📚 Chapter 4 Index
Chapter 4: Describing Motion Around Us
🎯 Learning Objectives
- Describe the position of an object using a reference point, distance, and direction
- Differentiate between distance travelled and displacement, and identify when their magnitudes are equal
- Define and calculate average speed, average velocity, and average acceleration
- Interpret position-time and velocity-time graphs to identify the nature of motion (uniform, non-uniform, accelerated)
- Calculate velocity and acceleration from the slope of a graph, and displacement from the area under a velocity-time graph
- Derive and apply the three kinematic equations of motion for constant acceleration
- Explain uniform circular motion and understand why it is accelerated motion even at constant speed
🤔 Think It Over
- How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?
- Does this distance depend upon the speed with which we are moving?
- Everything in nature is in motion — from massive astronomical objects to subatomic particles
- Variety in motion in nature: flitting butterflies, slithering snakes, hopping hares, galloping horses, tendrils of climbers twinning around a support, closing of flytraps, dancing dust particles in a sunbeam, smoke particles moving in air, rising and falling of ocean tides, gathering clouds
- To explore a complex phenomenon, scientists first study it in its idealised simplified forms
- Such types of motion are: linear, circular, and oscillatory
- In this chapter: linear motion (motion in a straight line) and uniform circular motion
- Earlier learnt: distance, time, speed
- Now will learn: displacement, average velocity, average acceleration
- Motion will be described in words, numbers, equations, and graphs
4.1 Motion in a Straight Line
- When an object moves in a straight line, its motion is called linear motion
- It can also be called motion in a straight line
- It is the simplest kind of motion
- Examples: children in a swimming race, a vertically falling ball, a car moving along a straight stretch of a highway, a train moving on a straight track
Fig. 4.1: Objects in a straight line motion
- To discuss the motion of an object, we first need to describe its position at various instants of time
4.1.1 Describing position
- To describe the position of an object, we first need to specify a fixed point as the reference point
- The distance and direction of the object with respect to the reference point, at any instant of time, describes the position of the object at that instant of time
- Apart from the distance, we also specify the direction from the reference point in which the object is located
- If the position of the object with respect to the reference point changes with time, the object is said to be in motion
- The object is said to be at rest if its position with respect to the reference point does not change with time
Fig. 4.2: An athlete running on a straight track
- Example: athlete running on a straight track — her starting point is taken as the reference point
- A straight line is made with distances marked on it, and the reference point is marked on it as the origin ‘O’
- The athlete starts running from O, and her positions at two instants of time are marked by points B and A
Fig. 4.3: Reference point and positions of the athlete at different instants of time on a straight line
- To describe the position of an object, we also need to specify its direction
- For an object moving in a straight line, the object can move only in one of two directions — forward and backward
- Direction is represented by plus (+) and minus (–) signs
- Positions to the right of the reference point O are generally taken as positive, and to the left of O as negative
📝 Questions — Reference Point & Position
LOTS
What is a reference point, and why is it needed to describe the position of an object?
Show Answer
A reference point is a fixed point chosen to measure the distance and direction of an object. It is needed because position can only be described relative to something fixed — without it, we cannot say where an object is.
Medium
Why do we need to specify direction along with distance to describe the position of an object?
Show Answer
Distance alone only tells us how far the object is from the reference point, but not where it is. Two objects can be the same distance away but on opposite sides — direction tells us which side of the reference point the object is on.
HOTS
Two students choose different points as the origin while describing the same athlete’s motion. Will their descriptions of the athlete’s position differ? Will their conclusion about whether the athlete is in motion or at rest differ?
Show Answer
Yes, the numerical position values will differ since they depend on the chosen origin. However, whether the athlete is in motion or at rest will not change — that depends only on whether position changes with time, which is independent of where the origin is placed.
HOTS
A bus is moving, but a passenger sitting inside it appears “at rest” to another passenger inside the same bus. Explain why this happens using the idea of a reference point.
Show Answer
If the reference point is taken as another seat inside the same bus, the passenger’s position relative to that reference point does not change — so they appear at rest. But relative to a reference point outside the bus (like a tree on the road), the passenger’s position is constantly changing, so they are in motion. Motion is always described relative to a chosen reference point.
4.1.2 Distance travelled and displacement
- Suppose an athlete starts running from point O at time t = 0 s, reaches point B at t = 4 s, then reaches point A at t = 10 s, then runs back along the same path till point B reaching there at t = 16 s
- Total distance travelled by the athlete = OA + AB = 100 m + 60 m = 160 m
- Distance between starting and stopping positions = OB = 40 m (different from total distance travelled)
Fig. 4.4: Reference point and positions of athlete at different instants of time
- Displacement is the net change in the position of an object between two given instants of time
- A complete description of physical quantities like displacement requires specifying both a direction and its numerical value (with units)
- The numerical value (with units) of such a physical quantity is called its magnitude
- The magnitude of displacement is the distance between the object’s positions at the two instants
- The direction of displacement is specified from the position at the first instant towards the position at the second instant
- To describe the total distance travelled, only the numerical value (with units) is required, NOT the direction of motion
- The SI unit for both distance travelled and displacement is the metre (m)
- Example: between t = 0 s and t = 16 s, total distance travelled by the athlete = 160 m, but her displacement = 40 m in the positive direction
- The total distance travelled and the magnitude of displacement are not equal in this case
📝 Questions — Distance & Displacement
LOTS
Define displacement. What is its SI unit?
Show Answer
Displacement is the net change in the position of an object between two given instants of time. Its SI unit is the metre (m), same as distance.
Medium
A person walks 300 m east and then 300 m west, returning to the starting point. Find the total distance travelled and the displacement.
Show Answer
Total distance travelled = 300 m + 300 m = 600 m. Displacement = 0 m, since the person returns to the starting point (net change in position is zero).
HOTS
Can the magnitude of displacement ever be greater than the total distance travelled? Justify your answer with reasoning.
Show Answer
No. Displacement is the shortest, straight-line change in position between two points, while distance travelled accounts for the actual path taken, which may be longer (especially if the object changes direction). So the magnitude of displacement can be less than or equal to distance travelled, but never greater.
HOTS
A car’s odometer (which measures distance) shows it has travelled 50 km after a round trip, but the car is back at its starting garage. Explain this situation using the concepts of distance and displacement, and explain why a car’s fuel gauge depends on one of these quantities and not the other.
Show Answer
The odometer correctly shows distance travelled = 50 km, since the car actually covered that much road. But since the car returns to its starting point, its displacement = 0. Fuel consumption depends on distance travelled, not displacement, because the engine uses fuel for every metre actually covered on the road, regardless of the net change in position.
Note:
- For motion in a straight line, the total distance travelled and the magnitude of displacement are equal if the object moves without turning back, i.e., if it moves in one direction
📦 Activity 4.1: Let us analyse
- As shown in Fig. 4.5, a ball is thrown vertically upwards from O. It moves up straight till B and then falls back to O. Can this be considered a motion in a straight line?
- For this motion, fill up the values in Table 4.1
Fig. 4.5: A ball in vertical motion (two separate lines are shown only for clarity; in reality, the object goes up and falls back in the same straight line)
Table 4.1: Distance travelled and displacement of the ball
- S.No. 1 — Position O — Total distance travelled from O: 0 cm — Displacement from O: 0 cm
- S.No. 2 — Position A — Total distance travelled from O: 40 cm — Displacement from O: 40 cm in upward direction
- S.No. 3 — Position B — Total distance travelled from O: ______ — Displacement from O: ______
- S.No. 4 — Position C — Total distance travelled from O: ______ — Displacement from O: ______
- S.No. 5 — Position O — Total distance travelled from O: ______ — Displacement from O: ______
- Analyse the data filled in Table 4.1 and choose which of the following is true for displacement:
- (i) It is never zero
- (ii) Its magnitude can be greater than the total distance travelled
- (iii) Its magnitude is less than or equal to the total distance travelled
- (iv) Its magnitude is less than the total distance travelled in all cases
⏸️▶️ Pause and Ponder
- In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?
- Fuel used up in a vehicle depends on which of the following? Justify your answer.
- (i) Total distance travelled
- (ii) Displacement
- A ball rolls down an inclined track as shown in Fig. 4.6. Is its motion, a straight line motion? Assuming the starting point of the ball (O) to be the origin, can its motion from O to D be depicted using a horizontal line as shown in Fig. 4.3? Are the values of total distance travelled and magnitude of displacement from O equal or different at positions A, B, C and D?
Fig. 4.6: A ball rolling down an inclined track
4.1.3 Average speed and average velocity
- Average speed tells us how fast or slow an object moves
- The average speed of an object is the total distance travelled divided by the time interval during which this distance is covered
average speed = total distance travelled / time interval (4.1)
- Since distance travelled has no direction (only a numerical value), average speed also has no direction, only a numerical value
- If an object moving in a straight line travels equal distances in equal intervals of time (for all possible choices of time intervals), it is said to be in uniform motion in a straight line — the object moves at a constant speed
- If the object travels unequal distances in equal intervals of time, it is in non-uniform motion in a straight line — speed is increasing, decreasing, or a combination of both
- If the distances travelled in successive intervals of time are increasing, its speed is increasing
🇮🇳 India’s Scientific Contributions
- The concept that an object’s speed is the distance travelled divided by the time taken is well-established, dating back to ancient times even in India, as seen in the treatise Aryabhatiya (5th century CE)
- The following problem, based on this concept, is from a comprehensive mathematical text, the Ganitakaumudi (14th century CE)
Example 4.1: Consider two postmen. They start walking towards each other from a distance of 210 yojanas (a unit of distance used in ancient India). One travels 9 yojanas per day and the other covers 5 yojanas per day. Can you determine in how many days they will meet each other?
Show Answer
- Total distance covered by each postman in one day = 9 yojanas + 5 yojanas = 14 yojanas
- To meet each other, the postmen need to cover 210 yojanas together
- Time taken = 210 / 14 = 15 days
- (In 15 days, first postman covers 135 yojanas and second postman covers 75 yojanas)
- Speed tells us how fast an object is moving but provides no information about the direction of motion
- Many real-world situations need direction along with speed for a complete picture
Average velocity
- The average velocity of an object in a time interval is the change in position (or displacement) divided by the time interval in which the change occurs
average velocity = change in position / time interval = displacement / time interval (4.2a)
vav = s / t (4.2b)
- To express average velocity, we need to specify its magnitude as well as direction
- The direction of velocity is the same as the direction of displacement, indicated by a ‘+’ or ‘–’ sign
- The SI unit of average speed and average velocity is the same: metre per second (m s⁻¹ or m/s); also commonly measured in kilometre per hour (km h⁻¹)
- The ratio of change in one quantity to the corresponding change in time is called the rate of change
- Average velocity is the average rate of change of position of an object with respect to time
Note:
- For motion in a straight line, the average speed and the magnitude of average velocity in a time interval are equal if the object moves in one direction
Example 4.2: Sarang takes 50 seconds to swim from one end to the other end and back in the swimming pool shown in Fig. 4.7. Find his average speed and average velocity within the time interval of 50 s.
Fig. 4.7: Swimming pool (25 m)
Answer: (Find the total distance and displacement first, then use the formulas above)
Show Answer
- Total distance travelled by Sarang in 50 s = 50 m
- Displacement of Sarang in 50 s = 0 m
- average speed = total distance travelled / time interval = 50 m / 50 s = 1 m s⁻¹
- average velocity = displacement / time interval = 0 m / 50 s = 0 m s⁻¹
- During the 50 s time interval, the average speed of Sarang is approximately 1 m s⁻¹ while his average velocity is 0 m s⁻¹
⏸️▶️ Pause and Ponder
- During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.
- Under what condition(s) is the:
- (i) magnitude of average velocity of an object equal to its average speed?
- (ii) magnitude of average velocity of an object zero while its average speed is not zero?
📦 Ready to Go Beyond
- What we simply called ‘velocity at an instant’ is known as ‘instantaneous velocity’
- As the time interval around an instant is made progressively smaller, the change in average velocity gets smaller and smaller
- When the time interval becomes infinitesimally small, the average value of velocity approaches a fixed value called the instantaneous velocity
- This will be learnt in higher grades
- Throughout this chapter, the term ‘velocity’ means the velocity of an object at a particular instant
- The velocity of an object can be constant or it can change with time
📝 Questions — Speed & Velocity
LOTS
What is the formula for average speed, and what is its SI unit?
Show Answer
Average speed = total distance travelled / time interval. Its SI unit is metre per second (m s⁻¹).
Medium
A car travels 100 km in 2 hours along a straight road without changing direction. Find its average speed and average velocity.
Show Answer
Since the car moves in one direction without turning back, distance = displacement = 100 km. Average speed = average velocity = 100/2 = 50 km h⁻¹.
HOTS
Two cyclists cover the same distance of 20 km. Cyclist A travels in a straight line, while Cyclist B travels along a winding path that covers the same straight-line distance but a longer total path. If both take the same time, whose average speed is higher, and whose average velocity is higher?
Show Answer
Cyclist B’s average speed is higher (since B covers a greater total distance in the same time). Both cyclists have the same average velocity (since both have the same displacement — 20 km in a straight line — over the same time).
HOTS
Is it possible for an object’s average velocity to be zero while its average speed is not zero? Give an example and explain why this happens.
Show Answer
Yes. Example: Sarang swimming one length of a pool and back to the start in 50 s. His average speed is 1 m s⁻¹ (since he covered actual distance), but his average velocity is 0 m s⁻¹ (since his displacement is zero — he returns to the starting point). This happens whenever an object returns to its starting position after travelling some path.
4.1.4 Average acceleration
- When a vehicle suddenly moves from rest or suddenly stops, we feel a noticeable jolt — this captures the feeling of change in velocity
- The average acceleration of an object over a time interval is the change in its velocity divided by the time interval
average acceleration = change in velocity / time interval (4.3a)
average acceleration = (final velocity − initial velocity) / time interval (4.3b)
- If the velocity of an object changes from an initial value u at time t₁ to the final value v at time t₂, the average acceleration a is:
a = (v − u) / (t₂ − t₁) (4.3c)
- The SI unit of average acceleration is m s⁻² or m/s²
- Like displacement and velocity, we need to specify both the magnitude and direction of acceleration
- For motion in a straight line: if the magnitude of velocity is increasing, the average acceleration is in the direction of velocity
- If the magnitude of velocity is decreasing, the average acceleration is opposite to the direction of velocity
Fig. 4.8: Direction of average acceleration when magnitude of velocity is (a) increasing, and (b) decreasing
🧵 Threads of Curiosity
- The reading of the speedometer of a vehicle is nearly (but not exactly) the same as the magnitude of the velocity at an instant
- The direction of the tyres gives the direction of velocity at that instant
- The average acceleration can result from a change in the magnitude of velocity, or a change in its direction, or both
- For an object moving in a straight line in the same direction: if the magnitude of velocity increases or decreases by equal amounts in equal intervals of time (for all possible choices), the acceleration of the object is constant
📦 Activity 4.2: Let us calculate
- The magnitude of average acceleration of cars is generally specified as the time taken by the car to go from 0 km h⁻¹ to 100 km h⁻¹. Look it up on the internet and find this time for various cars, and record those in Table 4.2.
- Calculate the magnitude of average acceleration for each car.
Table 4.2: The magnitude of average acceleration in a time interval
- Car type — Time interval (0 to 100 km h⁻¹) — Magnitude of average acceleration (m s⁻²)
- ______ — ______ — ______
- ______ — ______ — ______
- ______ — ______ — ______
Example 4.3: A bus is moving on a long straight highway (Fig. 4.9) with a velocity of 36 km h⁻¹. The driver presses the accelerator for a time interval of 10 s and velocity of the bus increases to 54 km h⁻¹. For some time, the bus moves at a constant velocity. Then, the driver notices an obstacle on the road ahead and presses the brake. The bus comes to a stop in a time interval of 5 s. Find the average acceleration in the two time intervals, (i) when the accelerator was pressed, and (ii) when the brakes were pressed.
Fig. 4.9: A bus moving on a long straight highway
Answer: (Use a = (v−u)/t for each interval separately)
Show Answer
- (i) When the driver presses the accelerator
- u = 36 km h⁻¹ = 10 m s⁻¹, v = 54 km h⁻¹ = 15 m s⁻¹, t = 10 s
- a = (15 − 10) / 10 = 5/10 = 0.5 m s⁻²
- Since the magnitude of velocity is increasing, the acceleration acts in the direction of velocity
- (ii) When the driver presses the brake
- u = 54 km h⁻¹ = 15 m s⁻¹, v = 0 m s⁻¹, t = 5 s
- a = (0 − 15) / 5 = −15/5 = −3 m s⁻²
- The minus sign indicates the acceleration acts opposite to the direction of velocity (since the magnitude of velocity is decreasing)
Note:
- An object can be moving very fast and yet have zero acceleration
- Acceleration depends not on how fast an object is moving, but on how quickly its velocity is changing
- Example: a bus moving on a straight highway at constant velocity has zero acceleration, even though its velocity may be high
- The average acceleration during different time intervals can either be constant or changing
- For an object moving in a straight line in the same direction, if the magnitude of its velocity increases or decreases by equal amounts in equal intervals of time (for all possible choices of time intervals), the acceleration of the object is constant
Example 4.4: As we learnt earlier, when an object is dropped from a height, it takes a straight vertical path downwards before touching the ground. While coming down, the velocity of the object increases as shown in Fig. 4.10 at different instants. Find the magnitude of the average acceleration of the object in every successive interval of a second. Is the average acceleration constant across all intervals? What is the direction of this average acceleration?
Fig. 4.10: An object dropped from a height
Answer: (Calculate acceleration for each 1-second interval using a = (v−u)/t)
Show Answer
- Between 0 s and 1 s: (9.8 − 0) / (1 − 0) = 9.8 m s⁻²
- Between 1 s and 2 s: (19.6 − 9.8) / (2 − 1) = 9.8 m s⁻²
- Between 2 s and 3 s: (29.4 − 19.6) / (3 − 2) = 9.8 m s⁻²
- Between 3 s and 4 s: (39.2 − 29.4) / (4 − 3) = 9.8 m s⁻²
- The average acceleration is constant and equal to 9.8 m s⁻²
- As the velocity is increasing in the direction of motion, the acceleration is in the direction of motion
- This acceleration is called the acceleration due to gravitational force by the Earth, denoted by g
- In Fig. 4.5, the origin was chosen at the ground level, while in Fig. 4.10, the origin was chosen at the point from where the object is dropped
- We can choose the origin and positive direction as per our convenience — however, once chosen, it should not be changed while solving a problem
- Just like we can specify the velocity of an object at an instant, we can specify the acceleration of an object at an instant
Note:
- In this chapter, we will consider only the cases where the acceleration is constant
📦 Ready to Go Beyond
- Similar to ‘velocity at an instant’, ‘acceleration at an instant’ is known as ‘instantaneous acceleration’
- This will be learnt in higher grades
📝 Questions — Acceleration
LOTS
What is the SI unit of acceleration?
Show Answer
The SI unit of acceleration is metre per second squared (m s⁻²).
Medium
A scooter’s velocity decreases from 20 m s⁻¹ to 5 m s⁻¹ in 5 seconds. Find its acceleration and state its direction relative to the velocity.
Show Answer
a = (v − u)/t = (5 − 20)/5 = −15/5 = −3 m s⁻². The negative sign shows the acceleration is opposite to the direction of velocity (since the scooter is slowing down).
HOTS
A car moving at a constant 80 km h⁻¹ on a straight highway has its speedometer needle perfectly still. Does this mean the car has zero acceleration? What additional information would you need to be sure?
Show Answer
Not necessarily from the speedometer alone — the speedometer only shows the magnitude of velocity, not direction. If the car is moving on a straight road with constant speed, acceleration is indeed zero. But if the road were curving (even at constant speed), the car would still be accelerating because the direction of velocity would be changing. We would need to know whether the path is straight or curved.
HOTS
Two balls are dropped from the same height — one on Earth and one (hypothetically) on a planet with stronger gravity. Both start from rest. Which ball will have a greater velocity after 2 seconds, and why?
Show Answer
The ball on the planet with stronger gravity will have a greater velocity after 2 seconds. Since v = u + at, and u = 0 for both, a larger acceleration (g) due to stronger gravity directly produces a larger final velocity (v) for the same time interval.
4.2 Graphical Representation of Motion
- Graphical representation provides a visual representation of how position, velocity, and acceleration change with time
- Such graphs help in: comparing the motion of two objects, calculating physical quantities, and identifying whether the motion is uniform or non-uniform
- To describe motion, graphs show the dependence of one physical quantity (such as position, velocity or acceleration) on another quantity (such as time)
Note:
- All the graphs discussed in this chapter are for motion in a straight line in one direction only
- In this special case, distance travelled and magnitude of displacement are equal, and speed and magnitude of velocity are also equal
- If position is zero at time zero, then the position-time graph is the same as the distance-time graph, and the velocity-time graph is the same as the speed-time graph
4.2.1 Plotting graph
- To plot a graph, use the data given in Table 4.3 for a vehicle moving on a straight road
Table 4.3: Positions of vehicle at different instants of time
- Time: 0 s — Position: 0 m
- Time: 1 s — Position: 20 m
- Time: 2 s — Position: 40 m
- Time: 3 s — Position: 60 m
- Time: 4 s — Position: 80 m
- Time: 5 s — Position: 100 m
- Time: 6 s — Position: 120 m
📦 Activity 4.3: Let us plot a graph
- Take a sheet of graph paper, pre-divided into small squares, making it easier to plot data accurately (Fig. 4.11a)
- On the graph paper, draw two lines perpendicular to each other. Their point of intersection is known as origin O. Mark the horizontal line as OX — the X-axis. Mark the vertical line as OY — the Y-axis
- Decide which quantity (time or position) to show along each axis — here, time along the X-axis and position along the Y-axis
- Determine a suitable scale for each quantity:
- X-axis: 5 divisions = 1 s
- Y-axis: 5 divisions = 20 m
- Use the chosen scale to mark values for time (1 s, 2 s, …) along the X-axis, and position (20 m, 40 m, …) along the Y-axis (Fig. 4.11b)
- Begin plotting points to represent each set of time and position values from Table 4.3:
- (i) At time 0 s, the position is also 0 m — the point is the origin itself
- (ii) At 1 s, the position is 20 m. Draw a line parallel to the Y-axis at this point on X-axis, and a line parallel to X-axis at 20 m on Y-axis — their intersection represents (1 s, 20 m)
- (iii) Similarly, plot all points corresponding to positions at different instants of time
- Once all points are plotted, connect them to create the position-time graph (Fig. 4.11c) — it is a straight line for the data given in Table 4.3
Fig. 4.11(a): Marking origin, X and Y axes on graph paper
Fig. 4.11(b): Plotting points on the graph
Fig. 4.11(c): Making a graph
Note:
- A graph is not a route map. It does not show the route but how the position of the object changes with time with respect to the origin
📦 Ready to Go Beyond
- The intermediate points represent possible values for position of vehicle at intermediate times
- They are correct if the vehicle is moving with a constant speed
Example 4.5: For a vehicle starting from rest and speeding up, the data for position and time are given in Table 4.4. Plot the position-time graph corresponding to it.
Table 4.4: Positions of vehicle at different instants of time
- Time: 0 s — Position: 0 m
- Time: 2 s — Position: 1 m
- Time: 4 s — Position: 4 m
- Time: 6 s — Position: 9 m
- Time: 8 s — Position: 16 m
- Time: 10 s — Position: 25 m
- Time: 12 s — Position: 36 m
Answer: (Plot the points using the procedure of Activity 4.3 and observe the shape)
Show Answer
- Choosing the scale: X-axis: 5 divisions = 2 s; Y-axis: 5 divisions = 5 m
- Following the procedure of Activity 4.3, all points are marked
- Unlike Fig. 4.11c, the points do not fall on a straight line — they are joined by a curve (Fig. 4.12)
Fig. 4.12: Position-time graph of the vehicle
📝 Questions — Plotting Graphs
LOTS
On a position-time graph, which quantity is plotted on the X-axis and which on the Y-axis?
Show Answer
Time is plotted on the X-axis and position is plotted on the Y-axis.
Medium
Why is it important to choose a proper scale before plotting a graph?
Show Answer
A proper scale allows the data to be represented effectively and conveniently while making full use of the available graph paper space — without a good scale, the graph could be too cramped or too spread out to read clearly.
HOTS
In Example 4.5, the points for a vehicle’s position did not fall on a straight line. What does this tell you about the vehicle’s motion, and how is this different from the straight-line graph in Activity 4.3?
Show Answer
Since the points form a curve instead of a straight line, the vehicle’s velocity is not constant — it is speeding up over time (accelerated motion). In Activity 4.3, the straight-line graph showed the vehicle moving at a constant velocity, covering equal distances in equal time intervals.
HOTS
If you used a smaller scale (fewer divisions per unit) while plotting the same data, would the underlying motion of the object change? Explain.
Show Answer
No, the actual motion of the object does not change. Changing the scale only changes how the graph looks (how spread out or compressed it appears on paper) — it does not alter the real values of position, time, velocity, or acceleration of the object.
4.2.2 Position-time graphs
- The position-time graph represents the motion of an object, i.e., the change in its position with time
What does the shape of the position-time graph indicate about the nature of motion?
Fig. 4.13: Position-time graph of a moving object
- (a) In equal intervals of time (say, 2–3 s and 5–6 s), the magnitudes of displacements are equal (20 m) → magnitude of velocity is constant
- (b) In equal intervals of time (say, 4–6 s and 10–12 s), the magnitudes of displacements are increasing → magnitude of velocity is increasing
- A straight line position-time graph indicates the object is moving with a constant velocity
- A curved position-time graph indicates the velocity is not constant, and the object is in accelerated motion
Which physical quantities can be obtained from a position-time graph?
- From the position-time graph, we can find the position of an object at each instant of time
- We can also calculate how fast the position is changing, i.e., the magnitude of velocity of the object
📦 Activity 4.4: Let us calculate
- In the position-time graph plotted (Fig. 4.11c), consider a part (say, AB) as shown in Fig. 4.14. From A, draw a line parallel to X-axis and another line parallel to Y-axis. Repeat the same from B
- Extend the horizontal line from A — a triangle ABC is formed. BC represents the change in position (s₂ − s₁), and CA represents the change in time (t₂ − t₁)
- By dividing the change in position (BC) by the change in time (CA), we get the average velocity: v = (s₂ − s₁)/(t₂ − t₁) = BC/CA
- By extracting values from the graph, the magnitude of average velocity = (80 m − 40 m)/(4 s − 2 s) = 40 m / 2 s = 20 m s⁻¹
Fig. 4.14: Calculating velocity from a position-time graph
- Geometrically, BC/CA is called the slope of line AB — the slope of a line is the steepness of the line
- The slope of a graph gives information about the rate of change of the quantity on the Y-axis with respect to the quantity on the X-axis
📦 Ready to Go Beyond
- For the case of a curve also, the velocity at any instant can be calculated geometrically from the position-time graph
- This will be learnt in higher grades
Example 4.6: What does the graph shown in Fig. 4.15 indicate about the nature of motion of the vehicle?
Fig. 4.15: Position-time graph of a vehicle
Answer: (Think about what a horizontal line on a position-time graph means)
Show Answer
- The position of the vehicle is 40 m from the origin and is not changing with time
- The vehicle is at rest at 40 m from the origin
- A straight line parallel to the time axis on a position-time graph represents a stationary object (note: in this case, the position-time graph is not the distance-time graph)
Example 4.7: The position-time graphs of two objects A and B are given in Fig. 4.16a. The magnitude of average velocity of which object is higher?
Fig. 4.16(a): Position-time graph of two objects
Answer: (Compare the slopes of the two lines using Fig. 4.16b)
Show Answer
- By making lines parallel to axes as shown in Fig. 4.16b, the displacement of object B is more than object A for the same time interval
- The slope of the line for B is steeper than the slope for line A
- Thus, the velocity of B is higher than that of A
Fig. 4.16(b): Comparing displacement of two objects from position-time graph
📝 Questions — Position-Time Graphs
LOTS
What does the slope of a position-time graph represent?
Show Answer
The slope of a position-time graph represents the velocity of the object.
Medium
A position-time graph is a straight horizontal line parallel to the time axis. What does this tell you about the object’s motion?
Show Answer
The object is stationary (at rest) — its position is not changing with time, since the line shows zero slope.
HOTS
Two position-time graphs are both straight lines through the origin, but one is steeper than the other. What does the steeper line tell you about that object’s motion compared to the other?
Show Answer
The steeper line represents a higher velocity. Since both are straight lines, both objects move at constant velocity, but the object with the steeper slope covers more distance in the same time, meaning it is moving faster.
HOTS
Can a position-time graph ever show a vertical line (a line parallel to the position axis)? What would it physically mean if it did?
Show Answer
No, this is not physically possible. A vertical line would mean the object’s position changes instantly (zero time interval) — which would require infinite velocity. Since no real object can do this, a position-time graph can never be a vertical line.
4.2.3 Velocity-time graphs
- The velocity-time graph of an object in motion represents the change in its velocity with time
- Velocity-time graphs can be plotted following the steps of Activity 4.3, in a manner similar to the position-time graph
- Example: a car moving in the same direction on a straight stretch of highway at a steady velocity of 72 km h⁻¹ (20 m s⁻¹)
- Another car starting from rest, with velocity increasing as shown in Table 4.5
- Another car moving with a velocity of 15.0 m s⁻¹, decreasing with time as shown in Table 4.6
Table 4.5: Increasing velocity
- 0 s — 0 m s⁻¹
- 5 s — 2.5 m s⁻¹
- 10 s — 5.0 m s⁻¹
- 15 s — 7.5 m s⁻¹
- 20 s — 10.0 m s⁻¹
- 25 s — 12.5 m s⁻¹
- 30 s — 15.0 m s⁻¹
Table 4.6: Decreasing velocity
- 0 s — 15.0 m s⁻¹
- 5 s — 12.5 m s⁻¹
- 10 s — 10.0 m s⁻¹
- 15 s — 7.5 m s⁻¹
- 20 s — 5.0 m s⁻¹
- 25 s — 2.5 m s⁻¹
- 30 s — 0 m s⁻¹
What does the shape of the velocity-time graph indicate about the nature of motion?
Fig. 4.17: Velocity-time graph of a moving car
- (a) Velocity is not changing with time → velocity is constant → acceleration is zero
- (b) In equal intervals of time, velocity is increasing by equal amounts → acceleration is constant and in the direction of velocity
- (c) In equal intervals of time, velocity is decreasing by equal amounts → acceleration is constant and opposite to the direction of velocity
- When the velocity of the car is constant, the velocity-time graph is a straight line parallel to the X-axis, and acceleration is zero
- A straight line velocity-time graph sloping upward indicates velocity increasing with constant acceleration (in the direction of velocity)
- A straight line velocity-time graph sloping downward indicates velocity decreasing with constant acceleration (opposite to the direction of velocity)
Which physical quantities can be obtained from a velocity-time graph?
Slope of the straight line on graph
- Velocity can be determined from the slope of the straight line on a position-time graph
- The slope of the line on a velocity-time graph gives how fast the velocity is changing with time, i.e., the acceleration
- For the graph in Fig. 4.17a, there is no change in height of the line — slope is zero, so acceleration is zero
Fig. 4.17(d): Calculating acceleration from a velocity-time graph
- Consider a part of the line, AB. At time t₁ (point A), velocity = u. At time t₂ (point B), velocity = v
- By dividing the change in velocity (BC) by the change in time (CA): a = (v − u)/(t₂ − t₁) = BC/CA
- Substituting values from the graph, the magnitude of average acceleration between 10 s and 20 s: a = (10 − 5)/(20 − 10) = 5/10 = 0.5 m s⁻²
Fig. 4.17(e): Calculating acceleration from a velocity-time graph
- Following the same steps for Fig. 4.17e, the average acceleration = −0.5 m s⁻²
- The minus sign indicates the direction of acceleration is opposite to the direction of velocity (since velocity is decreasing)
Area enclosed by the line on graph and time axis
- The graphs in Fig. 4.18 are the same as Fig. 4.17, except some parts between the line and the time axis are shaded
- In Fig. 4.18a, OA is the magnitude of velocity and OC is the time interval
- Area of rectangle OABC = OA × OC = velocity × time interval (since velocity is constant here, it equals average velocity)
- Using Eq. (4.2a): area of rectangle OABC = displacement
Fig. 4.18(a): Velocity-time graph of an object moving with constant velocity
- Substituting values: displacement between 0 s and 6 s = area of OABC = 20 m s⁻¹ × 6 s = 120 m
- This is the displacement of the car from the origin in 6 seconds
- The area enclosed by the velocity-time graph and the time axis for a desired time interval is equal to the displacement in that time interval
- For an object moving with a constant acceleration (Fig. 4.18b), the displacement between 10 s to 20 s = area of ABDE = area of rectangle ACDE + area of triangle ABC
- = (CD × DE) + (½ × CA × BC)
- Substituting values: displacement = (5 m s⁻¹ × 10 s) + (½ × 10 s × 5 m s⁻¹) = 50 m + 25 m = 75 m
- This is the displacement of the car between 10 s and 20 s
Fig. 4.18(b): Velocity-time graph of an object moving with changing velocity with constant acceleration
- By finding the slope and area from a velocity-time graph for motion with constant acceleration, we can determine acceleration and displacement, respectively
📝 Questions — Velocity-Time Graphs
LOTS
What does the area enclosed between a velocity-time graph and the time axis represent?
Show Answer
It represents the displacement of the object during that time interval.
Medium
A velocity-time graph is a straight line sloping downward. What does this tell you about the object’s acceleration?
Show Answer
The object has a constant (uniform) acceleration, and it acts opposite to the direction of velocity — since the velocity is decreasing, the object is slowing down (decelerating).
HOTS
A velocity-time graph shows a car at 10 m/s for the first 5 seconds, then steadily increasing to 20 m/s over the next 5 seconds. Describe in words how you would calculate the total displacement using this graph, without doing the actual calculation.
Show Answer
Split the area under the graph into two parts: a rectangle for the first 5 seconds (constant velocity 10 m/s) and a trapezium or rectangle+triangle for the next 5 seconds (increasing velocity from 10 to 20 m/s). Calculate each area separately and add them together to get the total displacement.
HOTS
Why can the slope method (for finding acceleration) and the area method (for finding displacement) both be applied to the same velocity-time graph? What does this tell you about how much information a single graph can hold?
Show Answer
A velocity-time graph encodes two independent pieces of geometric information: its steepness (slope) and the space it encloses (area). These correspond to two different physical quantities — acceleration and displacement respectively — showing that a single, well-constructed graph can represent multiple physical quantities of motion at once.
4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration
- Let us consider the special case of motion with constant acceleration and derive equations for analysing such motions
- Since acceleration is constant, acceleration at each instant equals the average acceleration over any time interval
- Using the definition of average acceleration: a = (v − u) / t, where u is initial velocity at t = 0 s, v is final velocity at time t
- Rearranging: at = v − u
v = u + at (4.4a)
- This equation allows us to calculate velocity (v) at all times if initial velocity (u) and acceleration (a) are known
- The velocity-time graph of Eq. (4.4a) is shown in Fig. 4.19 — similar to Fig. 4.18 but now the initial velocity is not zero
- The initial velocity (AO) is u, and the final velocity at time t (EO) is v
- The graph is a straight line indicating velocity changes with constant acceleration; the slope gives the acceleration
Fig. 4.19: Velocity-time graph where the initial velocity of object is not zero
- The displacement s during time interval t is given by the area enclosed within OABD
- s = area of rectangle OACD + area of triangle ABC = (AO × DO) + (½ × CA × BC)
- Substituting AO = u, DO = CA = t, and BC = (v − u):
- s = ut + ½ × t × (v − u)
- Substituting (v − u) = at from Eq. (4.4a):
s = ut + ½at² (4.4b)
- Eq. (4.4a) and Eq. (4.4b) are the primary equations. Combining them in different ways gives three more equations
- Deriving one by eliminating t in Eq. (4.4b):
- From Eq. (4.4a): t = (v − u)/a
- Substituting into Eq. (4.4b) and simplifying step by step:
- s = u[(v−u)/a] + ½ × a × [(v−u)/a]²
- s = (uv − u²)/a + (u² + v² − 2uv)/2a
- s = (2uv − 2u² + u² + v² − 2uv) / 2a
- s = (−u² + v²) / 2a
- 2as = −u² + v²
v² = u² + 2as (4.4c)
- For the motion of an object in a straight line with constant acceleration, the five physical quantities — displacement (s), time interval (t), initial velocity (u), final velocity (v) and acceleration (a) — are related by the following set of equations:
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ut + ½at²
v² = u² + 2as
- These are known as kinematic equations. They provide a mathematical description of how the motion of an object changes with time
- Using these equations, it is possible to predict the position or velocity of the object at a future time
Note:
- These kinematic equations are valid only when the acceleration is constant
- For motion in a straight line in one direction, distance travelled = magnitude of displacement, and speed = magnitude of velocity
- For motion in both directions, the sign of u, v, a, and s in these equations tells us about the direction of that particular quantity
Example 4.8: Suppose a car is moving on a highway and brakes are applied, which cause an acceleration of −4 m s⁻². How much will be the distance travelled by the car before coming to a stop, if the car was moving with a velocity of (i) 54 km h⁻¹, and (ii) 108 km h⁻¹ when the brakes were applied?
Answer: (Given: a = −4 m s⁻², v = 0 m s⁻¹. Use v² = u² + 2as to find s)
Show Answer
- (i) u = 54 km h⁻¹ = 15 m s⁻¹
- (ii) u = 108 km h⁻¹ = 30 m s⁻¹
- Using Eq. (4.4c): v² = u² + 2as → 0 = u² − 8s → s = u²/8
- Substituting values: (i) s = 28.1 m, and (ii) s = 112.5 m
🌍 Bridging Science and Society
- When brakes are applied to a moving vehicle, it moves some distance before coming to a stop
- The distance travelled depends upon: the velocity of the vehicle when brakes were applied, the road surface (wet/dry, etc.), the braking capacity of the vehicle (negative acceleration caused by brakes), and the driver’s reaction time
- This is why it is important to maintain a safe distance from the vehicle moving ahead, adjusted according to your initial velocity
- There is a vehicle-to-vehicle (V2V) communication technology, now being developed in many countries including India, which allows vehicles to exchange signals and warns drivers of possible collisions
Fig. 4.20: Safe distance between two moving vehicles
📝 Questions — Kinematic Equations
LOTS
Write the three kinematic equations of motion for constant acceleration.
Show Answer
v = u + at s = ut + ½at² v² = u² + 2as
Medium
A bike starts from rest and accelerates at 2 m/s² for 4 seconds. Find its final velocity.
Show Answer
Using v = u + at: v = 0 + (2 × 4) = 8 m/s.
HOTS
Why are the kinematic equations valid only for constant acceleration, and not for an object whose acceleration keeps changing?
Show Answer
These equations were derived assuming acceleration at each instant equals the average acceleration over the whole time interval — this is only true when acceleration is constant. If acceleration changes with time, the average over an interval would not represent the value at every instant, so the equations would give incorrect results.
HOTS
A driver doubles their car’s speed before braking with the same deceleration as before. Using v² = u² + 2as, explain how the stopping distance changes — does it also just double?
Show Answer
No, it does not just double — it quadruples. Since 0 = u² − 2as gives s = u²/2a, stopping distance is proportional to the square of the initial velocity (u²). So doubling u makes s four times as large, not twice. This is why high speeds are far more dangerous when braking.
- Till now, we have discussed motion in a straight line, also called motion in one dimension. Let us now explore motion in a plane
4.4 Motion in a Plane
- Motion in a plane — such as a vehicle overtaking another, the path of a kicked ball, or a satellite moving in a circular path — is called motion in two dimensions
Fig. 4.21: Motion in a plane
4.4.1 Uniform circular motion
- When an object moves in a circular path, its motion is called circular motion
- Example: a child sitting on a moving merry-go-round, moving on a circular path from A to B to C
- The distance travelled by the child is the curve ABC; the displacement is the straight line AC — these are not equal
Fig. 4.22: Top view of merry-go-round in circular motion
- Distance travelled in making one revolution (going round the circle once) = circumference of the circle
- If the radius of the circular path is R, distance travelled in one revolution = 2πR
- Displacement is zero after one revolution, since the child returns to its original position
- If an object takes time T to make one revolution, its average speed vav will be:
vav = 2πR / T (4.5)
- The average velocity during the time interval T will be 0, since the displacement is 0
Note:
- The average speed (Eq. 4.5) was calculated for one revolution, but for uniform circular motion, since the speed is constant, its value is the same at every point on the circle
- When an object moves in a circular path with constant (uniform) speed, its motion is called uniform circular motion
- In case of uniform circular motion, the speed is constant — but what about the direction of velocity at an instant? Is it changing?
- Imagine an athlete running along a closed path:
- On a rectangular track, the athlete changes direction 4 times to complete one round
- On a hexagonal track, the athlete changes direction 6 times to complete one round
- As the number of sides increases indefinitely, the athlete has to take turns more and more frequently
- Finally, the track approaches a circle, with each side decreasing to a point and the direction of the athlete’s velocity changing continuously
Fig. 4.23: An athlete running along (a) a rectangular track, (b) a hexagonal track, and (c) a circular track
📦 Activity 4.5: Let us investigate
- Take a ring, such as an adhesive tape ring, and one marble
- Place the ring flat on a smooth surface and throw the marble inside the ring in a way that it rotates along the inner boundary of the ring (Fig. 4.24)
- Predict what will happen if you lift the ring while the marble is moving
- After one or two complete revolutions of the marble, pick up the ring without disturbing the motion of the marble. What do you observe? Does the marble continue moving in a circular motion, or does it move in some other manner?
- Repeat the activity multiple times to confirm the result
Fig. 4.24: A marble moving inside a ring
- When the marble is released by lifting the ring, it moves in a straight line
- This happens because once the marble is released, it continues to move in the direction it has been moving at the instant the ring was removed (the reason for this will be learnt in a later chapter)
📦 Ready to Go Beyond
- The velocity at a point is along the tangent to the circle at that point, in the direction of motion
- A straight line that meets the circle at one and only one point is called a tangent to the circle at that point
Fig. 4.25: Tangent to a circle at point B
Note:
- In uniform circular motion, the speed is constant at every point on the circle — it is only the direction of velocity that changes
- Acceleration is non-zero if the velocity of an object changes — velocity changes if either its magnitude or direction, or both, changes
- In uniform circular motion, the motion is accelerated because the direction of its velocity continuously changes
Note:
- In everyday life, we say a vehicle is accelerating when the magnitude of its velocity is changing — but we often fail to recognise that there can be acceleration when there is only a change in the direction of velocity
📦 Ready to Go Beyond
- Motion in space — such as a car climbing up a mountain road, a bird flying in the sky, or an aircraft moving through air — is called motion in three dimensions
- This will be learnt in higher grades
Fig. 4.26: A mountain road
- In the real world, the conditions for uniform circular motion — constant speed and a circular path — are often not met
- So, uniform circular motion is an idealised model of real-world situations
- Still, it is a useful model, serving as the foundation for more complex real-world situations, such as the motion of planets revolving around the Sun, or a vehicle making a circular turn
📝 Questions — Uniform Circular Motion
LOTS
What is the displacement of an object after it completes one full revolution in a circular path?
Show Answer
The displacement is zero, since the object returns to its original starting position.
Medium
A wheel of radius 7 m completes one revolution in 4 seconds. Find its average speed. (Use π ≈ 22/7)
Show Answer
vav = 2πR/T = (2 × 22/7 × 7)/4 = 44/4 = 11 m/s.
HOTS
A student says, “If the speed is constant in circular motion, the object cannot be accelerating.” Is the student correct? Explain using the concept of velocity.
Show Answer
The student is incorrect. Velocity has both magnitude (speed) and direction. Even though speed stays constant in uniform circular motion, the direction of velocity changes continuously as the object moves around the circle. Since velocity is changing (in direction), the object is accelerating.
HOTS
In Activity 4.5, when the ring is lifted and the marble is released, it moves in a straight line instead of continuing in a circle. What does this tell you about what was causing the marble to move in a circle in the first place?
Show Answer
This shows that the marble’s circular path was only maintained because the ring’s inner boundary was continuously pushing the marble inward, changing the direction of its velocity. Once that boundary (the ring) is removed, there is nothing to keep changing the marble’s direction, so it continues moving in a straight line in whatever direction it was moving at that instant.
📋 At a Glance
- The distance and direction of an object with respect to the reference point, at any instant of time, describes the position of the object at that instant of time
- If the position of the object with respect to a reference point changes with time, the object is said to be in motion
- Displacement is the net change in the position of the object between two given instants of time
- The average speed of an object is the total distance travelled divided by the time interval during which this distance is covered
- The average velocity of an object in a time interval is the change in position (also known as displacement) divided by the time interval in which the change in position (or displacement) occurs
- The average acceleration of an object over a time interval is the change in its velocity divided by the time interval
- For the motion of an object in a straight line with constant acceleration, the five physical quantities — displacement (s), time interval (t), initial velocity (u), final velocity (v) and acceleration (a) — can be related by the kinematic equations: v = u + at; s = ut + ½at²; v² = u² + 2as
- When an object moves in a circular path with constant (uniform) speed, its motion is called the uniform circular motion
Revise, Reflect, Refine
- 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?
- 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) their displacement from the starting point.
- 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?
- A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
- A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
- Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
Fig. 4.27 - A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
- (i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions
- (ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time
- (iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds
- (iv) The average speed of A over the 10 s time interval is greater than that of B since B’s speed is lower than A’s in some segments
Fig. 4.28 - A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
Fig. 4.29 - A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
- A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle?
- A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.
- The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity, and (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.
Fig. 4.30 - A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
Fig. 4.31 - On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
- Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).
- Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity. The length of the minute’s hand is 7 cm (Fig. 4.32).
Fig. 4.32
🚀 The Journey Beyond
- Take a cardboard disc (radius ~ 8 cm) (Fig. 4.33). Write numbers 1 to 12 on the outer part (7 cm from the centre) and the letters ‘ABCDEF’ on the inner part (4 cm from the centre), using the same font size. Spin the disc slowly, then faster, and observe how the numbers and letters appear. Why do the numbers fade or disappear while the letters remain visible? Are the speeds of the numbers and letters the same or different? Justify your answer.
Fig. 4.33 - Many smartphones have an inbuilt accelerometer that can detect very small accelerations. Install an app, such as Phyphox (phyphox.org) and open ‘Accelerometer (without g)’. Note the readings when (i) the phone is on an outstretched palm, and (ii) the phone is kept on the floor. What differences do you observe? What does this tell you about motion and acceleration in real situations? (Such tiny, involuntary movements are also studied in medical research, for example, in movement disorders) This activity is recommended to be performed as a classroom group activity facilitated by teacher.
- For motion in a straight line with constant acceleration, we derived two primary equations given by Eq. (4.4a) and (4.4b). Using these two equations, three more equations can be derived, out of which we derived one given in Eq. (4.4c). Derive the remaining two equations given below:
s = vt − ½at² s = ½(u + v)tIn mathematics, you have learnt the formula for calculating the area of a trapezium. Using that formula, derive the second equation given above.
- Plot graphs for data given in Table 4.4, using different X and Y scales, on different graph papers. Compare the graphs to find how the appearance of graph is affected by the choice of scales and decide which scale is better and why. Now repeat this with any graph plotting app. Such apps generally automatically adjust the axes to fit the data well on the screen.
- Talk to a motor mechanic about how a vehicle’s braking or stopping distance is affected by: (i) wet roads, (ii) worn-out tyres, (iii) higher vehicle mass, (iv) driving at night, (v) fog, (vi) severe weather (rain, snow, storm), and (vii) driver reaction time. Using this information, design safety posters for your school and prepare a short skit to present it in the assembly.