Chapter 3.3 - Average velocity and Average speed

In the previous section we saw some applications of the position-time graph. In this section we will see average speed and average velocity.

• In the previous section we saw an object moving with a uniform velocity. 
• But in day to day life, we often see objects moving with non-uniform velocity.
■ Consider the object that we saw in the previous section. 
• We turned the stop watch on only when the object reached P. 
    ♦ From P onwards, it travelled with a uniform velocity of 12 ms^-1. 
• But the object started it's journey from O. It started from rest at O.
■ If started from rest at O, how did it attain a velocity of 12 ms^-1?

We can put this question in another way using the following 4 statements:

(i) At the instant when the object started from rest, the velocity will be zero
(ii) When it is at P, the velocity is 12 ms^-1
(iii) So the velocity increased from zero to 12 ms^-1
(iv) How did this increase happen?
• To find the answer, we will do another experiment. This time, we observe the motion of a car
• Also this time, will turn on the stop watch at the very instant when the car starts it's motion from O
• We will need the help of the odometer or trip meter of the car
(Trip meter is more preferable because, it can be reset to zero)  
Let us begin:
(i) At the instant when the car begins it's motion, t = 0, x = 0
(ii) When t = 1 s, the trip meter reading is 1.8 m
• That means, in the first second, the car moved 1.8 m. 
• Let this new position be A. 
    ♦ Then OA = 1.8 m
This is shown in fig.3.29 below:
 

Fig.3.29

(iii) When t = 2 s, the trip meter reading is 7.2 m
• That means, in the second second, the car moved (7.2 - 1.8) = 5.4 m. 
• Let this new position be B. Then: 
    ♦ AB = 5.4 m 
    ♦ OB = 7.2 m
This is shown in fig.3.23 above
(iv) When t = 3 s, the trip meter reading is 16.2 m
• That means, in the third second, the car moved (16.2 - 7.2) = 9 m. 
• Let this new position be C. Then: 
    ♦ BC = 9 m 
    ♦ OC = 16.2 m
This is shown in fig.3.23 above
(v) When t = 4 s, the trip meter reading is 28.8 m
• That means, in the fourth second, the car moved (28.8 - 16.2) = 12.6 m. 
• Let this new position be D. Then: 
    ♦ CD = 12.6 m
    ♦ OD = 28.8 m
This is shown in fig.3.23 above
(vi) When t = 5 s, the trip meter reading is 45 m
• That means, in the fifth second, the car moved (45 - 28.8) = 16.2 m. 
• Let this new position be E. Then: 
    ♦ DE = 16.2 m 
    ♦ OE = 45 m
This is shown in fig.3.23 above
(vii) When t = 6 s, the trip meter reading is 64.8 m
• That means, in the sixth second, the car moved (64.8 - 45) = 19.8 m. 
• Let this new position be E. Then: 
    ♦ EF = 19.8 m 
    ♦ OF = 64.8 m
This is shown in fig.3.23 above
(viii) When t = 7 s, the odometer reading is 88.2 m
• That means, in the seventh second, the car moved (45 - 28.8) = 23.4 m. 
• Let this new position be E. Then: 
    ♦ EG = 23.4 m 
    ♦ OG = 88.2 m

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From the above values, the following two points become very clear: 
(i) In the previous experiment, 
    ♦ The time durations are all same: 1 s
    ♦ The distance travelled in each of these durations are the same: 12 m
• That means the object travelled with a uniform velocity
(ii) In the present experiment 
    ♦ The time durations are all same: 1 s
    ♦ But the distance travelled in each of these durations are not the same
• That means, the object travelled with non-uniform velocity
Let us plot the position-time graph. It is shown in fig.3.30 below:

Fig.3.30

• To avoid congestion, coordinates of only alternate points are written.
■ We see that, the graph is a curve

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• We have seen that, slope of the position-time graph will give the velocity
• If the object moves with uniform velocity, the position-time graph will be a straight line
• Then we can easily calculate the slope. That slope will be same at all points
• But here, the graph is a curve. The slope will be different at different points.
• That means, the velocity is different at different points
• In such a situation, we use average velocity.
We can explain this using an example:
1. Consider the points D and F
• The car passed through both D and F 
2. Suppose someone wants to know the 'velocity with which the car travelled between D and F'
• We cannot give a definite answer. Because the velocity continuously changed
3. In such a situation we take the ratio: ^{Displacement from D to F}⁄_{Time required for the travel from D to F}
• We know that:
    ♦ Displacement from D to F = Δx = (x₂-x₁) = (y coordinate of F - y coordinate of D) 
    ♦ = (64.8 - 28.8) = 36
Time required to travel from D to F = Δt = (t₂-t₁) = (x coordinate of F - x coordinate of D) 
    ♦ = (6 - 4) = 2
• So the ratio is ³⁶⁄₂ = 18 ms^-1.
4. This velocity of 18 ms^-1 is the average velocity with which the object travelled from D to F. 
We can write the definition:
■ Average velocity is defined as the displacement (Δx) divided by the time duration (Δt) in which the displacement occurs
• The formula is:

    ♦ The bar over v is a standard notation used to indicate average quantities
5. We can check whether this formula is dimensionally correct:
• On the left side we have velocity [LT^-1]
• On the right side, 
    ♦ in numerator we have distance [L]
    ♦ in denominator we have distance [T]
• Thus we get [LT^-1] on both sides

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We have seen that, if the velocity is uniform, we can easily calculate the velocity as the slope
■ Can we relate average velocity to slope?
Let us try:
• The portion between D and F is enlarged and shown in fig.3.31 below:

Fig.3.31

• We know that DF is a curve. But now, a straight line is draw between them in cyan colour
• Also a triangle DFF' is formed
• Now take the ratio: ^altitude⁄_base
• We get: ^altitude⁄_base = ^FF'⁄_DF' = ^(64.8-28.8)⁄_(6-4) = ³⁶⁄₂ = 18 ms^-1
So we can write:
■ Average velocity is the slope of the line joining the initial and final positions of the object in the position-time graph

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• We have seen the details about average velocity. 
• Average speed can also be calculated in a similar way. 
• But in the numerator, instead of 'displacement', we use 'path length'. Thus we write:
Average speed =
^{Total path length between two points}⁄_{Time required for the travel between the two points}

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■ Average velocity is a vector quantity. It has both magnitude and direction 
■ Average speed is a scalar quantity. It has magnitude but no direction 
The difference between the two can be understood using an example. We will write it in steps:
1. Fig.3.32 below shows the positions of an object at various times shown by a stop watch

Fig.3.32

From the fig., the following information can be obtained:

(i) The object moves along the x axis

(ii) It started it's journey from the origin

    ♦ At that instant the stop watch was turned on

(iii) It travelled in the +x direction

(iv) After 18 seconds it reached P

    ♦ P is 360 m away from O

(v) Then it began to travel in the -x direction

(vi) After 6 more seconds, it reached Q

    ♦ Q is 240 m away from O

2. We are asked to find the following two items:

(i) The average velocity with which the object travelled from O to Q

(ii) The average speed with which the object travelled from O to Q 

Solution:

Part 1: To find average velocity: 

1. To get a better understanding of the problem, we will see the position-time graph of the motion of an object: 

Fig.3.32

• The graphs are curves. So it is clear that, the object travelled with non-uniform velocity
• But that does not affect this problem because we are asked to find average velocity and speed
2. Average velocity depends only on initial and final positions and also the time. 
• It does not matter whether it is uniform or non-uniform velocity
• We have:

x₂ = position of the final point Q = 240 
x₁ = position of the initial point O = 0 
• So displacement = Δx = (x₂-x₁) = (240-0) = 240
t₂ = time at which the final point Q is reached = 24 s 
t₁ = time at which the object is at the initial position = 0 s 
• So duration required for the displacement = Δt = (t₂-t₁) = (24-0) = 24 s
3. Thus average velocity = ²⁴⁰⁄₂₄ = 10 ms^-1
Another method:
1. We have seen that, average velocity is the slope of the line joining initial and final points. 
• This line is shown in fig.3.33 below:

Fig.3.33

2. Clearly, altitude = QQ' = 240
• Base = OQ' = 24
• Thus we get: Slope = ^altitude⁄_base = ^QQ'⁄_OQ' = ²⁴⁰⁄₂₄ = 10 ms^-1
Part 2:
To find average speed:
1. For this, we have to take the total path length
• Total path length = (Distance from O to P + Distance from P to Q) = (360+120) = 480
2. Total time - 24 s
3. We have: Average speed =
^{Total path length between two points}⁄_{Time required for the travel between the two points} = ⁴⁸⁰⁄₂₄ = 20 ms^-1s
■ So we find that average velocity and average speed need not be the same

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Now we will see a solved example. The link is given below:
Solved example 3.1

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In the next section, we will see instantaneous velocity.

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Chapter 3.2 - Applications of Position-Time graph

In the previous section we saw the position-time graph. In this section we will some applications of the graph.
1. Consider an object moving with a uniform velocity of 12 ms^-1.
• Uniform velocity of 12 ms^-1 means this:
The object is travelling with the same velocity of 12 ms^-1 during it's entire journey
2. Let us move parallel to the object in another vehicle. 
• At the instant when it passes a point P, we turn on the stop watch. 
• So at P, the time t = 0. This is shown in fig.3.14 below:

Fig.3.14

• Note that at P, the time t = 0
    ♦ But the distance x is not equal to zero
• we can have 'x = 0' only at the origin O. 
    ♦ Point 'P' is away from the origin
(i) The object started it's journey (from rest) at O. 
(ii) Then it began to travel with a uniform velocity of 12 ms^-1
(iii) We began our experiment only at the instant when the object passed the point P
(iv) The distance already travelled by the object to reach P (that is., OP) is not required for our discussion for the time being. We will consider it later.
3. Let us continue our experiment:
• Consider a 'duration of one second'. 
    ♦ It may be a small duration. But it is important for our experiment
• Consider this duration just after 'the instant when the object pass P'. 
• Obviously, at the end of this duration, the object will be 12 m away from P
■ This is because, 12 ms^-1 means that, the object is travelling 12 m in every second
• Also, at the end of this duration, the stop watch will show t = 1 s
• Let the new position be Q. It is marked in fig.3.15 below:

Fig.3.15

4. Consider 'the duration of 1 s' just after 'the instant when the object pass Q'. 
• Obviously, at the end of this duration, the object will be 12 m away from Q
• Again, this is because, 12 ms^-1 means that, the object is travelling 12 m in every second
• At the end of this duration, the stop watch will show t = 2 s
• Let the new position be R. It is marked in fig.3.16 below:

Fig.3.16

5. Consider 'the duration of 1 s' just after 'the instant when the object pass R'. 
• Obviously, at the end of this duration, the object will be 12 m away from R
• Again, this is because, 12 ms^-1 means that, the object is travelling 12 m in every second
• At the end of this duration, the stop watch will show t = 3 s
• Let the new position be S. It is marked in fig.3.17 below:

Fig.3.17

6. In this way, we can mark any number of points we like.
• But for our discussion, this much is sufficient
• Let us say, we want to know the distance travelled at the end of 2 s
    ♦ From fig.3.17, we can see that, the required distance = (12+12) = (12×2) = 24 m
• Let us say, we want to know the distance travelled at the end of 3 s
    ♦ From fig.3.17, we can see that, the required distance = (12+12+12) = (12×3) = 36 m
7. In general, if we want to know the distance travelled at the end of 't' seconds, all we need to do is this:
• Multiply 'velocity' by 't'
    ♦ Distance is denoted by the letter 'x'
    ♦ Velocity is denoted by the letter 'v'
    ♦ Time is denoted by the letter 't'
■ So we can write: x = vt
• Note that, when we multiply, the units must be compatible
    ♦ If velocity is given in ms^-1, we must multiply it by 'time in s'
    ♦ If velocity is given in kmph, we must multiply it by 'time in h'
An example:
An object travelled with a uniform velocity of 15 ms^-1 for 7 seconds. How much distance did it travel during this time?
Solution:
• Given: v = 15 ms^-1, t = 7 s
• So distance travelled = x = vt = 15 × 7 = 105 m
8. In the above experiment, the following information were known to us:
(i) The velocity with which the object travelled
(ii) The time duration for which distance is required
• So both the quantities on the right side of the equation were known to us.
■ But another situation can arise:
• We are given the following information:
(i) The distance travelled by the object
(ii) The time required to travel this distance
• We are asked to find the velocity with which the object travelled 
9. In this situation, we rearrange the equation x = vt as: v = ^x⁄_t
• So now we know all the quantities on the right side. We can easily calculate v
An example:
An object travelling with an uniform velocity, covered 108 m in 12 seconds. Find out the uniform velocity with which it travelled
Solution:
Given: x = 108 m. t = 12 s
so velocity v = ^x⁄_t = ¹⁰⁸⁄₁₂ = 9 ms^-1

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• Now let us draw the position-time graph of the above experiment. 
• To draw the graph we need coordinates. So first let us write the coordinates:
1. Coordinates of P:
(i) The x coordinate of P is the time at which the object is at P. 
    ♦ From the fig.3.17 above, it is clear that, x coordinate is 0
(ii) The y coordinate of P is the distance of P from the origin.
    ♦ But we are not given the distance of P from the origin. 
    ♦ We just turned the stop watch on at the instant when the object passed P
(iii) Let us assume the distance to be say 20 m. 
    ♦ Assuming a value for this initial distance will not affect our present discussion   
(iv) So the coordinates of P are: (0,20)
This is shown in fig.3.18 below:

Fig.3.18

2. Coordinates of Q:
(i) The x coordinate of Q is the time at which the object is at Q. 
    ♦ From the fig.3.18 above, it is clear that, x coordinate is 1
(ii) The y coordinate of Q is the distance of Q from the origin. 
    ♦ It will be equal to: (y coordinate of P) + (distance PQ) = 20 + 12 = 32
(iii) So the coordinates of Q are: (1,32) 
3. Coordinates of R:
(i) The x coordinate of R is the time at which the object is at R. 
    ♦ From the fig.3.18 above, it is clear that, x coordinate is 2
(ii) The y coordinate of R is the distance of R from the origin. 
    ♦ It will be equal to: (y coordinate of Q) + (distance QR) = 32 + 12 = 44
(iii) So the coordinates of R are: (2,44)
4. Coordinates of S:
(i) The x coordinate of S is the time at which the object is at S. 
    ♦ From the fig.3.18 above, it is clear that, x coordinate is 3
(ii) The y coordinate of S is the distance of S from the origin. 
    ♦ It will be equal to: (y coordinate of R) + (distance RS) = 44 + 12 = 56
(iii) So the coordinates of S are: (3,56)
5. Now we can draw the graph.
• The following scales would be appropriate:  
    ♦ x axis: 4 units = 1 s
    ♦ y axis: 1 unit = 10 m
It is shown in fig.3.19 below:

Fig.3.19

6. Like in our previous experiment, here also, we can answer some special questions.
An example:
Consider the following two instants:
• Instant at which the stop watch showed 1.2 s 
• Instant at which the stop watch showed 2.7 s 
How much distance did the object travel between these two instants?
Solution:
(i) Draw a vertical dashed line through t = 1.2 s
This is shown in fig.3.20 below. 
• Let it cut the yellow graph at A
• Draw a horizontal dashed line through A
• It is cutting the y axis at s = 52.4 m
(The reader may verify this in his/her own graph)

Fig.3.20

(ii) Draw a vertical dashed line through t = 2.7 s
• Let it cut the yellow graph at B
• Draw a horizontal dashed line through B
• It is cutting the y axis at s = 52.4
(iii) So the vertical distance between the two horizontal dashed lines = (52.4 - 34.4) = 18 m
• This is the distance travelled by the object during the interval between t = 1.2 and t = 2.7 s
7. In the graph in the above fig.,3.20, there is a triangle ABC 
(i) It is a right triangle, right angled at C
• It's altitude is 18 m
• It's base = (2.7 - 1.2) = 1.5 s
(ii) Take the ratio: ^altitude⁄_base
• We get: ^altitude⁄_base = ¹⁸⁄_1.5 = 12
• But '12' is the uniform velocity with which the object travelled
(iii) From what we have learned in our maths classes, the ratio '^altitude⁄_base' is the slope of the yellow line
So we can write:
■ The slope of the position-time graph is equal to the velocity of the object
• We can form any number of right triangles like ABC
• For each of them, we can calculate '^altitude⁄_base'
• We will find that all of them are '12' 
• The reader may verify this by forming different right triangles and calculating the ratio in each case

Mathematical explanation:
■ Why do we get velocity as the 'slope of the graph'?
A mathematical explanation can be written as follows:
(i) We have seen that v = ^x ⁄_t.
• In the numerator we have distance
• In the denominator we have time
(ii) Now consider the ratio : ^altitude⁄_base
• The 'altitude' in the numerator is in fact 'distance'
    ♦ (distance at t = 2.7 s - distance at t = 1.2 s) = a distance of 18 m
• The 'base' in the denominator is in fact time
    ♦ a time duration of 1.5 s
• So when we take the ratio '^altitude⁄_base', we are in fact dividing distance by time
(iii) But 'dividing distance by time' gives velocity (∵ v = ^x⁄_t)
■ So the ratio '^altitude⁄_base', which is the slope, is same as velocity
• Also note that, for a single straight line, the slope will always be the same, wherever we draw the right triangle

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Relation to coordinate geometry
1. In our coordinate geometry classes in maths, we have seen that:
• Equation of a straight line is: y = mx + c
    ♦ Where m is the slope of the line
    ♦ c is the intercept made by the line on the y axis
2. In the graph in fig.3.20 above, we have the same situation.
    ♦ The slope of the yellow line is 12, which is the velocity 'v'
    ♦ y intercept is 20
3. On the x axis, time (t) is plotted
    ♦ so 'x' in y = mx + c should be replaced by t
• On the y axis, distance s is plotted
    ♦ so 'y' in y = mx + c should be replaced by x
• Thus the equation of the yellow line is:
x = vt + 20
4. Using this equation, we can calculate the position of the object at any given time 't'.
For example:
• when t = 15 s, we have:
x = [(12 × 15)] + 20 = [180] + 20 = 200 m 
•  So, we can write:
(i) The stop watch was turned on when the object passed P
(ii) When the watch showed 15 s, the object is at a distance of (12 × 15) = 180 m from P
(iii) But before reaching P, it had already travelled 20 m from the origin O 
(iv) So the total distance from O = (180+20) = 200 m
Another example:
• when t = 22 s, we have:
x = [(12 × 22)] + 20 = [264] + 20 = 284 m 
•  So, we can write:
(i) The stop watch was turned on when the object passed P
(ii) When the watch showed 22 s, the object is at a distance of (12 × 22) = 264 m from P
(iii) But before reaching P, it had already travelled 20 m from the origin O 
(iv) So the total distance from O = (264+20) = 284 m
5. So 20 m is the initial distance. 
• If this value is changed, the distances will change. 
• But the slope of the line will not change. 
    ♦ That means, the velocities will not change. 
    ♦ This is shown in fig.3.21 below:

Fig.3.21

• Consider the topmost cyan line
It's y intercept is 35. So we can write:
(i) The stop watch was turned on when the object passed P
(ii) But the object had already travelled a distance of 35 m from O
• Consider the bottom most red line
It's y intercept is 10. So we can write:
(i) The stop watch was turned on when the object passed P
(ii) But the object had already travelled a distance of 10 m from O
■ Note that the 'O' mentioned above is the 'O' in the path of the object (fig.3.18). 
It is not the 'O' in the graph
6. We have seen that the equation of the yellow line is: x = 12t + 20
• Since there is no change in slope. the equation of cyan line will be: x = 12t + 35
• Similarly, the equation of red line will be x = 12t + 10
■ In general, if v is the velocity and x₀ the initial distance, 
then the total distance travelled in 't' seconds is given by: x = vt + x₀.

Object travelling in the opposite direction
1. Consider an object moving with a uniform velocity of 15 ms^-1.
• Uniform velocity of 15 ms^-1 means this:
The object is travelling with the same velocity of 15 ms^-1 during it's entire journey
• Like in the previous experiment, here also the object is moving along the x axis
• But this time the object is moving (from a far way point on the positive side of x axis) towards the origin
2. Let us move parallel to the object in another vehicle. 
• At the instant when it passes a point U, we turn on the stop watch. 
• So at U, the time t = 0. 
• Also, let U be at a distance of 90 m from the origin. This is shown in fig.3.22 below:

Fig.3.22

• Note that at U, the time t = 0
    ♦ But the distance x is not equal to zero
• we can have 'x = 0' only at the origin O. 
    ♦ Point 'U' is away from the origin
(i) The object started it's journey (from rest) at some far way point on the positive side of x axis. 
(ii) Then it began to travel (with a uniform velocity of 15 ms^-1) towards the origin 
(iii) We began our experiment only at the instant when the object passed the point U
(iv) The distance already travelled by the object to reach U is not required for our present discussion.
3. Let us continue our experiment:
• Consider a 'duration of one second'. 
    ♦ It may be a small duration. But it is important for our experiment
• Consider this duration just after 'the instant when the object pass U'. 
• Obviously, at the end of this duration, the object will be 15 m away from U
■ This is because, 15 ms^-1 means that, the object is travelling 15 m in every second
• Also, at the end of this duration, the stop watch will show t = 2 s
• The object is now nearer to the origin by 15 m. So it's distance from the origin = (90-15) = 75 m
• Let the new position be V. It is marked in fig.3.23 below:

Fig.3.23

4. Consider 'the duration of 1 s' just after 'the instant when the object pass V'. 
• Obviously, at the end of this duration, the object will be 15 m away from V
• Again, this is because, 15 ms^-1 means that, the object is travelling 15 m in every second
• At the end of this duration, the stop watch will show t = 3 s
• Let the new position be W. It is marked in fig.3.24 below:

Fig.3.24

5. Consider 'the duration of 1 s' just after 'the instant when the object pass W'. 
• Obviously, at the end of this duration, the object will be 15 m away from W
• Again, this is because, 15 ms^-1 means that, the object is travelling 15 m in every second
• At the end of this duration, the stop watch will show t = 4 s
• Let the new position be T. It is marked in fig.3.25 below:

Fig.3.25

6. In this way, we can mark any number of points we like. We can mark points even beyond the origin, if the object continues it's journey in the negative x direction 
• But for our present discussion, this much is sufficient
• Let us plot the position-time graph. It is shown in fig.3.26 below:

Fig.3.26

7. The graph is a straight line. So, here also, we can calculate the velocity as slope
(i) Consider the ratio ^altitude⁄_base.
• We know that, 'altitude' in the numerator is displacement. 
(ii) But in our present experiment, displacement is in the negative direction. So it is negative.
• Thus the ratio as a whole becomes negative. So we get negative velocity
(iii) The velocity of the object in our present experiment is indeed negative because, it is travelling towards the negative side of the x axis
■ So now we know the following two items:
• Shape of the graph when an object travels with uniform velocity towards the positive side of x axis  
• Shape of the graph when an object travels with uniform velocity towards the negative side of x axis
8. Consider the instant when the object is at T. How much more time will be required for the object to reach O?
(i) Distance to be travelled more = (90-45) = 45 m
(ii) We have: v = ^x ⁄_t.
⟹ t = ^x ⁄_v. = ⁴⁵ ⁄₁₅ = 3 s
(iii) So if the object travels for 3 more seconds, it will reach O
• That means, starting from the initial point, if it travels for 6 seconds, it will reach O
• This can be verified by extending the graph beyond T, using a dotted line. This is shown in fig.3.27 below:

Fig.3.27

Relation to coordinate geometry:
• We see that the above graph is a 'decreasing graph'. 
• In coordinate geometry, such graphs have a negative slope
• That is., in the equation 'y = mx +c', the slope m will be a negative value

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An interesting case:
• Consider an object at rest at a distance of 40 m from O. 
• Whatever be the reading in the stop watch, the position will be the same 40 m. 
• So the position time graph of that object will be a horizontal line. This is shown in fig.3.28 below:

Fig.3.28

■ In general, the position time graph of an object at rest will be a line parallel to the 'time axis'   

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In the next section, we will see average velocity and average speed.

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Chapter 3.1 - Path length and Displacement

In the previous section we saw how to write the position of a object in the x axis. In this section we will see some basics about distances. Later in this section, we will see the position-time graph

Path length

Consider fig.3.6 that we saw in the previous section. For convenience it is shown again below:

Fig.3.6

We will write the steps to describe path length:
1. Positions of the car:
• Let the initial position of the car be at O
• After some time, we find that, the car is at P
• After some more time, we find that, the car is at Q
2. For the time being, we are not interested in the time that was required to travel to these different points 
• For the time being, we are also not interested in the speed with which the car travelled to these points
• Those will be discussed later
■ At present we are interested in these:
• The car was initially at O
• After some time, it was at P
• After some more time, it was at Q
3. We want to know the distance which the car travelled
• It is easy to calculate:
(i) Distance travelled from O to P = length of OP = 360 m
(ii) Distance travelled from P to Q = length of PQ = (360-240) = 120 m
(iii) Total distance travelled = (360+120) = 480 m
■ This total distance is called Path length
4. Another example of path length:
• Let the initial position be O
• After some time, the car is seen at Q
• After some more time, the car is seen at R
• Then path length = Distance travelled by the car = (OQ + QR) = (240 + 360) = 600 m

Displacement

1. 'Displacement' is the difference between the initial and final positions
• Mathematically, it can be written as Δx = (x₂-x₁)
• We use the Greek letter delta (Δ) to denote a 'change in quantity'
• So displacement is denoted as Δx
• x₂ and x₁ are mere coordinates. So it is easy to calculate 'Δx'
2. An example:
(i) The car was initially at O
(ii) After some time, it was at P
(iii) After some more time, it was at Q
(iv) Then displacement of the car from O to P = (x₂-x₁) = (360-0) = +360
(v) Displacement of the car from P to Q = (x₂-x₁) = (240-360) = -120
(vi) Displacement of the car from O to Q = (x₂-x₁) = (240-0) = +240
• If a displacement is -ve, That -ve sign should not be ignored because, displacement is a vector quantity.
♦ That is., it has both magnitude and direction
■ +ve sign indicates that the 'direction of displacement' is from left to right
♦ That is., the travel is towards the positive side of the x axis  
■ -ve sign indicates that the 'direction of displacement' is from right to left
♦ That is., the travel is towards the negative side of the x axis
3. A comparison between path length and displacement:
(i) Consider the travel from O to P:
• Path length = OP = 360 m
• Displacement = Δx = (x₂-x₁) = (360-0) = 360
(ii) So in this case, path length = 'magnitude of displacement'
Another example:
(i) Consider the travel from O to P and then back to Q:
• Path length = (OP+PQ) = (360+120) = 480 m
• Displacement from O to Q = Δx = (x₂-x₁) = (240-0) = 240
(ii) So in this case, path length is not equal to 'magnitude of displacement'
One more example:
(i) Consider the travel from O to P and then back to O:
• Path length = (OP+PO) = (360+360) = 720 m
• Displacement from O to O = Δx = (x₂-x₁) = (0-0) = 0
(ii) So in this case also, path length is not equal to 'magnitude of displacement'
(iii) Also note that even after travelling much distances, 'magnitude of displacement' can become zero. But path length cannot be zero

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• Now we know 'how to specify the positions of any object' on a straight line.
• And, from those positions, we can calculate distances (path lengths and displacements)
• But the method that we learned is applicable to only those objects which are at rest. For example, 
    ♦ When the car was at rest at P (360 m away from O), we noted down it's position
    ♦ When the car was at rest at Q (240 m away from O), we noted down it's position
■ Suppose the car is moving continuously, what position will we mark?
• We are unable to mark any single point because, the position is changing continuously
• To solve this problem, we introduce 'time' also into our reference frame 
• Let us learn how to connect 'time' and 'position':
• We will learn it using an example:
1. Let a car start from rest from the origin O
■ Let us turn on a stop watch at the same instant when the car begins to move
So we can write:
• At the instant when the car begins to move, it's position is O. 
    ♦ That is., x = 0 
• At the instant when the car begins to move, time shown by the stop watch is '0 seconds'. 
    ♦ That is., t = 0
2. So we have two information: x = 0 and t = 0.
• They are marked on the x axis in fig.3.7 below:

Fig.3.7

3. Let us travel parallel to the car in another vehicle. The stop watch is running.
• We see a mark P on the path. 
• At the instant when the car passes P, the reading shown by the watch is 9 s. Note it down.   
• It is shown in fig.3.8 below:

Fig.3.8

4. We see a mark Q on the path. 
• At the instant when the car passes Q, the reading shown by the watch is 16 s. Note it down.
• It is shown in fig.3.9 below:

Fig.3.9

5. We see a mark R on the path. 
• At the instant when the car reaches R, the reading shown by the watch is 21 s. Note it down
• It is shown in fig.3.10 below:

Fig.3.10

• In this way, we can note down the readings at any number of convenient points
6. Next step is to measure the distances. We want the following distances:
    ♦ Distance of P from O. That is., OP
    ♦ Distance of Q from O. That is., OQ
    ♦ Distance of R from O. That is., OR
• Those distances are measured and marked in the fig.3.11 shown below:
• Note that, these are not distances between points. These are distances from O

Fig.3.11

• The field work is complete. Next we have to do some office work. 
■ Note that, points P, Q and R were pre-marked. There is no rule regarding 'where to mark the points'. We can make marks at any convenient points.
7. Now we begin the office work.
• On a fresh graph paper, mark the x and y axes. 
    ♦ The x axis shows time (s)
    ♦ The y axis shows distance (m)
• Choose any convenient scale. The following scales would be appropriate:
    ♦ For x axis: 1 cm = 1 s
    ♦ For y axis: 1 cm = 10 m
• The resulting graph is shown in fig.3.12 below:

Fig.3.12

8. We obtained the above graph using the following 3 steps:
(i) Plot the points O(0,0), P(9,63), Q(16,112) and R(21,147)
(ii) Join the 4 points 
(iii) The lines connecting the four points is the required graph
9. We find that it is a single straight line. 
• That is., if we draw a line connecting O and the last point R, the other two points P and Q will lie on that line. (The reader may check this in his/her own graph)
• Why do we get the graph as a single line?
    ♦ We will see the answer in the next section
10. At present we are more concerned about 'specifying position of a moving object'
• If someone asks us about position, we are ready to give answers
■ So what is the position of the car?
• The answer will consist of the following 5 statements:
(i) The car moved along the x axis during it's entire journey 
(ii) It started from O when t = 0
(iii) After 9 seconds, it reached P which is 63 m away from O
(iv) After 7 more seconds, it reached Q which is 112 m away from O
(v) After 5 more seconds, it reached R which is 147 m away from O
11. Another type of question may arise:
An example:
■ What is the position of the car when it travelled for 12 seconds?
• To answer this question, we draw a vertical dashed line through t = 12
This is shown in fig.3.13 below:

Fig.3.13

• This dashed line meets the yellow graph at S.
• Through S, draw a horizontal dashed line. Let it meet the y axis at T
• The y coordinate of T is 84 (The reader may check this in his/her own graph)
• So we can write: After 12 s, the car is 84 m away from O    
12. Thus we are able to specify the position of a moving object. 
• We are able to do so, with the help of the graph in fig.3.12. 
• This graph is called the position-time graph.

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In the next section, we will see some applications of the position-time graph.

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