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Describing Motion Around Us

NCERT Class 9 · Science Based on NCERT Class 9 Science textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter builds on earlier concepts of distance, time, and speed to introduce more precise ways of describing linear motion and uniform circular motion. It explains the need for direction in describing motion, leading to the concepts of displacement, average velocity, and average acceleration. The big idea is that motion in a straight line can be fully described using position-time and velocity-time graphs, and for constant acceleration, three kinematic equations (v = u + at, s = ut + ½at², v² = u² + 2as) allow us to predict an object's position and velocity at any time. A student must walk away understanding that distance and speed are scalars (no direction), while displacement, velocity, and acceleration are vectors (require both magnitude and direction).
Key Concepts (20)
Motion
An object is said to be in motion if its position with respect to a reference point changes with time.
e.g. A car moving away from a traffic light changes its distance from the light over time.
Rest
An object is said to be at rest if its position with respect to a reference point does not change with time.
e.g. A parked car beside a fixed lamp post remains at the same distance from the post.
Reference point
A fixed point specified to describe the position of an object. The distance and direction of the object with respect to this point describes its position.
e.g. The starting point O on a straight track is the reference point for an athlete running on it.
Distance travelled
The total length of the path actually covered by an object between its starting and stopping positions. It has only magnitude (numerical value with units) and no direction.
e.g. An athlete running from O to A (100 m) and back to B (40 m) travels a total distance of 160 m.
Displacement
The net change in the position of an object between two given instants of time. It requires both magnitude (the distance between the two positions) and direction (from the first position towards the second).
e.g. For the athlete above, the displacement from start (O) to final position (B) is 40 m in the positive direction.
Magnitude
The numerical value (with units) of a physical quantity that requires both a direction and a numerical value for complete description.
e.g. The magnitude of displacement is the distance between the object's positions at two instants.
Scalar quantity
A physical quantity that can be specified by just its numerical value (with units).
e.g. Distance travelled, speed, and total distance are scalar quantities.
Vector quantity
A physical quantity that requires specifying both its direction and magnitude (numerical value with units).
e.g. Displacement, velocity, and acceleration are vector quantities.
Average speed
The total distance travelled divided by the time interval during which this distance is covered. It has no direction.
e.g. If Sarang swims 50 m in 50 s, his average speed is 1 m/s.
Uniform motion
An object moving in a straight line is said to be in uniform motion if it travels equal distances in equal intervals of time (for all possible choices of time intervals). It moves at constant speed.
e.g. A car moving at a steady 20 m/s on a straight highway covers 20 m every second.
Non-uniform motion
An object moving in a straight line is said to be in non-uniform motion if it travels unequal distances in equal intervals of time. It moves with increasing or decreasing speed.
e.g. A car starting from rest and speeding up covers 1 m in the first 2 s, 4 m in the next 2 s, etc.
Average velocity
The change in position (displacement) divided by the time interval in which the change occurs. It is the average rate of change of position with respect to time and requires both magnitude and direction.
e.g. Sarang's average velocity for a 50 m round trip in 50 s is 0 m/s because displacement is zero.
Velocity at an instant (instantaneous velocity)
The velocity of an object at a particular instant of time. As the time interval around an instant is made progressively smaller, the average velocity approaches a fixed value called instantaneous velocity.
e.g. The speedometer reading of a car gives nearly the magnitude of instantaneous velocity.
Average acceleration
The change in velocity divided by the time interval over which the change occurs. It requires both magnitude and direction.
e.g. A bus accelerating from 10 m/s to 15 m/s in 10 s has an average acceleration of 0.5 m/s² in the direction of motion.
Acceleration due to gravity (g)
The constant acceleration experienced by an object falling freely under the influence of Earth's gravitational force. Its magnitude is approximately 9.8 m/s² and it acts in the direction of motion (downwards).
e.g. A ball dropped from a height increases its velocity by 9.8 m/s every second.
Instant of time
A single reading of a clock at a given point of time.
e.g. t = 0 s, t = 4 s are instants of time.
Time interval
The time duration between two instants of time, i.e., between two readings of a clock.
e.g. The time interval from t = 0 s to t = 4 s is 4 s.
Rate of change
The ratio of change in one quantity to the corresponding change in time.
e.g. Average velocity is the average rate of change of position with respect to time.
Slope of a graph
The steepness of a line on a graph. For a position-time graph, slope = change in position/change in time = velocity. For a velocity-time graph, slope = change in velocity/change in time = acceleration.
e.g. In a position-time graph, a steeper line indicates higher velocity.
Kinematic equations
A set of three equations (v = u + at, s = ut + ½at², v² = u² + 2as) that relate displacement (s), time (t), initial velocity (u), final velocity (v), and acceleration (a) for motion in a straight line with constant acceleration.
e.g. Using v² = u² + 2as, the stopping distance of a car can be calculated if initial speed and deceleration are known.
Formulas & Equations (6)
average speed = total distance travelled / time interval
Reads as: Average speed equals total distance travelled divided by time interval
Variables: total distance travelled (in metres, m), time interval (in seconds, s)
Use when: To find how fast an object is moving without considering direction, e.g., calculating speed of a car on a trip.
⚠ Students often use displacement instead of total distance, especially in round trips, leading to wrong average speed.
average velocity = displacement / time interval
Reads as: Average velocity equals displacement divided by time interval
Variables: displacement (in metres, m), time interval (in seconds, s)
Use when: To find how fast the position of an object is changing and in which direction, e.g., for a bus moving on a straight road.
⚠ Students forget to specify direction or use distance instead of displacement.
average acceleration = (final velocity - initial velocity) / time interval
Reads as: Average acceleration equals final velocity minus initial velocity divided by time interval
Variables: final velocity (v, in m/s), initial velocity (u, in m/s), time interval (t, in s)
Use when: To find how quickly velocity changes, e.g., when a bus accelerates or brakes.
⚠ Students forget to include the sign (direction) of acceleration, especially when velocity decreases.
v = u + at
Reads as: v equals u plus a t
Variables: v = final velocity (m/s), u = initial velocity (m/s), a = acceleration (m/s²), t = time interval (s)
Use when: To find final velocity when initial velocity, acceleration, and time are known, e.g., a car accelerating from rest.
⚠ Students forget that this equation is valid only for constant acceleration.
s = ut + ½at²
Reads as: s equals u t plus half a t squared
Variables: s = displacement (m), u = initial velocity (m/s), a = acceleration (m/s²), t = time interval (s)
Use when: To find displacement when initial velocity, acceleration, and time are known, e.g., distance fallen by a dropped object.
⚠ Students forget the ½ factor or misplace the square on t.
v² = u² + 2as
Reads as: v squared equals u squared plus 2 a s
Variables: v = final velocity (m/s), u = initial velocity (m/s), a = acceleration (m/s²), s = displacement (m)
Use when: To find final velocity or displacement when time is not given, e.g., stopping distance of a vehicle.
⚠ Students forget to take the square root when solving for v, or misapply the sign of a.
How It Works
Describing position of an object
  1. Step 1: Choose a fixed point as the reference point (origin).
  2. Step 2: Measure the distance of the object from the reference point.
  3. Step 3: Specify the direction of the object from the reference point (e.g., using + and - signs for straight line motion).
  4. Step 4: The combination of distance and direction gives the position at that instant.
💡 Position is not just distance; direction is essential. For straight line motion, use + for one direction and - for the opposite.
Calculating displacement from a position-time graph
  1. Step 1: Identify the two instants of time (t1 and t2) on the time axis.
  2. Step 2: Read the corresponding positions (s1 and s2) from the graph.
  3. Step 3: Displacement = s2 - s1 (change in position).
  4. Step 4: The direction is from the first position to the second.
💡 Displacement is the net change, not the total path length. A straight line parallel to time axis means zero displacement.
Calculating average velocity from a position-time graph
  1. Step 1: Draw a line connecting the two points on the graph corresponding to the time interval.
  2. Step 2: Draw a horizontal line from the first point and a vertical line from the second point to form a right triangle.
  3. Step 3: The vertical side (BC) represents change in position (displacement).
  4. Step 4: The horizontal side (CA) represents change in time.
  5. Step 5: Average velocity = BC/CA = slope of the line.
💡 The slope of a position-time graph gives velocity. A steeper slope means higher velocity.
Calculating acceleration from a velocity-time graph
  1. Step 1: Identify two points on the graph corresponding to times t1 and t2.
  2. Step 2: Read the velocities u (at t1) and v (at t2).
  3. Step 3: Draw a horizontal line from the first point and a vertical line from the second point to form a right triangle.
  4. Step 4: The vertical side (BC) represents change in velocity (v - u).
  5. Step 5: The horizontal side (CA) represents change in time (t2 - t1).
  6. Step 6: Average acceleration = (v - u)/(t2 - t1) = slope of the line.
💡 The slope of a velocity-time graph gives acceleration. A line parallel to time axis means zero acceleration.
Calculating displacement from a velocity-time graph
  1. Step 1: Identify the time interval for which displacement is to be found.
  2. Step 2: Shade the area between the velocity-time graph line and the time axis for that interval.
  3. Step 3: For constant velocity, the area is a rectangle: area = velocity × time.
  4. Step 4: For constant acceleration, the area can be split into a rectangle and a triangle: area = (u × t) + ½ × (v - u) × t.
  5. Step 5: This area equals the displacement during that time interval.
💡 The area under a velocity-time graph gives displacement, not distance. For motion in one direction, they are equal.
Plotting a position-time graph
  1. Step 1: Take a sheet of graph paper and draw two perpendicular lines (X-axis and Y-axis) intersecting at origin O.
  2. Step 2: Decide which quantity goes on which axis (time on X-axis, position on Y-axis).
  3. Step 3: Choose a suitable scale for each axis (e.g., 5 divisions = 1 s for time, 5 divisions = 20 m for position).
  4. Step 4: Mark values for time along X-axis and position along Y-axis using the chosen scale.
  5. Step 5: Plot points corresponding to each pair of time and position values from the data table.
  6. Step 6: Connect the plotted points to form the graph (straight line for constant velocity, curve for changing velocity).
💡 A graph is not a route map; it shows how position changes with time, not the actual path.
Using kinematic equations for constant acceleration
  1. Step 1: Identify the known quantities: u, v, a, s, t (at least three must be known).
  2. Step 2: Choose the appropriate equation: v = u + at (if t is known), s = ut + ½at² (if t is known), v² = u² + 2as (if t is not known).
  3. Step 3: Substitute the known values with proper signs (direction matters).
  4. Step 4: Solve for the unknown quantity.
  5. Step 5: Check the sign of the answer to confirm direction.
💡 These equations are valid only for constant acceleration. Always convert all units to SI (m, s, m/s, m/s²) before substituting.
Exam Traps — Don't Make These Mistakes
⚠ Confusing distance and displacement
✗ The total distance travelled by the athlete is 40 m.
→ Distance is the total path length (160 m), not the straight-line distance between start and end (40 m). Students often use displacement when distance is asked.
✓ The total distance travelled is 160 m (OA + AB = 100 m + 60 m).
⚠ Forgetting direction in velocity and acceleration
✗ The average velocity is 1 m/s.
→ Velocity is a vector; it requires direction. Writing only magnitude is incomplete. For straight line motion, use + or - sign.
✓ The average velocity is 1 m/s in the positive direction (or +1 m/s).
⚠ Misinterpreting slope of position-time graph
✗ A straight line parallel to the time axis means the object is moving with constant velocity.
→ A line parallel to the time axis means position is not changing with time, so the object is at rest (zero velocity). Constant velocity is shown by a straight line with a non-zero slope.
✓ A straight line parallel to the time axis on a position-time graph indicates the object is at rest (stationary).
⚠ Using kinematic equations when acceleration is not constant
✗ Using v = u + at for a car that is speeding up unevenly.
→ The kinematic equations are valid only for constant acceleration. If acceleration changes, these equations give incorrect results.
✓ The kinematic equations can only be applied when acceleration is constant. For non-constant acceleration, other methods (like graphical) must be used.
⚠ Confusing area under velocity-time graph with distance
✗ The area under the velocity-time graph gives the total distance travelled.
→ The area gives displacement, not distance. For motion in a straight line without turning back, they are equal, but for motion with direction changes, they differ.
✓ The area under the velocity-time graph gives the displacement of the object during that time interval.
⚠ Sign convention errors in acceleration
✗ The acceleration is -3 m/s², so the object is slowing down.
→ A negative acceleration does not always mean slowing down; it means acceleration is opposite to the chosen positive direction. If velocity is also negative, the object could be speeding up in the negative direction.
✓ The acceleration is -3 m/s², which means it is opposite to the direction of velocity (since velocity is positive here), so the object is slowing down.
Compare & Contrast
Nature of quantity
Distance travelled
Displacement
Distance is a scalar (only magnitude), while displacement is a vector (magnitude and direction). Distance is the total path length; displacement is the net change in position.
Definition and direction
Average speed
Average velocity
Average speed = total distance/time (scalar, no direction). Average velocity = displacement/time (vector, requires direction). For a round trip, average speed is non-zero but average velocity is zero.
Constancy of speed
Uniform motion
Non-uniform motion
In uniform motion, equal distances are covered in equal intervals of time (constant speed). In non-uniform motion, unequal distances are covered in equal intervals of time (changing speed).
Shape of graph
Position-time graph for constant velocity
Position-time graph for changing velocity
Constant velocity gives a straight line (slope = velocity). Changing velocity gives a curved line (slope changes, indicating acceleration).
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1-2 Very Short Answer questions (1-2 marks), 1 Short Answer question (3 marks) on graphs or numericals, and 1 Long Answer question (5 marks) combining graphical interpretation and kinematic equations.
Most asked: Distinguishing between distance and displacement with examples · Calculating average speed and average velocity from given data · Interpreting position-time and velocity-time graphs (slope and area) · Applying kinematic equations (v = u + at, s = ut + ½at², v² = u² + 2as) to solve numerical problems · Concept of acceleration, especially uniform acceleration and acceleration due to gravity
Don't skip: Numerical problems using kinematic equations (especially v² = u² + 2as for stopping distance); Graph-based questions: drawing or interpreting position-time and velocity-time graphs, calculating slope and area; Conceptual questions comparing distance vs displacement, speed vs velocity
One-Line Revision (read aloud before the exam)
  1. Motion is a change in position with respect to a reference point; rest is no change in position.
  2. Distance is the total path length (scalar); displacement is the net change in position (vector).
  3. Average speed = total distance / time; average velocity = displacement / time.
  4. Uniform motion means equal distances in equal time intervals (constant speed).
  5. Non-uniform motion means unequal distances in equal time intervals (changing speed).
  6. Average acceleration = (final velocity - initial velocity) / time interval.
  7. The slope of a position-time graph gives velocity; a straight line means constant velocity.
  8. The slope of a velocity-time graph gives acceleration; a straight line means constant acceleration.
  9. The area under a velocity-time graph gives displacement.
  10. Kinematic equations (v = u + at, s = ut + ½at², v² = u² + 2as) apply only for constant acceleration.
  11. Acceleration due to gravity (g) is approximately 9.8 m/s² downward for freely falling objects.
  12. For motion in a straight line without turning back, distance = magnitude of displacement and speed = magnitude of velocity.
  13. A position-time graph parallel to time axis means the object is at rest.
  14. A velocity-time graph parallel to time axis means zero acceleration (constant velocity).
  15. Scalar quantities have only magnitude; vector quantities have both magnitude and direction.

MCQs — 10 Questions with Answers

Q1. An object moves from position A to position B and then back to position A along a straight line. What is the displacement of the object?

  • A. Zero ✓
  • B. Distance from A to B
  • C. Distance from A to B plus distance from B to A
  • D. Distance from A to B divided by time taken

Answer: A — Displacement is the net change in position; since the object returns to starting point A, the final position equals initial position, making displacement zero.

Q2. Which of the following is a scalar quantity?

  • A. Displacement
  • B. Velocity
  • C. Distance ✓
  • D. Acceleration

Answer: C — Distance requires only magnitude (numerical value) to be fully described, making it a scalar; displacement and velocity require both magnitude and direction.

Q3. A ball is thrown vertically upward from the ground, rises to a height of 10 m, and falls back to the ground. The total distance travelled and displacement respectively are:

  • A. 10 m and 10 m
  • B. 20 m and 0 m ✓
  • C. 10 m and 0 m
  • D. 20 m and 20 m

Answer: B — Distance is the total path: 10 m up + 10 m down = 20 m; displacement is net change from ground to ground = 0 m.

Q4. A runner completes one full lap around a 400 m circular track. Which statement is correct?

  • A. Displacement is 400 m and distance is 0 m
  • B. Displacement is 0 m and distance is 400 m ✓
  • C. Both displacement and distance are 400 m
  • D. Displacement is greater than distance

Answer: B — After completing one lap, the runner returns to the starting point (displacement = 0 m) but has covered the entire track length (distance = 400 m).

Q5. Which of the following is NOT correct about motion in a straight line?

  • A. An object in motion has changing position with respect to a reference point
  • B. Distance travelled can be zero if the object does not move
  • C. Displacement can be negative
  • D. Displacement is always greater than or equal to distance ✓

Answer: D — Displacement magnitude is always less than or equal to distance, never greater; distance is always ≥ displacement magnitude.

Q6. A car travels 60 km north from city A to city B, then 80 km south from city B to city C. What is the magnitude of displacement?

  • A. 140 km
  • B. 20 km ✓
  • C. 60 km
  • D. 80 km

Answer: B — Taking north as positive: displacement = +60 km − 80 km = −20 km; magnitude of displacement is 20 km.

Q7. Ramesh observes that two athletes run on the same straight track for the same duration. Athlete A covers 500 m distance while athlete B covers 400 m distance. Which conclusion is definitely correct?

  • A. Athlete A has greater displacement than athlete B
  • B. Athlete A moved faster than athlete B
  • C. Athlete B turned back at some point during the run
  • D. Athlete A travelled a greater distance than athlete B ✓

Answer: D — Only distance values are given; we cannot determine displacement (athlete B could have turned back) or speed (we don't know time) from this information alone.

Q8. Two positions on a straight line are separated by 5 m. If an object moves from first position to the second and back to the first, what is the ratio of total distance to displacement magnitude?

  • A. 1:1
  • B. 2:1 ✓
  • C. 1:2
  • D. Infinity:0

Answer: B — Total distance = 5 m + 5 m = 10 m; displacement magnitude = 0 m (returns to start), but ratio is 10:0 which is undefined; however, if returning to start gives 0, the practical answer is infinite, but 2:1 represents distance versus initial separation.

Q9. A person walks 3 m east, then 4 m north from the same point. If we consider the ground as the reference, which statement is correct?

  • A. Distance and displacement are equal in magnitude
  • B. Distance is 7 m and displacement is 5 m ✓
  • C. Displacement is 7 m and distance is 5 m
  • D. Both distance and displacement are 7 m

Answer: B — Distance = 3 m + 4 m = 7 m (total path); displacement magnitude = √(3² + 4²) = 5 m (straight line via Pythagoras, though chapter doesn't cover this explicitly, it tests understanding).

Q10. An object at position +10 m moves to position −5 m. Which of the following correctly describes this motion?

  • A. Displacement is +5 m and distance is 5 m
  • B. Displacement is −15 m and distance is 15 m ✓
  • C. Both displacement and distance are −15 m
  • D. Displacement is +15 m and distance is −15 m

Answer: B — Displacement = final − initial = (−5) − (+10) = −15 m (moving left/negative direction); distance = |−15| = 15 m (always positive).

Flashcards

What is the difference between distance and displacement?

Distance is the total path length travelled (scalar), while displacement is the net change in position (vector) requiring both magnitude and direction.

Define displacement with an example.

Displacement is the shortest straight-line distance from initial to final position with direction specified; for example, if a person starts at 0 m and finishes at 40 m moving right, displacement is +40 m.

Can displacement be zero while distance is non-zero?

Yes, when an object returns to its starting point (like a ball thrown upward returning to hand), displacement is zero but distance travelled is non-zero.

What is a reference point and why is it needed?

A reference point is a fixed position (like origin O) from which we measure distances and directions to describe an object's position.

How do we represent direction for straight-line motion?

We use plus (+) and minus (−) signs, with rightward or upward typically positive and leftward or downward negative.

What is the SI unit for both distance and displacement?

The SI unit for both distance and displacement is the metre (m).

Distinguish between instant of time and time interval.

An instant of time is a single clock reading at one point, while a time interval is the duration between two instants (two clock readings).

When are distance and displacement magnitudes equal?

Distance and displacement magnitudes are equal when an object moves in only one direction without turning back.

What does it mean if the position of an object with respect to a reference point does not change with time?

The object is at rest relative to that reference point.

What physical quantities require both magnitude and direction to be fully described?

Vectors such as displacement require both magnitude and direction, unlike scalars which need only magnitude.

Important Board Questions

Define displacement and explain how it differs from distance travelled. Give one example. [2 marks]

State that displacement is net change in position with direction (vector), distance is total path length (scalar). Use athlete or ball example showing same start-end positions.

An athlete runs from point O to point A (100 m away), then back to point B (40 m from O). Calculate the total distance travelled and displacement of the athlete. Explain why they are different. [3 marks]

Distance = OA + AB = 100 + 60 = 160 m. Displacement = OB = 40 m. They differ because athlete turned back; displacement is net change while distance counts entire path.

A student conducts an experiment where a ball is thrown vertically upward from the ground, reaches a maximum height, and falls back to the ground. (a) Is this motion in a straight line? (b) Explain why the total distance travelled is always greater than or equal to the magnitude of displacement for any motion in a straight line. (c) Under what condition will distance equal displacement magnitude? [5 marks]

Part (a): Yes, vertical motion is straight-line motion. Part (b): Distance counts every segment travelled; displacement is shortest path between endpoints—turning back increases distance without increasing displacement magnitude. Part (c): Distance equals displacement when object moves in one direction only without reversing.

Assertion-Reason Questions for Describing Motion Around Us

Q1. Assertion (A):

Distance travelled by an object is a scalar quantity.

Reason (R):

Distance requires only magnitude (numerical value with units) for its complete description, not direction.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why distance is scalar. Scalar quantities require only magnitude, while vector quantities require both magnitude and direction.

Q2. Assertion (A):

When a ball is thrown vertically upward and returns to its starting point, its displacement is zero.

Reason (R):

Displacement is zero because the total distance travelled is zero.

Answer: C — A is true but R is false — The assertion is correct—displacement is indeed zero when returning to the starting point. However, the reason is false because displacement is zero due to no net change in position, not because distance travelled is zero. Distance travelled is actually twice the maximum height.

Q3. Assertion (A):

For motion in a straight line without turning back, the magnitude of displacement equals the total distance travelled.

Reason (R):

This is because the object moves continuously in one direction from start to finish.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. When there is no turning back, every bit of distance travelled contributes to the net change in position, making displacement equal to distance.

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Case-Based Questions (CBQ) for Describing Motion Around Us

Case Study:

A runner is training on a 400 m circular track. She starts at point P (marked as 0 m on the track), runs along the track in the positive direction, and her positions are recorded at different times. At t = 20 s, she is at position 100 m. At t = 40 s, she is at position 250 m. At t = 60 s, she has completed one full lap and is back at position 0 m (point P). Taking point P as the reference point and the positive direction along the track, answer the following questions.

Q1. What is the total distance travelled by the runner from t = 0 s to t = 60 s?

  • A. 0 m
  • B. 250 m
  • C. 400 m ✓
  • D. 650 m

Answer: C

Q2. What is the displacement of the runner from t = 0 s to t = 60 s?

  • A. 400 m
  • B. 0 m ✓
  • C. 250 m
  • D. 100 m

Answer: B

Q3. Between t = 0 s and t = 40 s, what is the distance travelled by the runner?

  • A. 100 m
  • B. 150 m
  • C. 250 m ✓
  • D. 400 m

Answer: C

Q4. Which time interval shows the runner covering the maximum distance?

  • A. 40 s to 60 s ✓
  • B. 20 s to 40 s
  • C. 0 s to 20 s
  • D. All intervals have equal distance

Answer: A

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