by Describing Motion Around Us
How can a car’s motion be described? How far will it travel before stopping? Can an object accelerate even when its speed remains constant?
Motion is all around us. This chapter explains how to describe it using position, distance, displacement, speed, velocity, acceleration, graphs, and equations. Everyday examples and solved questions make the concepts easier to understand.
Think It Over
- How much distance should be maintained from a truck ahead to avoid a collision if it suddenly applies the brakes?
- Does this safe distance depend on the speed of the vehicle?
To answer these questions, motion must first be described precisely. The concepts in this chapter provide the tools for doing so.
What Will Be Learned?
Motion in a Straight Line
Position, reference point, distance, displacement, average speed, average velocity, and acceleration.
Graphs of Motion
Position-time and velocity-time graphs, their slopes, and the area beneath a velocity-time graph.
Kinematic Equations
Three equations for motion in a straight line with constant acceleration.
Uniform Circular Motion
Why velocity changes in circular motion even when speed is constant.
4.1 Motion in a Straight Line
When an object moves along a straight path, its motion is called linear motion. Examples include a car travelling along a straight road, a train moving on a straight track, and a ball falling vertically.
4.1.1 Describing Position
To describe an object’s position, a fixed point called a reference point must be chosen. The object’s distance and direction from this point specify its position.
An object is in motion when its position relative to the reference point changes with time. It is at rest when that position does not change relative to the chosen reference point.
4.1.2 Distance Travelled and Displacement
Distance travelled is the total length of the path covered by an object.
Displacement is the net change in position from the starting point to the final point, including direction.
Distance
Scalar quantity: magnitude only.
SI unit: metre (m).
It cannot be negative.
Displacement
Vector quantity: magnitude and direction.
SI unit: metre (m).
It can be zero, even when distance travelled is not zero.
Example: An athlete runs from O to A, 100 m away, and then returns 60 m towards O.
- Total distance = 100 + 60 = 160 m.
- Displacement = 100 − 60 = 40 m towards A.
4.1.3 Average Speed and Average Velocity
Average speed is a scalar quantity. Average velocity is a vector quantity and has the direction of displacement.
Example: A swimmer covers 50 m in 50 s while swimming to the other end of a pool and back to the starting end.
Total distance = 50 m, if the trip described is the full out-and-back journey only when both lengths total 50 m. More generally, use the actual total distance supplied in a question.
4.1.4 Average Acceleration
Average acceleration describes how quickly velocity changes over a time interval.
Where:
- u = initial velocity
- v = final velocity
- t = time interval
- a = average acceleration
The SI unit of acceleration is m/s². If an object speeds up in the positive direction, acceleration is positive. If it slows down while moving in the positive direction, acceleration is negative.
Interactive Motion Simulation
Change the initial velocity and acceleration, then start the simulation. The car’s position is calculated using the equations of motion. Click the car or use Pause to pause or resume.
This is a simplified visual simulation, not a road-safety calculator.
4.2 Graphical Representation of Motion
Graphs help describe how an object’s position or velocity changes with time. Time is generally placed on the horizontal axis, while position or velocity is placed on the vertical axis.
4.2.1 Position-Time Graphs
A position-time graph shows the position of an object at different instants of time.
- A horizontal line indicates that the object is at rest.
- A straight inclined line indicates constant velocity.
- A curved line indicates changing velocity and therefore accelerated motion.
- The slope of a straight position-time graph gives velocity.
4.2.2 Velocity-Time Graphs
A velocity-time graph represents how velocity changes with time.
- A horizontal line indicates constant velocity and zero acceleration.
- A straight line sloping upward indicates constant positive acceleration.
- A straight line sloping downward indicates constant negative acceleration when the object is moving in the positive direction.
- The slope of a velocity-time graph gives acceleration.
- The signed area under a velocity-time graph gives displacement.
4.3 Kinematic Equations for Constant Acceleration
For motion in a straight line with constant acceleration, initial velocity, final velocity, acceleration, time, and displacement are related by three important equations.
Where:
- u is initial velocity, in m/s.
- v is final velocity, in m/s.
- a is constant acceleration, in m/s².
- t is time interval, in seconds.
- s is displacement, in metres.
Solved Example
Question: A car moves at 15 m/s and brakes with an acceleration of −4 m/s². How far does it travel before stopping?
Given: u = 15 m/s, v = 0, a = −4 m/s².
Use the third equation:
v² = u² + 2as 0² = 15² + 2(−4)s 0 = 225 − 8s s = 225 / 8 = 28.125 mAnswer: 28.125 m.
4.4 Motion in a Plane and Uniform Circular Motion
Motion in a plane occurs in two dimensions. A vehicle overtaking another vehicle, a kicked ball, and an object moving in a circular path are examples of motion in a plane.
What Is Uniform Circular Motion?
When an object moves along a circular path with constant speed, its motion is called uniform circular motion.
Distance in one revolution = 2πR Average speed for one revolution = 2πR / THere, R is the radius of the circular path and T is the time taken for one complete revolution.
After one complete revolution, the object returns to its starting position. Therefore, its displacement over one complete revolution is zero and its average velocity over that interval is zero.
Why Does an Object Move in a Straight Line When Released?
A marble moving around the inside of a ring follows a circular path while the ring guides it. When the ring is lifted away, the marble moves along a straight line in the direction of its velocity at the instant of release, assuming other forces can be neglected.
The reason for this is explored further through Newton’s laws in the chapter on force and its effects on motion.
At a Glance: Important Formulas
| Quantity | Formula | SI unit |
|---|---|---|
| Average speed | Total distance / Time | m/s |
| Average velocity | Displacement / Time | m/s |
| Average acceleration | (v − u) / t | m/s² |
| First kinematic equation | v = u + at | — |
| Second kinematic equation | s = ut + ½at² | — |
| Third kinematic equation | v² = u² + 2as | — |
| Circular path, one revolution | 2πR | m |
| Average speed, one revolution | 2πR / T | m/s |
Revise, Reflect, Refine — All 16 Textbook Questions
Practise first, then open the solution. All answers below follow the exercise questions and figures in the uploaded Class 9 chapter. Graph-based answers use the textbook graph values where readable.
Q1. My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
Home → shop → home → shop → home.
There are four one-way journeys, each of 250 m.
Step 2: Calculate total distance. Distance = 4 × 250 = 1000 m Step 3: Calculate displacement.The final position is the same as the starting position.
Displacement = 0 mAnswer: Total distance = 1000 m; displacement = 0 m.
Q2. A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find: (i) the total vertical distance travelled, and (ii) the displacement from the starting point.
Given: Height of each floor = 3 m.
Step 1: Distance travelled upward. 4 × 3 = 12 m Step 2: Distance travelled downward.From the fourth floor to the second floor is two floors.
2 × 3 = 6 m Step 3: Total distance. 12 + 6 = 18 m Step 4: Displacement.The final position is the second floor, which is 6 m above ground.
Displacement = 6 m upwardAnswer: Distance = 18 m; displacement = 6 m upward.
Q3. A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and, if so, how?
Yes. Acceleration is the rate of change of velocity, and velocity includes both speed and direction.
If the scooter moves along a circular path at constant speed, its direction of motion continuously changes. Therefore, its velocity changes and the scooter is accelerated.
Answer: Yes, acceleration can occur even when speed is constant, because the direction of velocity can change.
Q4. A car starts from rest and its velocity reaches 24 m/s in 6 s. Find the average acceleration and the distance travelled in these 6 s.
Given: u = 0 m/s, v = 24 m/s, t = 6 s.
Step 1: Calculate acceleration. a = (v − u) / t a = (24 − 0) / 6 = 4 m/s² Step 2: Calculate distance. s = ut + ½at² s = 0 × 6 + ½ × 4 × 6² s = 2 × 36 = 72 mAnswer: Average acceleration = 4 m/s²; distance = 72 m.
Q5. A motorbike moving with initial velocity 28 m/s and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
Given: u = 28 m/s, v = 0, s = 98 m.
Step 1: Use v² = u² + 2as. 0² = 28² + 2 × a × 98 0 = 784 + 196a a = −784 / 196 = −4 m/s² Step 2: Use v = u + at. 0 = 28 + (−4)t 4t = 28 ⇒ t = 7 sAnswer: Acceleration = −4 m/s²; stopping time = 7 s. The negative sign indicates acceleration opposite to the initial direction of motion.
Q6. Figure 4.27 shows a position-time graph of two objects A and B moving along parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify the answer.
Answer: No.
The slope of a position-time graph gives velocity. The two straight lines have different slopes, so their velocities are different. They may have the same position at the instant the lines cross, but that does not mean they have equal velocities.
Q7. Figure 4.28 shows position-time graphs of objects A and B moving in a straight line from 0 to 10 s. Choose the correct option(s).
- (i) Their average velocities are equal because they have the same initial and final positions.
- (ii) Their average speeds are equal because both cover equal distance in equal time.
- (iii) Average speed of A is lower than that of B because A covers a shorter distance.
- (iv) Average speed of A is greater than that of B.
Correct options: (i) and (ii).
(i) Correct: Both objects have the same initial and final positions, so their displacements over 10 s are equal. Their average velocities are therefore equal.
(ii) Correct: The graphs show that both objects move continuously in the same direction, from the same initial to the same final position. Therefore, their total distances and average speeds over the interval are equal.
(iii) and (iv) Incorrect: Neither object covers a shorter total distance than the other over this interval.
Q8. A truck driver travelling at 54 km/h notices a speed-limit sign of 40 km/h for trucks. The driver slows down to 36 km/h in 36 s. What distance does the truck travel during this time? Assume constant acceleration.
Given: u = 54 km/h, v = 36 km/h, t = 36 s.
Step 1: Convert speeds into m/s. u = 54 × 5/18 = 15 m/s v = 36 × 5/18 = 10 m/s Step 2: Calculate acceleration. a = (v − u) / t = (10 − 15) / 36 = −5/36 m/s² Step 3: Calculate displacement. s = (u + v)t / 2 s = (15 + 10) × 36 / 2 = 450 mAnswer: 450 m.
Q9. A car starts from rest and accelerates uniformly to 20 m/s in 5 s. It then travels at 20 m/s for 10 s and finally brakes to a stop in 6 s with uniform acceleration. Find the total distance travelled.
u = 0, v = 20 m/s, t = 5 s.
s₁ = (u + v)t / 2 = (0 + 20) × 5 / 2 = 50 m Phase 2: Constant velocity.v = 20 m/s, t = 10 s.
s₂ = vt = 20 × 10 = 200 m Phase 3: Braking to rest.u = 20 m/s, v = 0, t = 6 s.
s₃ = (u + v)t / 2 = (20 + 0) × 6 / 2 = 60 m Total distance: s = 50 + 200 + 60 = 310 mAnswer: 310 m.
Q10. A bus is travelling at 36 km/h when the driver sees an obstacle 30 m ahead. The driver takes 0.5 s to react before braking. Once the brakes are applied, the velocity decreases with constant acceleration of 2.5 m/s². Will the bus stop before reaching the obstacle?
Step 1: Convert speed into m/s.
u = 36 × 5/18 = 10 m/sStep 2: Calculate distance travelled during reaction time.
sᵣ = vt = 10 × 0.5 = 5 mStep 3: Calculate braking distance.
Given u = 10 m/s, v = 0 and a = −2.5 m/s².
v² = u² + 2as 0 = 100 + 2(−2.5)s s = 100/5 = 20 mStep 4: Total stopping distance.
Total = 5 + 20 = 25 mThe obstacle is 30 m ahead, and the stopping distance is 25 m.
Answer: Yes. Under the stated assumptions, the bus stops 5 m before the obstacle.
Q11. A student says, “The Earth moves around the Sun.” In this context, discuss whether an object kept on the Earth can be considered to be at rest.
Rest and motion depend on the reference point.
- Relative to the Earth: An object kept on the ground can be considered at rest if its position relative to the Earth does not change.
- Relative to the Sun: The same object is moving because the Earth revolves around the Sun.
Answer: The object can be at rest relative to the Earth while being in motion relative to the Sun.
Q12. The velocity-time graph from 0 s to 120 s for a cyclist is shown below. Identify the areas representing displacement (i) while the cyclist is moving with constant velocity and (ii) while the velocity is decreasing. Calculate total displacement and average acceleration during the 120 s interval.
(i) Constant velocity: Shade the rectangular area under the horizontal section from 20 s to 100 s.
(ii) Decreasing velocity: Shade the trapezoidal area under the downward-sloping section from 100 s to 120 s.
Step 1: Area of the first triangle. s₁ = ½ × 20 × 3 = 30 m Step 2: Area of the rectangle. s₂ = (100 − 20) × 3 = 80 × 3 = 240 m Step 3: Area of the final trapezium. s₃ = ½ × (3 + 2) × 20 = 50 m Step 4: Total displacement. s = 30 + 240 + 50 = 320 m Step 5: Average acceleration.Initial velocity = 0 m/s; final velocity = 2 m/s; total time = 120 s.
a = (2 − 0) / 120 = 0.0167 m/s² (approx.)Answer: Displacement = 320 m; average acceleration ≈ 0.0167 m/s².
Q13. A girl is preparing for her first marathon by running on a straight road. Her smartwatch records the velocity-time graph shown in Figure 4.31. Estimate the distance she ran based on the graph.
Distance travelled is the area under the velocity-time graph. Because the graph provides an estimate, its area can be approximated by dividing it into trapezia and rectangles.
Using approximate readings from the plotted curve gives an average speed of roughly 7 km/h over approximately 6.5 h.
Estimated distance ≈ Average speed × Time ≈ 7 × 6.5 = 45.5 kmAnswer: Approximately 46 km. This is an estimate from the graph, not an exact distance.
Q14. On entering a state highway, a car moves with constant velocity of 6 m/s for 2 minutes and then accelerates at 1 m/s² for 6 seconds. Find the displacement in the total time of 2 min 6 s using a velocity-time graph.
Given v = 6 m/s and t = 2 min = 120 s.
s₁ = vt = 6 × 120 = 720 m Phase 2: Accelerated motion.Given u = 6 m/s, a = 1 m/s², t = 6 s.
s₂ = ut + ½at² s₂ = 6 × 6 + ½ × 1 × 6² s₂ = 36 + 18 = 54 m Total displacement. s = 720 + 54 = 774 mAnswer: 774 m.
Q15. Two cars A and B start from rest and move in a straight line with constant acceleration. Car A reaches 5 m/s in 5 s; car B reaches 3 m/s in 10 s. Plot their velocity-time graphs on the same graph and calculate the displacement in the two time intervals.
Car A:
aA = (5 − 0) / 5 = 1 m/s²Car B:
aB = (3 − 0) / 10 = 0.3 m/s² Step 2: Prepare points for plotting.| Car A: Time (s) | Velocity (m/s) | Car B: Time (s) | Velocity (m/s) |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | 1 | 2 | 0.6 |
| 2 | 2 | 4 | 1.2 |
| 3 | 3 | 6 | 1.8 |
| 4 | 4 | 8 | 2.4 |
| 5 | 5 | 10 | 3.0 |
For car A, area of triangle:
sA = ½ × 5 × 5 = 12.5 mFor car B, area of triangle:
sB = ½ × 10 × 3 = 15 mAnswer: Car A travels 12.5 m in 5 s; car B travels 15 m in 10 s.
Q16. Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute hand of a wall clock. During the interval, calculate (i) distance travelled, (ii) displacement, (iii) average speed, and (iv) average velocity. The length of the minute hand is 7 cm.
Given: Radius r = 7 cm; time interval = 90 minutes = 5400 s.
(i) Distance travelledThe minute hand completes 1.5 revolutions in 90 minutes.
Distance = 1.5 × 2πr = 1.5 × 2 × π × 7 = 21π cm ≈ 66 cm (ii) DisplacementThe initial tip is at 12 o’clock and the final tip is at 6 o’clock. The points are opposite ends of a diameter.
Displacement magnitude = 2r = 14 cmDirection: vertically downward.
(iii) Average speed Average speed = Distance / Time = 66 / 5400 ≈ 0.0122 cm/s (iv) Average velocity Average velocity = Displacement / Time = 14 / 5400 ≈ 0.00259 cm/s downwardAnswers: Distance ≈ 66 cm; displacement = 14 cm downward; average speed ≈ 0.0122 cm/s; average velocity ≈ 0.00259 cm/s downward.
From Describing Motion to Understanding Force
This chapter has established the language of motion: position, displacement, velocity, acceleration, graphs, and equations. These tools tell us how an object moves and how its motion changes.
The natural next question is: What causes a change in velocity? Why does a ball accelerate when kicked, why does a vehicle slow down when brakes are applied, and why does a moving object tend to continue moving in the absence of a net force?
The companion chapter on Force and How Forces Affect Motion connects these ideas through Newton’s laws. In particular, Newton’s Second Law links net force to acceleration:
Fnet = maThe acceleration studied in this chapter is therefore the bridge between describing motion and explaining its cause. With both chapters together, it becomes easier to understand everyday motion and solve physics numericals.
Revision tip: Before practising force numericals, revise the meaning of velocity, acceleration, graph slopes, and the equations of motion from this chapter.
