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:
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:
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:
• 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.
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:
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.
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
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
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:
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:
• 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:
• 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
In the next section, we will see a few more solved examples.
• 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 |
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 |
• 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 |
♦ 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.
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 |
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.
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
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
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:
Solution:
The required distance time graph is shown in fig.1.28 below:
 |
| Fig.1.28 |
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.
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 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 |
• 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
In the next section, we will see a few more solved examples.
