This chapter introduces the concept of motion, explaining how to describe motion using distance, displacement, speed, velocity, and acceleration. It covers uniform and non-uniform motion, graphical representation of motion (distance-time and velocity-time graphs), equations of motion for uniformly accelerated motion, and uniform circular motion. Understanding motion helps us describe and predict the movement of objects in our daily life and in the universe.
Introduction
In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries, and cars move. Atoms, molecules, planets, stars, and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time. However, there are situations where motion is inferred through indirect evidences. For example, we infer the motion of air by observing the movement of dust and leaves.
An object may appear to be moving for one person and stationary for another. For passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the roadside perceives the bus along with the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate.
7.1 Describing Motion
We describe the location of an object by specifying a reference point. For example, if a school in a village is 2 km north of the railway station, we have specified the position of the school with respect to the railway station. In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin.
7.1.1 MOTION ALONG A STRAIGHT LINE
The simplest type of motion is the motion along a straight line. Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point. Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B.
Example:
- Position O = 0 km
- Position C = 25 km
- Position B = 35 km
- Position A = 60 km
Total path length covered by the object is OA + AC = 60 km + 35 km = 95 km. This is the distance covered by the object.
Distance
Distance is the total path length covered by an object. To describe distance we need to specify only the numerical value and not the direction of motion. The numerical value of a physical quantity is its magnitude.
Displacement
The shortest distance measured from the initial to the final position of an object is known as the displacement.
In the above example:
- Distance of final position C from initial position O = 35 km (this is the displacement).
- But the total distance covered = 95 km.
Difference Between Distance and Displacement
Case 1: Motion from O to A
- Distance covered = 60 km
- Magnitude of displacement = 60 km (equal)
Case 2: Motion from O to A and back to B
- Distance covered = 60 km + 25 km = 85 km
- Magnitude of displacement = 35 km (from O to B)
- Thus, magnitude of displacement (35 km) is not equal to path length (85 km).
Case 3: Motion from O to A and back to O
- Final position coincides with initial position.
- Distance covered = 60 km + 60 km = 120 km
- Displacement = 0 (starting and ending at the same point)
Important: The magnitude of displacement for a course of motion may be zero but the corresponding distance covered is not zero.
Questions (7.1.1) and Solutions
Q1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Solution:
Yes, an object can have zero displacement even though it has moved through a distance.
Example:
- If a person walks from point A to point B and then returns back to point A, the distance covered = AB + BA (not zero).
- But the displacement = 0 because the final position is the same as the initial position.
- Another example: An athlete running one complete round on a circular track covers a distance equal to the circumference of the track, but displacement = 0.
Q2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Solution:
- Side of square field = 10 m
- Perimeter of square = 4 × 10 m = 40 m
- Time to complete one round = 40 s
- Total time = 2 minutes 20 seconds = 2 × 60 + 20 = 140 s
Number of rounds completed:
Number of rounds = 140 s ÷ 40 s = 3.5 rounds
After 3 complete rounds, the farmer returns to the starting position. Then he completes 0.5 round (half round).
After half round, the farmer is at the opposite corner of the square (diagonally opposite to starting point).
Displacement = Diagonal of the square
Using Pythagoras theorem:
Diagonal = √(10² + 10²) = √(100 + 100) = √200 = 10√2 m
= 10 × 1.414 = 14.14 m
Answer: The magnitude of displacement = 14.14 m or 10√2 m.
Q3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Solution:
Both statements are FALSE.
Explanation:
- Statement (a) is false because displacement can be zero when an object returns to its initial position.
- Statement (b) is false because the magnitude of displacement is always less than or equal to the distance travelled, never greater.
Correct statement: The magnitude of displacement can be zero, and it is always less than or equal to the distance travelled.
7.1.2 UNIFORM MOTION AND NON-UNIFORM MOTION
Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second, and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion.
In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example:
- A car moving on a crowded street
- A person jogging in a park
These are instances of non-uniform motion.
Table: Motion of Two Objects A and B
| Time | Distance travelled by Object A (m) | Distance travelled by Object B (m) |
|---|---|---|
| 9:30 am | 10 | 20 |
| 9:45 am | 20 | 19 |
| 10:00 am | 30 | 40 |
| 10:15 am | 40 | 23 |
| 10:30 am | 50 | 35 |
| 10:45 am | 60 | 37 |
| 11:00 am | 70 | 41 |
Observation:
- Object A: Covers approximately equal distances in equal time intervals → Uniform motion
- Object B: Covers unequal distances in equal time intervals → Non-uniform motion
7.2 Measuring the Rate of Motion
Different objects may take different amounts of time to cover a given distance. Some move fast and some move slowly. The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed.
Speed
Definition: Speed is the distance travelled by an object in unit time.
Formula:
textSpeed (v) = Distance (s) / Time (t)
SI Unit: metre per second (m s⁻¹ or m/s)
Other units:
- centimetre per second (cm s⁻¹)
- kilometre per hour (km h⁻¹)
To specify the speed of an object, we require only its magnitude (not direction).
Average Speed
The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed.
Formula:
textAverage speed = Total distance travelled / Total time taken
Example: A car travels a distance of 100 km in 2 h. Its average speed is 50 km h⁻¹. The car might not have travelled at 50 km h⁻¹ all the time. Sometimes it might have travelled faster and sometimes slower.
Example 7.1
An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
Solution:
Total distance travelled by the object = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed = Total distance travelled / Total time taken
= 32 m / 6 s
= 5.33 m s⁻¹
Answer: The average speed of the object is 5.33 m s⁻¹.
7.2.1 SPEED WITH DIRECTION
The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity.
Definition: Velocity is the speed of an object moving in a definite direction.
The velocity of an object can be uniform or variable. It can be changed by:
- Changing the object’s speed
- Changing the direction of motion
- Changing both
Average Velocity
When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity.
Formula (for uniformly changing velocity):
textAverage velocity (v_av) = (Initial velocity (u) + Final velocity (v)) / 2
Mathematically:
v_av = (u + v) / 2
SI Unit: metre per second (m s⁻¹ or m/s) – same as speed.
Example 7.2
A car travels 400 km in 8 hours. Calculate its average speed.
Solution:
Distance (s) = 400 km
Time (t) = 8 h
Average speed (v) = s / t = 400 km / 8 h = 50 km h⁻¹
Converting to m s⁻¹:
50 km/h = 50 × (1000 m / 3600 s) = 50000 / 3600 = 13.9 m s⁻¹
Answer: The average speed of the car is 50 km h⁻¹ or 13.9 m s⁻¹.
Example 7.3
Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution:
Total distance covered by Usha in 1 min = 180 m
Displacement of Usha in 1 min = 0 m (returns to starting point)
Total time = 1 min = 60 s
Average speed:
Average speed = Total distance / Total time
= 180 m / 60 s
= 3 m s⁻¹
Average velocity:
Average velocity = Displacement / Total time
= 0 m / 60 s
= 0 m s⁻¹
Answer:
- Average speed = 3 m s⁻¹
- Average velocity = 0 m s⁻¹
Questions (7.2) and Solutions
Q1. Distinguish between speed and velocity.
Solution:
| Speed | Velocity |
|---|---|
| Distance travelled per unit time | Displacement per unit time |
| Scalar quantity (has only magnitude) | Vector quantity (has both magnitude and direction) |
| Always positive | Can be positive, negative, or zero |
| Does not depend on direction | Depends on direction of motion |
| Example: A car travels at 60 km/h | Example: A car travels at 60 km/h towards north |
Q2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Solution:
The magnitude of average velocity is equal to average speed when:
- The object moves along a straight line in a single direction without changing direction.
- In other words, when displacement = distance travelled.
Example: A car moving on a straight road from point A to point B without reversing direction.
Q3. What does the odometer of an automobile measure?
Solution:
An odometer measures the total distance travelled by an automobile. It shows the cumulative distance covered from the time the vehicle was manufactured or from when the odometer was last reset. It does not measure displacement.
Q4. What does the path of an object look like when it is in uniform motion?
Solution:
When an object is in uniform motion:
- It covers equal distances in equal intervals of time.
- The path can be of any shape (straight line, circular, curved).
- On a distance-time graph, uniform motion is represented by a straight line.
For motion along a straight line with uniform speed: The path is a straight line.
Q5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light (3 × 10⁸ m s⁻¹).
Solution:
Speed of signal (speed of light) = 3 × 10⁸ m s⁻¹
Time taken = 5 minutes = 5 × 60 = 300 s
Using the formula:
textSpeed = Distance / Time
Distance = Speed × Time
= (3 × 10⁸ m s⁻¹) × (300 s)
= 9 × 10¹⁰ m
= 90,000,000,000 m
= 9 × 10⁷ km
Answer: The distance of the spaceship from the ground station was 9 × 10¹⁰ m or 9 × 10⁷ km (90 million km).
7.3 Rate of Change of Velocity
During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity for any time interval is zero.
However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity during any time interval is not zero.
To express the change in velocity of an object, we introduce another physical quantity called acceleration.
Acceleration
Definition: Acceleration is the measure of the change in velocity of an object per unit time.
Formula:
Acceleration (a) = Change in velocity / Time taken
= (Final velocity - Initial velocity) / Time
= (v - u) / t
SI Unit: metre per second squared (m s⁻²)
Types:
- Positive acceleration: When velocity increases (acceleration is in the direction of velocity).
- Negative acceleration (Retardation/Deceleration): When velocity decreases (acceleration is opposite to the direction of velocity).
Uniform and Non-Uniform Acceleration
Uniform Acceleration:
- If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, the acceleration is said to be uniform.
- Example: A freely falling body.
Non-Uniform Acceleration:
- If the velocity changes at a non-uniform rate, the acceleration is non-uniform.
- Example: A car moving on a crowded street where speed changes irregularly.
Example 7.4
Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s⁻¹ in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s⁻¹ in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
Solution:
In the first case:
- Initial velocity (u) = 0 m s⁻¹ (starting from rest)
- Final velocity (v) = 6 m s⁻¹
- Time (t) = 30 s
Using formula:
texta = (v - u) / t = (6 - 0) / 30 = 6 / 30 = 0.2 m s⁻²
In the second case:
- Initial velocity (u) = 6 m s⁻¹
- Final velocity (v) = 4 m s⁻¹
- Time (t) = 5 s
a = (v - u) / t = (4 - 6) / 5 = -2 / 5 = -0.4 m s⁻²
Answer:
- Acceleration in the first case = +0.2 m s⁻² (positive, velocity increasing)
- Acceleration in the second case = -0.4 m s⁻² (negative, velocity decreasing)
Questions (7.3) and Solutions
Q1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Solution:
(i) Uniform Acceleration:
A body is in uniform acceleration when:
- Its velocity changes by equal amounts in equal intervals of time.
- The rate of change of velocity is constant.
- Example: A freely falling body under gravity (neglecting air resistance).
(ii) Non-Uniform Acceleration:
A body is in non-uniform acceleration when:
- Its velocity changes by unequal amounts in equal intervals of time.
- The rate of change of velocity is not constant.
- Example: A car moving on a crowded road where the driver frequently changes speed.
Q2. A bus decreases its speed from 80 km h⁻¹ to 60 km h⁻¹ in 5 s. Find the acceleration of the bus.
Solution:
Initial velocity (u) = 80 km h⁻¹ = 80 × (5/18) = 22.22 m s⁻¹
Final velocity (v) = 60 km h⁻¹ = 60 × (5/18) = 16.67 m s⁻¹
Time (t) = 5 s
Using formula:
texta = (v - u) / t
= (16.67 - 22.22) / 5
= -5.55 / 5
= -1.11 m s⁻²
Answer: The acceleration of the bus is -1.11 m s⁻² (negative sign indicates deceleration or retardation).
Q3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h⁻¹ in 10 minutes. Find its acceleration.
Solution:
Initial velocity (u) = 0 (starting from rest)
Final velocity (v) = 40 km h⁻¹ = 40 × (5/18) = 11.11 m s⁻¹
Time (t) = 10 minutes = 10 × 60 = 600 s
Using formula:
a = (v - u) / t
= (11.11 - 0) / 600
= 11.11 / 600
= 0.0185 m s⁻²
Answer: The acceleration of the train is 0.0185 m s⁻² or approximately 0.02 m s⁻².
7.4 Graphical Representation of Motion
Graphs provide a convenient method to present basic information about a variety of events. To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity (such as distance or velocity) on another quantity (such as time).
7.4.1 DISTANCE-TIME GRAPHS
The change in the position of an object with time can be represented on the distance-time graph. In this graph:
- Time is taken along the x-axis
- Distance is taken along the y-axis
Distance-time graphs can be employed for objects moving with uniform speed, non-uniform speed, or at rest.
Graph for Uniform Motion
When an object travels equal distances in equal intervals of time, it moves with uniform speed. The distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, the distance-time graph is a straight line.
Determining Speed from Graph:
To determine speed from a distance-time graph:
- Consider a small part AB of the graph
- AC represents the time interval (t₂ – t₁)
- BC represents the distance (s₂ – s₁)
Speed (v):
textv = (s₂ - s₁) / (t₂ - t₁)
This is the slope of the distance-time graph.
Graph for Non-Uniform Motion
When an object covers unequal distances in equal intervals of time, it is in non-uniform motion. The distance-time graph for non-uniform motion is a curved line (not a straight line).
7.4.2 VELOCITY-TIME GRAPHS
The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph:
- Time is represented along the x-axis
- Velocity is represented along the y-axis
Graph for Uniform Velocity
If the object moves at uniform velocity, the height of the velocity-time graph will not change with time. It will be a straight line parallel to the x-axis.
Finding Distance from Velocity-Time Graph:
The area enclosed by the velocity-time graph and the time axis equals the magnitude of displacement (or distance for straight-line motion).
For uniform velocity:
Distance (s) = Velocity × Time
= Area of rectangle under the graph
Graph for Uniformly Accelerated Motion
For uniformly accelerated motion, the velocity-time graph is a straight line (not parallel to x-axis).
Finding Distance:
The distance travelled is given by the area under the velocity-time graph.
For a velocity-time graph showing uniform acceleration:
Distance (s) = Area of rectangle + Area of triangle
= (u × t) + (1/2 × a × t²)
where u = initial velocity, a = acceleration, t = time
Graph for Non-Uniform Acceleration
In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape (curved lines). The distance travelled is still the area under the curve.
Questions (7.4) and Solutions
Q1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Solution:
For Uniform Motion:
- The distance-time graph is a straight line.
- The slope of the line represents the constant speed.
For Non-Uniform Motion:
- The distance-time graph is a curved line.
- The slope changes at different points, indicating changing speed.
Q2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Solution:
If the distance-time graph is a straight line parallel to the time axis:
- The distance remains constant while time increases.
- This means the object is at rest (not moving).
- The speed of the object = 0.
Q3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Solution:
If the speed-time graph is a straight line parallel to the time axis:
- The speed remains constant while time increases.
- This means the object is moving with uniform speed (uniform motion).
- The acceleration of the object = 0.
Q4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Solution:
The area occupied below (under) the velocity-time graph represents the displacement (or distance travelled for motion in a straight line).
Formula:
Displacement = Area under velocity-time graph
7.5 Equations of Motion
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration, and distance covered by a set of equations known as the equations of motion.
Three Equations of Motion
1. First Equation of Motion (Velocity-Time Relation):
v = u + at
2. Second Equation of Motion (Position-Time Relation):
s = ut + (1/2)at²
3. Third Equation of Motion (Position-Velocity Relation):
2as = v² - u²
or
v² = u² + 2as
Where:
- u = initial velocity
- v = final velocity
- a = uniform acceleration
- t = time taken
- s = distance travelled (or displacement)
Example 7.5
A train starting from rest attains a velocity of 72 km h⁻¹ in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Solution:
Given:
- u = 0 (starting from rest)
- v = 72 km h⁻¹ = 72 × (5/18) = 20 m s⁻¹
- t = 5 minutes = 5 × 60 = 300 s
(i) Acceleration:
Using first equation: v = u + at
20 = 0 + a × 300
a = 20 / 300 = 1/15 m s⁻²
a = 0.067 m s⁻²
(ii) Distance travelled:
Using third equation: v² = u² + 2as
(20)² = (0)² + 2 × (1/15) × s
400 = (2/15) × s
s = 400 × 15 / 2 = 3000 m = 3 km
Answer:
- Acceleration = 1/15 m s⁻² or 0.067 m s⁻²
- Distance travelled = 3 km
Example 7.6
A car accelerates uniformly from 18 km h⁻¹ to 36 km h⁻¹ in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Solution:
Given:
- u = 18 km h⁻¹ = 18 × (5/18) = 5 m s⁻¹
- v = 36 km h⁻¹ = 36 × (5/18) = 10 m s⁻¹
- t = 5 s
(i) Acceleration:
a = (v - u) / t = (10 - 5) / 5 = 5 / 5 = 1 m s⁻²
(ii) Distance covered:
Using second equation: s = ut + (1/2)at²
s = 5 × 5 + (1/2) × 1 × (5)²
s = 25 + (1/2) × 25
s = 25 + 12.5 = 37.5 m
Answer:
- Acceleration = 1 m s⁻²
- Distance covered = 37.5 m
Example 7.7
The brakes applied to a car produce an acceleration of 6 m s⁻² in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Solution:
Given:
- a = -6 m s⁻² (negative because it’s opposite to motion)
- t = 2 s
- v = 0 m s⁻¹ (car stops)
Finding initial velocity (u):
Using first equation: v = u + at
0 = u + (-6) × 2
0 = u - 12
u = 12 m s⁻¹
Finding distance (s):
Using second equation: s = ut + (1/2)at²
s = 12 × 2 + (1/2) × (-6) × (2)²
s = 24 + (1/2) × (-6) × 4
s = 24 - 12 = 12 m
Answer: The car will move 12 m before it stops after the application of brakes.
Note: This is why drivers are cautioned to maintain some distance between vehicles while travelling on the road.
Questions (7.5) and Solutions
Q1. A bus starting from rest moves with a uniform acceleration of 0.1 m s⁻² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
Given:
- u = 0 (starting from rest)
- a = 0.1 m s⁻²
- t = 2 minutes = 2 × 60 = 120 s
(a) Speed acquired (v):
Using v = u + at
v = 0 + 0.1 × 120 = 12 m s⁻¹
(b) Distance travelled (s):
Using s = ut + (1/2)at²
s = 0 × 120 + (1/2) × 0.1 × (120)²
s = 0 + 0.05 × 14400 = 720 m
Answer:
- Speed acquired = 12 m s⁻¹
- Distance travelled = 720 m
Q2. A train is travelling at a speed of 90 km h⁻¹. Brakes are applied so as to produce a uniform acceleration of -0.5 m s⁻². Find how far the train will go before it is brought to rest.
Solution:
Given:
- u = 90 km h⁻¹ = 90 × (5/18) = 25 m s⁻¹
- v = 0 (train comes to rest)
- a = -0.5 m s⁻²
Using v² = u² + 2as
(0)² = (25)² + 2 × (-0.5) × s
0 = 625 - 1 × s
s = 625 m
Answer: The train will travel 625 m before coming to rest.
Q3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s⁻². What will be its velocity 3 s after the start?
Solution:
Given:
- u = 0 (starts from rest)
- a = 2 cm s⁻²
- t = 3 s
Using v = u + at
v = 0 + 2 × 3 = 6 cm s⁻¹
Answer: The velocity after 3 s will be 6 cm s⁻¹.
Q4. A racing car has a uniform acceleration of 4 m s⁻². What distance will it cover in 10 s after start?
Solution:
Given:
- u = 0 (starts from rest)
- a = 4 m s⁻²
- t = 10 s
Using s = ut + (1/2)at²
s = 0 × 10 + (1/2) × 4 × (10)²
s = 0 + 2 × 100 = 200 m
Answer: The distance covered will be 200 m.
Q5. A stone is thrown in a vertically upward direction with a velocity of 5 m s⁻¹. If the acceleration of the stone during its motion is 10 m s⁻² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
Given:
- u = 5 m s⁻¹ (upward)
- a = -10 m s⁻² (downward, so negative)
- v = 0 (at maximum height, velocity becomes zero)
Finding height (s):
Using v² = u² + 2as
(0)² = (5)² + 2 × (-10) × s
0 = 25 - 20s
20s = 25
s = 25/20 = 1.25 m
Finding time (t):
Using v = u + at
0 = 5 + (-10) × t
10t = 5
t = 5/10 = 0.5 s
Answer:
- Height attained = 1.25 m
- Time taken = 0.5 s
7.6 Uniform Circular Motion
When the velocity of an object changes, we say that the object is accelerating. The change in velocity could be due to:
- Change in its magnitude
- Change in its direction of motion
- Change in both
Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion?
Understanding Through an Example
Consider an athlete running along different tracks:
1. Rectangular Track:
- The athlete has to change direction 4 times to complete one round (at each corner).
2. Hexagonal Track:
- The athlete has to change direction 6 times to complete one round.
3. Octagonal Track:
- The athlete has to change direction 8 times to complete one round.
As the number of sides increases, the athlete has to take turns more often. If we keep increasing the number of sides indefinitely, the track approaches the shape of a circle and the length of each side decreases to a point.
If the athlete moves with a velocity of constant magnitude along the circular path, the only change in velocity is due to the change in direction of motion. Therefore, the motion of the athlete is an example of accelerated motion even though the speed remains constant.
Definition
Uniform Circular Motion: When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
Speed in Circular Motion
The circumference of a circle of radius r is given by 2πr.
If an object takes t seconds to go once around the circular path of radius r, the speed v is given by:
v = 2πr / t
Important Points
Demonstration Activity:
- Take a piece of thread and tie a small stone at one end.
- Move the stone to describe a circular path with constant speed by holding the thread at the other end.
- Now, let the stone go by releasing the thread.
- Observation: The stone moves along a straight line tangential to the circular path.
- Reason: Once released, the stone continues to move along the direction it was moving at that instant.
This shows that the direction of motion changed at every point when the stone was moving along the circular path.
Examples of Uniform Circular Motion
- Motion of the moon around the earth
- Motion of the earth around the sun
- A satellite in a circular orbit around the earth
- A cyclist on a circular track at constant speed
- An athlete throwing a hammer or discus (circular motion before release)
End-of-Chapter Exercises – Questions and Solutions
Q1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
Given:
- Diameter = 200 m, so Radius (r) = 100 m
- Circumference = 2πr = 2 × (22/7) × 100 = 628.57 m
- Time for one round = 40 s
- Total time = 2 min 20 s = 140 s
Number of rounds:
Number of rounds = 140 / 40 = 3.5 rounds
Distance covered:
Distance = 3.5 × 628.57 = 2200 m
Displacement:
- After 3 complete rounds, the athlete returns to the starting point.
- Then he completes 0.5 round (half round).
- After half round, the athlete is at the diametrically opposite point.
- Displacement = Diameter = 200 m
Answer:
- Distance covered = 2200 m or 2.2 km
- Displacement = 200 m
Q2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution:
(a) From A to B:
Distance = 300 m
Time = 2 min 30 s = 150 s
Displacement = 300 m (straight line from A to B)
Average speed:
Average speed = 300 / 150 = 2 m s⁻¹
Average velocity:
Average velocity = 300 / 150 = 2 m s⁻¹
(b) From A to C:
Total distance = 300 m (A to B) + 100 m (B to C) = 400 m
Total time = 150 s + 60 s = 210 s
Displacement = 300 – 100 = 200 m (from A towards B)
Average speed:
Average speed = 400 / 210 = 1.90 m s⁻¹
Average velocity:
Average velocity = 200 / 210 = 0.95 m s⁻¹
Answer:
- (a) Average speed = 2 m s⁻¹, Average velocity = 2 m s⁻¹
- (b) Average speed = 1.90 m s⁻¹, Average velocity = 0.95 m s⁻¹
Q3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h⁻¹. On his return trip along the same route, there is less traffic and the average speed is 30 km h⁻¹. What is the average speed for Abdul’s trip?
Solution:
Let the distance between home and school = d km
Time for onward journey:
textt₁ = d / 20 hours
Time for return journey:
t₂ = d / 30 hours
Total distance = 2d
Total time = t₁ + t₂ = d/20 + d/30
Total time = (3d + 2d) / 60 = 5d / 60 = d / 12 hours
Average speed:
Average speed = Total distance / Total time
= 2d / (d/12)
= 2d × 12 / d
= 24 km h⁻¹
Answer: Average speed for Abdul’s trip = 24 km h⁻¹
Q4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s⁻² for 8.0 s. How far does the boat travel during this time?
Solution:
Given:
- u = 0 (starting from rest)
- a = 3.0 m s⁻²
- t = 8.0 s
Using s = ut + (1/2)at²
s = 0 × 8 + (1/2) × 3 × (8)²
s = 0 + 1.5 × 64 = 96 m
Answer: The boat travels 96 m during this time.
Q5. A driver of a car travelling at 52 km h⁻¹ applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h⁻¹ in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
For Car 1:
- u = 52 km h⁻¹ = 52 × (5/18) = 14.44 m s⁻¹
- v = 0, t = 5 s
For Car 2:
- u = 3 km h⁻¹ = 3 × (5/18) = 0.83 m s⁻¹
- v = 0, t = 10 s
Plotting speed-time graphs:
- Car 1: Straight line from (0, 14.44) to (5, 0)
- Car 2: Straight line from (0, 0.83) to (10, 0)
Distance = Area under speed-time graph
For Car 1:
Distance = (1/2) × base × height
= (1/2) × 5 × 14.44 = 36.1 m
For Car 2:
Distance = (1/2) × 10 × 0.83 = 4.15 m
Answer: Car 1 travelled farther (36.1 m) compared to Car 2 (4.15 m) after the brakes were applied.
Q6. Fig 7.10 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
Solution: The object with the steepest slope on a distance-time graph is travelling the fastest. Object B has the steepest slope, so B is travelling the fastest.
(b) Are all three ever at the same point on the road?
Solution: Yes, all three objects are at the same point on the road at approximately 0.8 hours. This is where all three lines intersect.
(c) How far has C travelled when B passes A?
Solution: B passes A at approximately 0.4 hours. At this time, C has travelled approximately 4 km.
(d) How far has B travelled by the time it passes C?
Solution: B passes C at approximately 1.0 hour. By this time, B has travelled approximately 8 km.
Q7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground? After what time will it strike the ground?
Solution:
Given:
- u = 0 (dropped gently)
- a = 10 m s⁻²
- s = 20 m
Finding final velocity (v):
Using v² = u² + 2as
v² = 0 + 2 × 10 × 20 = 400
v = √400 = 20 m s⁻¹
Finding time (t):
Using v = u + at
20 = 0 + 10 × t
t = 20 / 10 = 2 s
Answer:
- Velocity with which it strikes the ground = 20 m s⁻¹
- Time taken = 2 s
Q8. The speed-time graph for a car is shown in Fig. 7.11.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
Solution:
From the graph:
- Speed increases uniformly from 0 to 6 m s⁻¹ in the first 4 seconds.
Distance = Area of triangle under the graph
Distance = (1/2) × base × height
= (1/2) × 4 × 6 = 12 m
The shaded area is a triangle from (0, 0) to (4, 6) to (4, 0).
(b) Which part of the graph represents uniform motion of the car?
Solution: The part of the graph where the line is parallel to the time axis (horizontal line) represents uniform motion. This is typically from 4 s to 6 s (or the flat portion of the graph).
Answer:
- Distance travelled in first 4 seconds = 12 m
- The horizontal portion of the graph represents uniform motion
Q9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity
Solution: Possible
Example: A ball thrown vertically upward has constant acceleration (g = 9.8 m s⁻² downward) throughout its motion. At the highest point, its velocity becomes zero momentarily, but it still has constant acceleration (downward).
(b) an object moving with an acceleration but with uniform speed
Solution: Possible
Example: An object moving in uniform circular motion. The speed remains constant, but the direction changes continuously, so there is acceleration (centripetal acceleration) towards the center of the circle.
(c) an object moving in a certain direction with an acceleration in the perpendicular direction
Solution: Possible
Example: A projectile motion (like a ball thrown horizontally from a height). The ball moves horizontally with initial velocity, but experiences vertical downward acceleration due to gravity, which is perpendicular to the initial direction of motion.
Q10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
Given:
- Radius (r) = 42250 km = 42,250,000 m
- Time (t) = 24 hours = 24 × 3600 = 86,400 s
Circumference of orbit:
textCircumference = 2πr = 2 × (22/7) × 42250 = 265,464.29 km
Speed:
Speed = Distance / Time
= 265,464.29 / 24
= 11,061 km h⁻¹
Or in m s⁻¹:
Speed = (2 × 3.14 × 42,250,000) / 86,400
= 3,072 m s⁻¹
≈ 3.07 km s⁻¹
Answer: The speed of the satellite is approximately 11,061 km h⁻¹ or 3.07 km s⁻¹ or 3072 m s⁻¹.
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