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
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
• 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
• 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:
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:
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:
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:
• 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
• 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:
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:
• 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.
Path length
Consider fig.3.6 that we saw in the previous section. For convenience it is shown again below: |
| Fig.3.6 |
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
• 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 |
• 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 |
• 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 |
• 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 |
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 |
■ 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 |
(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 |
• 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.
In the next section, we will see some applications of the position-time graph.
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