Chapter 4 - Work, Power and Energy

In the previous section we saw mass, weight, thrust and pressure. In this section we will see work, power and energy.
■ Consider some of our day to day activities. We walk, run, play, read, think. We engage in many such activities. 
• For performing those activities, we need energy. Some activities like running and playing requires more energy. 
• We get the required energy from the food we eat.
■ Consider animals. They also perform various activities. Like running, hunting, hiding from enemies etc.,  Also we humans make them do activities like pulling vehicles, carrying loads etc., 
• Animals also get the required energy from the food that they eat
■ Consider machines. They perform activities like pumping water, pulling trains, lifting heavy loads etc., 
• They get their energy from the fuels like petrol, coal, diesel etc.,

To define energy from a scientific point of view, we must first understand about another concept 'work'. In day to day life we regularly come across the term work. Let us see some examples:
• Two laborers worked hard to load the bricks into the truck
• Much work is required to arrange all the newly arrived books in the library
• A person is holding a heavy luggage on his shoulders. He may have to stand in that position for a long time. He is applying upward force through his shoulders. He may even get exhausted. But we would not say that he is 'working' to keep the luggage on his shoulders.
• Talking with friends for a long time about some academic topics may be considered as 'work'. It involves expenditure of energy too.
• At the same time, taking with friends about a movie may not be considered as work
■ Thus, even in common day do day conversations, we give a difference between work and energy. 
• In day to day conversations, we would not mind if they are interchanged. 
• From a scientific point of view, such interchanging is not allowed.

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■ In science, work is done only if two conditions are satisfied: 

• A force should act on an object
• The object must get displaced due to the force
Let us see some examples:
1. Push a pebble lying on the ground. The pebble moves through a distance
• A pushing force was applied on the object (the object here is pebble)
• The object got displaced
■ So work is done
2. A girl pulls a trolley. The trolley moves through a distance
• A pulling force was applied on the object (the object here is trolley)
• The object got displaced
■ So work is done
3. Lift a book through a height. For lifting the book, a force must be applied. The book rises up.
• A lifting force was applied on the object (the object here is book)
• The object got displaced
■ So work is done
4. Push a large rock on the ground. The rock does not move.
• A pushing force was applied on the object (the object here is rock)
• The object did not get displaced
■ So in this case, work is not done

Work done by constant force

We are now able to identify those situations in which a work is done. What we need next, is a method to calculate the amount of work done.
Let a constant force, F act on an object. Let the object be displaced through a distance, s in the direction of the force. This is shown in fig. 4.1 below:

Fig.4.1

Let W be the work done.
Then W = F × s
Thus, work done by a force is equal to the product of 
• The force and
• The displacement produced by the force
■ Next we need a unit for the force. 
• Unit for force is newton 
• Unit for distance is metre.
■ So unit for work is newton-metre. In short form, it is N m
• This 'N m' has another name: joule. In short form, it is J
• Let a force of 1 N move an object through a distance of 1 m. Then the amount of work done is obtained as follows:
W = Fs = 1 × 1 = 1N m = 1 joule
• Thus we can write:
1 N m or 1 joule of work is done on an object, if a force of 1 N moves it through a distance of 1 m along the line of action of the force.

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The unit 'joule' for work is named after James Prescott Joule. He was a British scientist who made significant contributions in the field of electricity and thermodynamics. 

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Now we will see a solved example:

Solved example 4.1

A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force. If the force acts on the object all through the displacement, calculate the work done.

Solution:

We have: W = F×s = 5 × 2 = 10 N m or 10 J

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When we defined work, we specified an important point. That is:

■ The displacement should be along the line of action of the force. 

Now consider the situation in fig.4.2 below:
1. A force F is acting on a body. The body is being displaced. The displacement is along the line of the force.

Fig.4.2

2. But the displacement is in the opposite direction of the force.
■ Let us see an example for such a situation:
• A body is moving with a uniform velocity u
• A retarding force F acts on it in a direction opposite to the direction of motion
• As a result of this retarding force, the body comes to rest. That is., final velocity v = 0 
• In this situation, how will we define force?
• We can use the same principle. But note that the displacement is opposite to the direction of the force. We know that displacement has both magnitude and direction. So it's direction should be taken as negative. This is to indicate that, it is opposite to the direction of the force.
So we get: W = F × (-s) = -Fs
■ Thus, the work can be either positive or negative
Another example:
1. Lift an object upwards. When we lift an object we are applying a force on the object
• The direction of movement of the object is same as the direction of the force
• So the work done is positive
2. When the object is lifted up, another force is also acting on the object. It is the gravitational force
• This force acts in a direction opposite to the direction of movement of the object
• So work done by the gravitational force is negative

Solved example 3.2
A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage. 
Solution:
1. Work done W = F×s
2. Here displacement s = 1.5 m
3. We need to find the force F. 
4. We know that the luggage which has a mass of 15 kg is acted upon by a force. This force is the 'Gravitational force' F_g exerted by the earth towards it's centre. 
5. We know that this force is equal to the weight mg. Where m is the mass and g the acceleration due to gravity.
So gravitational force of the earth = W = mg = 15 × 10 =150 N
6. To lift the object, we need to apply an equal and opposite force. That is., we need to apply 150 N in the upward direction. The displacement of 1.5 m is in the same direction of the lifting force. So we can take (+1.5) m
Thus work done = F×s = 150 ×1.5 = 225 J 

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In the next section, we will see Kinetic energy. 

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Chapter 1.9 - Solved examples on Motion of Objects

In the previous section, we completed the discussion on the motion of objects. In this section we will see some solved examples.

Solved example 1.16
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:
1. Diameter of the circular track = d = 200 m. So radius = r = 100 m
2. Perimeter of the track = 2πr = πd = 200π m
3. Time for completing one round = t = 40 s
4. Speed of the athlete = ^distance⁄_time  = ^πd⁄_t  = ^200π⁄₄₀ = 5π m s^-1.
5. So the athlete covers a distance of 5π m in 1 s. We have to find the distance that the athlete covers in 2 mins and 20 s
6. 2 mins 20 s = 140 s
7. So the distance covered in this time = 140 × 5π = 700π m
8. Now we have to find the displacement at the end of 140 s
9. When the athlete covers one perimeter 200π, from the starting point, he reaches back to his starting point. At that time, his displacement is zero
10. In this way, when he travels 700π, he reaches back the starting point 3 times. After the third time he travels a 'certain distance' to make the 700π. This 'certain distance is the displacement'. It can be calculated as follows:
11. 700π = 3 × 200π + 100π.
So the 'certain distance' = displacement = 100π.

Solved example 1.17
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:
Case 1. From A to B:
1. Distance = 300 m
2. Time = 2 mins 30 s = 150 s
3. Speed = ^distance⁄_time = ³⁰⁰⁄₁₅₀ = 2 m s^-1.
4. Velocity = ^displacement⁄_time 
• Here displacement = 300 m
• So velocity = ^displacement⁄_time = ³⁰⁰⁄₁₅₀ = 2 m s^-1.
Case 2. From A to C
1. Distance = (300 m + 100 m) = 400 m
2. Time = (2 mins 30 s + 1 min) = 3 mins 30 s = 210 s
3. Speed = ^distance⁄_time = ⁴⁰⁰⁄₂₁₀ = 1.904 m s^-1.
4. Velocity = ^displacement⁄_time 
• Here displacement = 200 m
• So velocity = ^displacement⁄_time = ²⁰⁰⁄₂₁₀ = 0.952 m s^-1.

Solved example 1.18
Abdul, while driving to school, computes the average speed for his trip to be 20 kmph. On his return trip along the same route, there is less traffic and the average speed is 30 kmph. What is the average speed for Abdul’s trip?  
Solution:
1. Let the distance from home to school be 's'
2. Let the time required in the morning be t1
3. Then Abdul calculated his average speed in the morning to be 20 kmph in the following way:
Speed = ^distance⁄_time = ^s⁄_t1 = 20 kmph
4. From this we get: t1 = ^s⁄₂₀ .
5. In the evening he travels the same distance 's'
6. Let the time required in the evening be t2
7. Then he calculated his average speed in the evening to be 30 kmph in the following way:
Speed = ^distance⁄_time = ^s⁄_t2 = 30 kmph
8. From this we get: t2 = ^s⁄₃₀ .
9. So total time required for the travel = (^s⁄₂₀ + ^s⁄₃₀) = ^5s⁄₆₀.
10. Total distance = s + s = 2s
11. So average speed = ^{total distance}⁄_{total time} = 2s ÷ ^5s⁄₆₀ = 2s × ⁶⁰⁄_5s = 24 kmph

Solved example 1.19
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:
1. The motorboat starts from rest. So initial velocity u = 0
• Time of travel = 8 s
• Acceleration = 3 m s^-2.
2. We want the distance s
The second equation of motion connects the above quantities. So we will use it:
3. s = ut + ¹⁄₂ at² ⇒ s = 0 × 300 + ¹⁄₂ × 3 × 8² ⇒ s = 96 m

Solved example 1.20
A driver of a car travelling at 52 kmph applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 30 kmph 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:
The fig. shows the speed-time graph for the two cars:

1. The yellow line AB is the speed-time graph of the first car. The purple line PQ is the speed-time graph of the second car.
Note that, the speeds are converted from kmph to m s^-1. 
2. The area enclosed by a speed-time graph and the time axis will give the distance travelled by the object. We saw the details here. The areas are shown shaded in the above fig.
3. First car:
Area = Area of the triangle OAB = ¹⁄₂ × 5 × 14.44 = 36.1 m
[ ∵ Base = 5 units and altitude = 14.44 units]
4. Second car : 
Area = Area of the triangle OPQ = ¹⁄₂ × 10 × 8.33 = 41.65 m
[ ∵ Base = 10 units and altitude = 8.33 units]

Solved example 1.21
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:
1. Height of fall = distance of travel = s = 20 m
2. Acceleration = a = 10 m s^-2.
3. The ball falls from rest. So it has an initial velocity u = 0
4. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
5. So we can write:
v² = u² + 2as ⇒ v² = 0² + 2 × 10 × 20 ⇒ v² = 400 ⇒ v  = 20 m s^-1.
6. To find the time, we can use the first equation of motion:
v = u + at ⇒ 20 = 0 + 10t ⇒ 20 = 10 t ⇒ t = 2 s

Solved example 1.22
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:
1. Radius of the orbit = r = 42250 km
2. Perimeter of the orbit = 2πr = 2 × π × 42250 = 84500π km = 84500000 m
3. Time for completing one round revolution = t = 24 hours = 24 × 60 × 60 = 86400 s
4. Speed of the satellite = ^distance⁄_time  =  ^84500000π⁄₈₆₄₀₀ = 3070.95 m s^-1

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So we have completed the discussion on the 'Motion of objects'. In the next chapter, we will see 'Forces acting on objects'. 

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Chapter 1.6 - Equations of Motion from Velocity-Time Graph

In the previous section, we determined the acceleration from a velocity-time graph. In this section we will determine the distance travelled. 
• Consider our original velocity-time graph (fig.1.19 in the previous section). 
• Suppose we want to know the distance travelled by the car in a duration of 8 seconds starting from 12. 
• Then our t₁ is 8 and t₂ is 20. 
• To find that distance, we use the following procedure:
1. Erect two perpendiculars from t₁ and t₂. Extend them upwards until they meet the velocity-time graph. This is shown in the fig.1.24 below:

Fig.1.24

2. The distance travelled by the car in the duration of 12 seconds (from t₁ = 8 to t₂ = 20) is the area of the quadrilateral ABCD. Let us see the proof:
3. It is easy to prove this if, the velocity-time graph is horizontal. We saw it earlier in fig.1.18 of the previous section. The quadrilateral we saw there is a rectangle. But here, the quadrilateral is a trapezium.
4. Consider a very small duration of time dt, after say 12 seconds (In 'dt', the 'd' stands for 'delta', the Greek symbol used to denote very small quantities'). It is shown shaded in the fig.1.25 below:

Fig.1.25

5. The shaded area is a small trapezium. Let us see it's details:
• The height of it's left vertical side gives the initial velocity (velocity just before the beginning of dt)
• The height of it's right vertical side gives the final velocity after the duration dt
• But dt is so small that, the increase in velocity during that duration can be ignored.
• So the height on the right side is same as that on the left side. That means, the car travelled at a constant velocity for a duration of dt.
• So, it is not a trapezium but a rectangle
6. We know that, if it is a rectangle, the distance travelled during dt will be equal to the area of the rectangle
7. So the distance travelled by the car for the duration of dt after 12 seconds, is the area of the rectangle, whose width is dt, shown shaded in the fig.1.25 above
8. We can divide the total area of the trapezium ABCD into so many such rectangles. All rectangles will have the same width of dt. But the heights will be different. See fig.1.26(a) below:

Fig.1.26

• Consider a rectangle of width dt near the left side in fig.1.26(a) 
    ♦ It's height will be less. 
    ♦ So it's area will be less 
    ♦ So the distance travelled during that dt will be less
• Consider a rectangle of width dt near the right side in fig.1.26(a) 
    ♦ It's height will be more.
    ♦ So it's area will be more
    ♦ So the distance travelled during that dt will be more
9. We divide the total duration between A and B into a number of dt's. This is shown in fig.1.26(b). (In the fig.b, only a few rectangles are shown. In reality, the space between A and B is completely filled with rectangles. And the width of all those rectangles is dt) 
10. The distance travelled in each of these dt's = Area of the rectangle in that dt
11. Sum of the distances travelled in all the dt's = Total distance travelled in the duration from t₁ to t₂
12. But left side of (11) is the 'sum of areas of all the rectangles'. So this can be included in (11)
13. So (11) becomes:
Sum of the distances travelled in all the dt's 
= Total distance travelled in the duration from t₁ to t₂ 
= Sum of areas of all the rectangles 
14. Consider the last part in the above: 'Sum of areas of all the rectangles'. This is the 'total area of the trapezium ABCD'.
15. So (13) becomes:
Sum of the distances travelled in all the dt's 
= Total distance travelled in the duration from t₁ to t₂ 
= Sum of areas of all the rectangles
= Total area of the trapezium ABCD
16. Take the second and fourth parts from (15). We get:
■ Total distance travelled in the duration from t₁ to t₂ = Total area of the trapezium ABCD 
This is the required proof.

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Now we will try to write a general equation for the area. So that, we will be able to calculate the distance  's' quickly. Consider the velocity-time graph given below:

Fig.1.27

1. We want the area of the trapezium ABCD in the fig.1.27
2. From the fig. it is clear that the height AD = u and height BC = v. Also AB = t = (t₂ - t₁)
3. We have: Area of a trapezium = ^h⁄₂ × (a+b)
Where a is the top parallel side, b is the bottom parallel side and h is the height Details here.
4. To use this formula in our present case, we must consider:
• AD as the top parallel side
• BC as the bottom parallel side
• AB as the height
5. So we get:
Area of trapezium ABCD = Ditance travelled s = ^t⁄₂ × (u+v)
6. But in the previous section we obtained Eq.1.6:
v = u + at 
7. Substituting this v in (5) we get:
s =  ^t⁄₂ × (u + u + at )
⇒ s =   ^t⁄₂ × (2u + at ) 
From this we get the equation below:
Eq.1.7:
s = ut + ¹⁄₂ at²
This is the second equation of motion.

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We have to derive one more equation. But it does not require the help of a graph. We can derive it from the first two.
1. While we were deriving the second equation of motion, we had reached step (5) above. In that step, we got: s =  ^t⁄₂ × (u+v)
2. But we can obtain the value of t from the first equation of motion: v = u + at
⇒ t = ^(v-u)⁄_a.
3. Substituting this in (1) we get:
s =  ^(v-u)⁄_2a × (u+v). 
But [(v+u)(v-u)] is an identity. We have:
[(v+u)(v-u)] = (v² - u²)     
4. So (3) becomes:
s = ^(v2-u2)⁄_2a.
5. From this we get the third equation of motion:
Eq.1.8:
v² = u² + 2as

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Let us write the three equations together:
■ First equation of motion: v = u + at 
■ Second equation of motion: s = ut + ¹⁄₂ at²
■ Third equation of motion: v² = u² + 2as

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Now we will see some solved examples:
Solved example 1.6
The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table below:

Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.
Solution:
The required distance time graph is shown in fig.1.28 below:

Fig.1.28

Interpretation:
1. At Station A, there are two red dots in the graph. 
• These red dots have the same y coordinates '0' because, the train remains at the same point (Station A) from 8 hours to 8:15 hours. 
• There is no change in distance during those 15 minutes. So the y coordinates does not change. 
• But the x coordinates change because the time passes from 8:00 hours to 8:15 hours  (Note that 0:15 hours is 15 minutes, which is 0.25 hours)
• At 8:15 hours it begins it travel to Station B. So the distance from Station A increases. This is the 'yellow sloping line' from (8.25,0) to (11.25,120)

2. At Station B, there are two red dots in the graph. 
• These red dots have the same y coordinates '120' because, the train remains at the same point (Station B) from 11:15 hours to 11:30 hours. 
• There is no change in distance during those 15 minutes. So the y coordinates does not change. 
• But the x coordinates change because the time passes from 11:15 hours to 11:30 hours  (Note that 0:30 hours is 30 minutes, which is 0.50 hours)
• At 11:30 hours it begins it travel to Station C. So the distance from Station A (The zero distance point) increases. This is the 'yellow sloping line' from (11.5,120) to (13,180)

3. At Station C, there are two red dots in the graph. 
• These red dots have the same y coordinates '180' because, the train remains at the same point (Station C) from 13:00 hours to 13:15 hours. 
• There is no change in distance during those 15 minutes. So the y coordinates does not change. 
• But the x coordinates change because the time passes from 13:00 hours to 13:15 hours  (Note that 0:15 hours is 15 minutes, which is 0.25 hours)
• After station C there no further travel is given in the table

Solved example 1.7
Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table below shows the distance travelled by them in different times.

Plot the distance-time graph for their motions on the same scale and interpret.
Solution:
The required distance time graph is shown in fig.1.29 below:

Fig.1.29

• The Distance-time graph of the travel made by Feroz is shown in yellow colour
• The Distance-time graph of the travel made by Sania is shown in cyan colour
• In the x-axis, only minutes are taken. If we take hours also, we will have to convert into decimal form as follows:
8:05 = 8 + 5/60 = 8 + 0.08333 = 8.08333
8:10 = 8 + 10/60 = 8 + 0.16667 = 8.16667
8:15 = 8 + 15/60 = 8 + 025 = 8.25
So on...
• So it is convenient to take minutes alone
• Once the scales on x and y axes are fixed, we can plot the graphs easily. The coordinates can be taken directly from the table. No calculations are required.

Interpretation:
• Feroz was travelling at a greater speed because he travelled 3.6 Kms in 20 minutes, while Sania took 25 minutes to travel the same distance
• After the plotting, if we examine carefully, we can find that, the points do not fall on straight lines. 
• This will be clear if we join the first and last points separately for the two graphs. It is shown in the fig.1.30 below:

Fig.1.30

• Note that, dashed lines are used to join the first and last points. This is to distinguish from the main graph.
• Since they are not straight lines, we can infer the following:
    ♦ Feroz did not travel at an uniform speed
    ♦ Sania did not travel at an uniform speed
Reason:
■ If the travel is at an uniform speed, we must get similar triangles for any random pairs of points that we take in a Distance-time graph. So that the ratio ^altitude⁄_base is always a constant. If it is not a straight line, we will not get similar triangles at any random pair we take. This we saw in fig.1.15
■ Also, from the readings in the given table, it is evident that the speeds were not uniform:
Feroz travelled 1 Km in the first 5 minutes
He travelled (1.9 - 1) = 0.9 Km in the next 5 minutes
He travelled (2.8 - 1.9) = 0.9 Km in the next 5 minutes
He travelled (3.6 - 2.8) = 0.8 Km in the next 5 minutes
■ Sania travelled 0.8 Km in the first 5 minutes
She travelled (1.6 - 0.8) = 0.8 Km in the next 5 minutes
She travelled (2.3 - 1.6) = 0.7 Km in the next 5 minutes
She travelled (3.0 - 2.3) = 0.7 Km in the next 5 minutes
She travelled (3.6 - 3.0) = 0.6 Km in the next 5 minutes

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In the next section, we will see a few more solved examples. 

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Chapter 1 - Objects in Motion - Displacement

In this chapter we will learn about the motion of objects. But first we will consider objects at rest. 
■ The most important property concerning an object at rest, is it's position. We must be able to say 'where' that object is. 
For that, we use references. Let us see some examples:
• Consider fig.1.1(a) below. The school is situated 2 km north of the railway station. Here, the railway station is used as the reference to specify the position of the school

Fig.1.1

• In fig.1.1(b), a car is parked 15 m to the left of the house. Here, the house is used as the reference to specify the position of the car.
■ With out a reference, we will not be able to tell the position of an object.

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Now let us see objects in motion. 

1. Consider fig.1.2(a) below. An object is initially at rest at the origin point 'O'.

Fig.1.2

2. From O, it travels in the positive direction of the x-axis for a distance of 4 m, and reaches 'A'. 

3. Then it travels in the positive direction of the y-axis for a distance of 3 m, and reaches 'B'. 

4. So B is the final position of the object. The object travelled a total distance of 3 + 4 = 7 m and reached B.

5. We can say this:
The object is displaced from O to B. The distance OB is 5 m. So the displacement of the object is 5m. 
6. Note that the object travelled a total distance of 7 m. But it's displacement is only 5 m. 
7. The line OB makes an angle of 30^o with the horizontal X axis. 
■ So we are able to specify the final position of the object. We do that as follows:
• The displacement of the object is 5 m
• This 5 m is along a line which makes an angle 30^o with the horizontal axis
8. Thus we see that, to specify a position of the object, we need two items:
(i) The magnitude of the displacement
    ♦ In our present case, it is '5m'.
(ii) The direction of the displacement
    ♦ In our present case, it is 'a line making 30^o with the horizontal'.
■ 'Magnitude of the displacement' is the 'shortest distance measured from the initial to the final position of an object'.
9. Note that, if instead of stopping at B, the object travels back to O along OB, the distance travelled will be 4+ 3 + 5 = 12 m. And the displacement will be zero. 

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Another example:
1. Consider fig.1.2(b) above. An object is initially at rest at the origin point 'O'. 

2. From O, it travels a distance of 7 m, and reaches 'P'. Then it travels 4 m and reaches 'Q'. Finally it travels 5 m and reaches 'R'. 

3. So R is the final position of the object. The object travelled a total distance of 7 + 4 + 5 = 16 m and reached R.

5. We can say this:
The object is displaced from O to R. The distance OR is 4.8 m. So the displacement of the object is 4.8 m. 
6. Note that the object travelled a total distance of 16 m. But it's displacement is only 4.8 m. 
7. The line OR makes an angle of 44^o with the horizontal X axis. 
■ So we are able to specify the final position of the object. We do that as follows:
• The displacement of the object is 4.8 m
• This 4.8 m is along a line which makes an angle 44^o with the horizontal axis
8. Thus, as in the prvious example, here also we see that, to specify a position of the object, we need two items:
(i) The magnitude of the displacement
    ♦ In our present case, it is '4.8m'.
(ii) The direction of the displacement
    ♦ In our present case, it is 'a line making 44^o with the horizontal'.
■ 'Magnitude of the displacement' is the 'shortest distance measured from the initial to the final position of an object'.
9. Note that, if instead of stopping at R, the object travels back to O along OR, the distance travelled will be 7+ 4 + 5 + 4.8 = 20.8 m. And the displacement will be zero.

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A special case:
1. Consider fig.1.3 below. An object is initially at rest at the origin point 'O'.

Fig.1.3

2. From O, it travels in the positive direction of the x-axis for a distance of 40 m, and reaches 'A'. 

3. Then it travels in the same direction for a distance of 30 m, and reaches 'B'.
4. From B, it travels in the opposite direction. That is in the negative direction of the x-axis. In this direction, it travels for a distance of 50 m, and reaches 'C'. This travel is indicated by the top most line between B and C in the fig.1.3 

5. So C is the final position of the object. The object travelled a total distance of 40 + 30 + 50 = 120 m and reached C.

6. We can say this:
The object is displaced from O to C. The distance OC is 20 m. So the displacement of the object is 20 m. 
7. Note that the object travelled a total distance of 120 m. But it's displacement is only 20 m. 
8. The line OC makes an angle of 0^o with the horizontal X axis. 
■ So we are able to specify the final position of the object. We do that as follows:
• The displacement of the object is 20 m
• This 20 m is along a line which makes an angle 0^o with the horizontal axis
9. Thus, here also we see that, to specify a position of the object, we need two items:
(i) The magnitude of the displacement
    ♦ In our present case, it is '20 m'.
(ii) The direction of the displacement
    ♦ In our present case, it is 'a line making 0^o with the horizontal'.
■ 'Magnitude of the displacement' is the 'shortest distance measured from the initial to the final position of an object'.
10. Note that, if instead of stopping at B, the object travels back to O along OC, the distance travelled will be 40 + 30 + 50 +20  = 140 m. And the displacement will be zero.
11. In this example, the direction of the displacement makes zero angle with the x-axis. This is because, all the travels were made along the x-axis. The object never left the x-axis.
12. A very interesting result:
• Let us take all the distances travelled towards the right (that is., positive direction of the x-axis) as positive
• And all the distances travelled towards left (that is., negative direction of the x-axis) as negative
• Now we add the distances:
OA + AB + BC = 40 + 30 -50 = 20
• '20' is the magnitude of the final displacement

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From the discussion so far in this chapter, we can write this:
■ To specify the position of an object, we need two items:
• Magnitude of the displacement
• Direction of the displacement

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In the next section, we will see Uniform and Non-uniform motion. 

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