble moves off in a straight line.
(b) It moves along the tangent to the circle at the point where it was released.
(c) This shows that the velocity in circular motion is directed along the tangent at each point.
(d) No — on a straight path at constant speed, and ignoring friction, the acceleration is zero.

(a) Car A: a = 5/5 = 1 m/s²; Car B: a = 3/10 = 0.3 m/s².
(b) Car A: ½ × 1 × 25 = 12.5 m; Car B: ½ × 0.3 × 100 = 15 m.
(c) Car A's line would be steeper, since a steeper slope means higher acceleration.
(d) By comparing the areas under their lines on the graph — the larger area corresponds to the greater displacement.

(a) The tip makes 1.5 revolutions in 1.5 hours.
(b) Distance = 1.5 × 2π × 7 cm ≈ 66 cm.
(c) After 1.5 revolutions the tip is diametrically opposite its starting point, so the displacement = 2 × 7 = 14 cm.
(d) Average speed = 66 cm / 5400 s ≈ 0.012 cm/s; the magnitude of average velocity = 14 cm / 5400 s ≈ 0.0026 cm/s.## Section I: 5 Mark Long Answer Questions

Introduction: To describe the motion of an object, we need to specify its position, track the distance it covers, and determine its displacement, speed and velocity.
Key Points:
1. Position and reference point — Position is the distance and direction of an object from a chosen reference point (origin) at any instant of time.
2. Distance travelled — The total path length covered, e.g., an athlete running from O to A (100 m) and back to B (40 m) travels 160 m.
3. Displacement — The net change in position between two instants; for the same athlete, displacement = 40 m, which can differ from distance.
4. Average speed — Total distance divided by time; for Sarang swimming 50 m in 50 s, average speed = 1 m/s.
5. Average velocity — Displacement divided by time; for Sarang returning to start, displacement = 0, so average velocity = 0 m/s.
Conclusion: Distance and speed are scalar quantities concerned with path length, while displacement and velocity are vector quantities that also consider direction.

Introduction: Average acceleration describes how quickly an object's velocity changes over a time interval, and its direction matters.
Key Points:
1. Definition and formula — Average acceleration is the change in velocity divided by the time interval: a = (v − u)/(t₂ − t₁), with SI unit m/s².
2. Speeding up — When the magnitude of velocity increases, acceleration is in the direction of velocity, e.g., a bus accelerating from 10 to 15 m/s.
3. Slowing down — When the magnitude of velocity decreases, acceleration is opposite to velocity, e.g., a bus braking from 15 m/s to 0 in 5 s gives a = −3 m/s².
4. Zero acceleration — An object at constant velocity has zero acceleration, even if it is moving very fast.
5. Free fall example — An object dropped from rest accelerates at g = 9.8 m/s², with velocity increasing by 9.8 m/s each second.
Conclusion: The sign and direction of acceleration tell us whether an object is speeding up, slowing down, or moving at constant velocity.

Introduction: Graphs provide a visual representation of how position, velocity and acceleration change with time, helping us analyse motion.
Key Points:
1. Position-time graph — Shows position vs time; a straight line means constant velocity, a curve means accelerated motion, and a line parallel to the time axis means rest.
2. Slope of position-time graph — The slope gives the velocity; a steeper slope means higher velocity, and comparing slopes compares velocities.
3. Velocity-time graph — Shows velocity vs time; a line parallel to the time axis means constant velocity, a slanted line means constant acceleration.
4. Slope of velocity-time graph — The slope gives the acceleration; a zero slope means zero acceleration (constant velocity).
5. Area under velocity-time graph — The area enclosed by the line and the time axis gives the displacement in that time interval.
Conclusion: Position-time and velocity-time graphs together provide a complete picture of motion, revealing velocity, acceleration and displacement.

Introduction: Kinematic equations describe the motion of an object moving in a straight line with constant acceleration.
Key Points:
1. First equation (v = u + at) — Derived from the definition of acceleration: a = (v − u)/t, rearranged to give final velocity.
2. Second equation (s = ut + ½at²) — Derived from the area under the velocity-time graph: rectangle (u × t) plus triangle (½ × at × t).
3. Third equation (v² = u² + 2as) — Derived by eliminating time t from the first two equations.
4. Five related quantities — The equations relate displacement (s), time (t), initial velocity (u), final velocity (v) and acceleration (a).
5. Validity and sign convention — Valid only for constant acceleration; the signs of u, v, a and s indicate direction for motion in a straight line.
Conclusion: These kinematic equations allow us to predict an object's future position and velocity, provided the acceleration remains constant.

Introduction: Uniform circular motion occurs when an object moves in a circular path with constant speed, but it is still accelerated.
Key Points:
1. Constant speed — The speed of the object remains the same at every point on the circular path.
2. Changing direction — The direction of velocity changes continuously as the object moves along the circle.
3. Velocity is not constant — Since velocity depends on both magnitude and direction, the changing direction means the velocity is changing.
4. Acceleration is non-zero — Because the velocity is changing (in direction), the motion is accelerated even though the speed is constant.
5. Tangent direction — The velocity at any point is along the tangent to the circle; when released, the object moves in a straight line along this tangent.
Conclusion: In uniform circular motion, acceleration arises from the continuous change in the direction of velocity, making it an accelerated motion despite constant speed.

Introduction: When an object is dropped from a height, it falls vertically downward with constant acceleration due to gravity.
Key Points:
1. Constant acceleration — The velocity increases by 9.8 m/s every second, giving a constant acceleration g = 9.8 m/s².
2. Velocity at each second — At t = 0, 1, 2, 3, 4 s, the velocities are 0, 9.8, 19.6, 29.4, 39.2 m/s respectively.
3. Using v = u + at — With u = 0 and a = g, the velocity at any time t is v = gt = 9.8t m/s.
4. Direction of acceleration — The acceleration is in the direction of motion (downward), so it is positive when downward is chosen as positive.
5. Origin choice — The origin can be at the point of release or at ground level; once chosen, it should not be changed during the problem.
Conclusion: Free fall is a straightforward application of kinematic equations with constant acceleration, demonstrating how velocity increases uniformly under gravity.

Introduction: When brakes are applied, a vehicle travels some distance before stopping, and this depends on several factors.
Key Points:
1. Stopping distance formula — Using v² = u² + 2as with v = 0, the stopping distance s = u²/(2a), where u is initial velocity and a is deceleration.
2. Velocity dependence — Doubling the initial velocity quadruples the stopping distance, since s is proportional to u².
3. Example — At 15 m/s with a = −4 m/s², stopping distance = 28.1 m; at 30 m/s, it is 112.5 m — four times greater.
4. Reaction time — The driver's reaction time adds extra distance before braking even begins.
5. Other factors — Road surface (wet/dry), braking capacity, tyre condition and weather all affect the total stopping distance.
Conclusion: Understanding how stopping distance depends on velocity and other factors is essential for maintaining safe following distances and preventing collisions.

Introduction: Distance and displacement, speed and velocity, are pairs of related but distinct quantities that describe motion.
Key Points:
1. Distance vs displacement — Distance is the total path length (scalar); displacement is the net change in position with direction (vector); e.g., an athlete travelling 160 m has a displacement of 40 m.
2. Speed vs velocity — Speed is distance ÷ time (scalar); velocity is displacement ÷ time (vector); Sarang's speed = 1 m/s but velocity = 0 m/s.
3. Equal only in one direction — Distance equals displacement magnitude, and speed equals velocity magnitude, only when the object moves in one direction without turning back.
4. Zero displacement, non-zero distance — An object completing one revolution has zero displacement but non-zero distance.
5. Practical example — Fuel consumption depends on distance, not displacement, as a vehicle returning to its start still consumes fuel for the full distance.
Conclusion: The distinction between scalar (distance, speed) and vector (displacement, velocity) quantities is crucial for accurately describing and analysing motion.

Introduction: Graphical representation of motion helps visualise how position, velocity and acceleration change with time.
Key Points:
1. Plotting procedure — Choose axes (time on X-axis, quantity on Y-axis), select a suitable scale, plot points from data, and connect them.
2. Position-time graph — Straight line = constant velocity; curve = accelerated motion; line parallel to time axis = stationary object.
3. Velocity-time graph — Line parallel to time axis = constant velocity (zero acceleration); slanted line = constant acceleration.
4. Extracting information — Slope of position-time graph gives velocity; slope of velocity-time graph gives acceleration; area under velocity-time graph gives displacement.
5. Graph is not a route map — A graph shows how position changes with time relative to the origin, not the physical path of the object.
Conclusion: Graphs are a powerful tool for analysing motion, allowing us to determine velocity, acceleration and displacement from visual representations.

Introduction: From a velocity-time graph, we can determine both the acceleration (from the slope) and the displacement (from the area).
Key Points:
1. Acceleration from slope — The slope of the line on a velocity-time graph equals the acceleration; slope = change in velocity ÷ change in time.
2. Displacement from area — The area enclosed by the line and the time axis equals the displacement in that time interval.
3. Constant velocity example — At 20 m/s for 6 s: slope = 0 (constant velocity); area = 20 × 6 = 120 m (displacement).
4. Changing velocity example — Velocity from 5 to 10 m/s over 10 s: a = 0.5 m/s²; displacement = rectangle (5 × 10) + triangle (½ × 5 × 10) = 50 + 25 = 75 m.
5. Decreasing velocity — When velocity decreases, the slope is negative (deceleration) but the area still gives the positive displacement.
Conclusion: The velocity-time graph is a versatile tool that allows us to calculate both acceleration and displacement from a single visual representation.