In the previous section, we completed the discussion on Distance-Time graph. In this section we will see another type of graph.
Let us look at the table in some detail:
1. The driver and the assistant starts the engine and begin the travel along a straight line
2. They continue their travel. The experiment has not begun yet
3. After some time the assistant starts his stop watch. At that instant, the experiment has begun
4. At that instant. the time is zero
5. So we see that, when the experiment began, time is zero. But velocity is not zero
• When the stop watch shows 5 s, the assistant notes down the reading of the speedometer. It is 40 kmph
• When the stop watch shows 10 s, the assistant notes down the reading of the speedometer. It is 40 kmph
• When the stop watch shows 15 s, the assistant notes down the reading of the speedometer. It is 40 kmph
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- - -
• The readings are taken upto 30 s. Thus we get Table 1.2 above
• We must not be discouraged on seeing 40 kmph every time. In science and engineering, all readings are important. The constant reading simply shows that the car was travelling at a constant speed of 40 kmph since the beginning of the experiment.
Let us make a graph. The table 1.2 above shows the recordings made in the field. As the time is in seconds, we will convert the velocity from kmph to m s-1. The modified table is given below:
The fig. below shows the graph based on table 1.3
• Time is taken along the x-axis and velocity is taken along the y-axis
• The reader may plot the above graph on a fresh graph paper. Any convenient scale can be used. The following scale is also adequate:
♦ X axis: 1 cm represents 2 s
♦ Y axis: 1 cm represents 1 m s-1
We see the following features:
1. All points fall on a straight line
2. This straight line is parallel to the x-axis
3. This straight line meets the y-axis at 11.1
■ So we can infer this:
When the velocity is a constant, the velocity-time graph is a straight line parallel to the x-axis.
Now let us see an application of the above graph:
1. Draw two vertical lines shown in purple colour in the fig.1.18 below:
• One through time t1 = 8 s
• One through time t2 = 22 s
2. Draw them upwards until they meet the horizontal yellow line. Now we get a rectangle which is shown shaded in the fig.1.18
• The height of the rectangle is equal to the constant velocity v, which is equal to 11.1 m s-1
• The width of the rectangle = (t2 – t1) = (22 – 8) = 14 s
• So area of the rectangle = 11.1 m s-1 × 14 s = 155.4 m
3. What is the peculiarity of this area?
Ans:
• To find the area, we have multiplied a velocity v with a time duration of 14 seconds
• We know that when we multiply a velocity with a time duration, we get the distance 's' travelled during that time duration. That is., v×t = s
• So the shaded area in fig.1.18 is the distance travelled by the car in a time duration of 14 seconds starting from t = 8 s
■ In this manner we can find the distance during any time interval by drawing a suitable rectangle.
■ Note that, we did not take any 'reading of distances' during the experiment. But after the experiment, we are able to find any required distance that we want, from the graph.
Let us repeat the experiment:
The procedure is the same. But this time, the velocities are different. The field recordings are shown below in table 1.4
The modified table with velocities converted into m s-1 is given below:
The resulting graph is:
• In the graph, we see that, as the time increases, the velocity is also increasing.
• Take any second from the graph. The velocity at that second will be greater than the velocity at the previous second.
• Such an increase in velocity can not be achieved with out acceleration.
• But no recordings related to acceleration were made at the field.
• In the graph also, no information seems to be available about the acceleration.
• But in fact, the acceleration is hiding inside the graph. If we want to know the acceleration, we will have to bring it out.
• Recall how we brought out the ‘speed’ which was hiding in the 'distance-time graph'. Details here. We will be using a similar method:
1. Consider the above graph in fig.1.19. Take any two convenient points on that graph. Let us take (10,10) and (20,15). Let us name them as C and D. This is shown in fig.1.20 below:
2. Draw a horizontal line through the lower point C
3. Draw a vertical line through the higher point D
4. These two lines will meet at a point. Name this point as G
5. So we get a triangle CGD. We want the base CG and altitude GD of this triangle. Let us find them:
6. G lies on the horizontal through C. Any point on the horizontal through C (10,10) will have the y coordinate 10
7. G lies on the vertical through D. Any point on the vertical through D (20,15) will have the x coordinate 20
8. The point G lies on both the above horizontal and vertical. So the coordinates of G are (20,10)
9. Consider the two points C and G.
• C is at a horizontal distance of 10 from the y-axis
• G is at a horizontal distance of 20 from the y-axis
• So the horizontal distance between C and G
= The base of the ⊿CGD
= 20 -10 = 10 s
10. Consider the two points G and D.
• D is at a vertical distance of 15 from the x-axis
• G is at a vertical distance of 10 from the x-axis
• So the vertical distance between G and D
= The altitude of the ⊿CGD
= 15 - 10 = 5 m s-1
11. Now take the ratio altitude⁄base
We get 5 ms-1⁄10 s = 0.5 m s-2
12. The car was given this much acceleration for a duration of 10 seconds from t = 10 s
Explanation:
(i) We formed the triangle CGD, and took the ratio altitude⁄base
(ii) Consider the altitude.
• We calculated the 'difference between the y coordinates' to find the altitude
• But the y coordinates are the 'velocities at t = 10 and t = 20' (We took them from table 1.5 above)
• So the difference is actually the 'change in velocity'
(iii) Now consider the base.
• We calculated the 'difference between the x coordinates' to find the base
• But the x coordinates are the 'times of travel' (We took them from table 1.5 above)
• So the difference is actually the duration of time in which change of velocity from 10 m s-1 to 20 m s-1 took place.
■ So, the altitude is the change in velocity
■ Base is the duration of time in which the change in velocity took place
(iv) Their ratio is the acceleration required for the change in velocity from 10 m s-1 to 20 m s-1 in 10 seconds (∵ acceleration = change in velocity⁄time)
• In the above calculations, we took two known points C and D. The x and y coordinates are already known to us because, they were recorded in the field.
2. Acceleration was not recorded
3. When the velocity-time graph was plotted, it was seen that all the points fall on a straight line
4. So the graph do not have any bends or curves
5. Because of the 'straight nature' of the curve, it was confirmed that, the object travelled at an uniform acceleration
6. And the uniform acceleration was calculated as 0.5 m s-2
■ Whenever we get a straight line for the velocity-time graph, we can confirm that the object moved with uniform acceleration
■ And this acceleration can be calculated using the formula:
Eq.1.5:
• Where a is the acceleration
• u is the initial velocity
• v is the final velocity
• t is the time taken for the velocity to increase from u to v
■ This formula is based on the fig.1.23 below:
• t1 and t2 are the times. Their difference 't' will give the duration in which the initial velocity u increases to the final velocity v
• base of the triangle is t = t2 - t1
• Altitude of the triangle is (v - u)
• Acceleration = altitude⁄base . Hence we get the Eq.1.5 above.
■ We have obtained Eq.1.5 above. From that equation we will get: (v-u) = at
From this we get the equation below:
Eq.1.6:
v = u + at
This is the first equation of motion.
• A body was moving with a uniform acceleration 'a'
• At time t1, it's velocity was u
• At time t2, it's velocity increased to v
♦ This increase was due to the uniform acceleration 'a'.
♦ This increase happened in a duration 't' = (t2-t1)
• In such a situation, we can calculate the final velocity v of the object using the Eq.1.6
In the next section, we will see another equation.
Velocity-Time Graph
Let us see another experiment. A car is being driven along a straight road. After some time, a person sitting next to the driver watches the speedometer of the car and notes down the velocity every 5 seconds. Here is what he got:
 |
| Table.1.2 |
1. The driver and the assistant starts the engine and begin the travel along a straight line
2. They continue their travel. The experiment has not begun yet
3. After some time the assistant starts his stop watch. At that instant, the experiment has begun
4. At that instant. the time is zero
5. So we see that, when the experiment began, time is zero. But velocity is not zero
• When the stop watch shows 5 s, the assistant notes down the reading of the speedometer. It is 40 kmph
• When the stop watch shows 10 s, the assistant notes down the reading of the speedometer. It is 40 kmph
• When the stop watch shows 15 s, the assistant notes down the reading of the speedometer. It is 40 kmph
- - -
- - -
• The readings are taken upto 30 s. Thus we get Table 1.2 above
• We must not be discouraged on seeing 40 kmph every time. In science and engineering, all readings are important. The constant reading simply shows that the car was travelling at a constant speed of 40 kmph since the beginning of the experiment.
Let us make a graph. The table 1.2 above shows the recordings made in the field. As the time is in seconds, we will convert the velocity from kmph to m s-1. The modified table is given below:
 |
| Table.1.3 |
• Time is taken along the x-axis and velocity is taken along the y-axis
 |
| Fig.1.17 |
♦ X axis: 1 cm represents 2 s
♦ Y axis: 1 cm represents 1 m s-1
We see the following features:
1. All points fall on a straight line
2. This straight line is parallel to the x-axis
3. This straight line meets the y-axis at 11.1
■ So we can infer this:
When the velocity is a constant, the velocity-time graph is a straight line parallel to the x-axis.
Now let us see an application of the above graph:
1. Draw two vertical lines shown in purple colour in the fig.1.18 below:
• One through time t1 = 8 s
• One through time t2 = 22 s
 |
| Fig.1.18 |
• The height of the rectangle is equal to the constant velocity v, which is equal to 11.1 m s-1
• The width of the rectangle = (t2 – t1) = (22 – 8) = 14 s
• So area of the rectangle = 11.1 m s-1 × 14 s = 155.4 m
3. What is the peculiarity of this area?
Ans:
• To find the area, we have multiplied a velocity v with a time duration of 14 seconds
• We know that when we multiply a velocity with a time duration, we get the distance 's' travelled during that time duration. That is., v×t = s
• So the shaded area in fig.1.18 is the distance travelled by the car in a time duration of 14 seconds starting from t = 8 s
■ In this manner we can find the distance during any time interval by drawing a suitable rectangle.
■ Note that, we did not take any 'reading of distances' during the experiment. But after the experiment, we are able to find any required distance that we want, from the graph.
Let us repeat the experiment:
The procedure is the same. But this time, the velocities are different. The field recordings are shown below in table 1.4
 |
| Table.1.4 |
 |
| Table.1.5 |
 |
| Fig.1.19 |
• Take any second from the graph. The velocity at that second will be greater than the velocity at the previous second.
• Such an increase in velocity can not be achieved with out acceleration.
• But no recordings related to acceleration were made at the field.
• In the graph also, no information seems to be available about the acceleration.
• But in fact, the acceleration is hiding inside the graph. If we want to know the acceleration, we will have to bring it out.
• Recall how we brought out the ‘speed’ which was hiding in the 'distance-time graph'. Details here. We will be using a similar method:
1. Consider the above graph in fig.1.19. Take any two convenient points on that graph. Let us take (10,10) and (20,15). Let us name them as C and D. This is shown in fig.1.20 below:
 |
| Fig.1.20 |
3. Draw a vertical line through the higher point D
4. These two lines will meet at a point. Name this point as G
5. So we get a triangle CGD. We want the base CG and altitude GD of this triangle. Let us find them:
6. G lies on the horizontal through C. Any point on the horizontal through C (10,10) will have the y coordinate 10
7. G lies on the vertical through D. Any point on the vertical through D (20,15) will have the x coordinate 20
8. The point G lies on both the above horizontal and vertical. So the coordinates of G are (20,10)
9. Consider the two points C and G.
• C is at a horizontal distance of 10 from the y-axis
• G is at a horizontal distance of 20 from the y-axis
• So the horizontal distance between C and G
= The base of the ⊿CGD
= 20 -10 = 10 s
10. Consider the two points G and D.
• D is at a vertical distance of 15 from the x-axis
• G is at a vertical distance of 10 from the x-axis
• So the vertical distance between G and D
= The altitude of the ⊿CGD
= 15 - 10 = 5 m s-1
11. Now take the ratio altitude⁄base
We get 5 ms-1⁄10 s = 0.5 m s-2
12. The car was given this much acceleration for a duration of 10 seconds from t = 10 s
Explanation:
(i) We formed the triangle CGD, and took the ratio altitude⁄base
(ii) Consider the altitude.
• We calculated the 'difference between the y coordinates' to find the altitude
• But the y coordinates are the 'velocities at t = 10 and t = 20' (We took them from table 1.5 above)
• So the difference is actually the 'change in velocity'
(iii) Now consider the base.
• We calculated the 'difference between the x coordinates' to find the base
• But the x coordinates are the 'times of travel' (We took them from table 1.5 above)
• So the difference is actually the duration of time in which change of velocity from 10 m s-1 to 20 m s-1 took place.
■ So, the altitude is the change in velocity
■ Base is the duration of time in which the change in velocity took place
(iv) Their ratio is the acceleration required for the change in velocity from 10 m s-1 to 20 m s-1 in 10 seconds (∵ acceleration = change in velocity⁄time)
• In the above calculations, we took two known points C and D. The x and y coordinates are already known to us because, they were recorded in the field.
• Let us take two points other than those recorded in the field. In such a case, we will have to find the coordinates ourselves
Consider fig.1.21 below.
1. Two random points U and W are marked on the graph. To find the coordinates of these points we do the following:
We get 6.25 ms-1⁄12.5 s = 0.5 m s-2
12. This was the acceleration at which the object travelled from U to W
The explanation was given in the previous example.
• In fig.1.21, we find that, the object travelled from U to W at an acceleration of 0.5 m s-2
• In fig.1.20, we find that, the object travelled from C to D at an acceleration of 0.5 m s-2
• We can take any pair of points we like on the graph.
♦ A first point and a second point.
♦ In all cases we will get the same result:
♦ The object travelled from the first point to second point at an acceleration of 0.5 m s-2
• The reader is advised to select random pairs, and do the calculations to confirm it.
■ Why do we get the same acceleration in all cases?
Ans:
• The graph is a straight line with no bends or curves.
• For every pair of points, the triangles formed will be similar.
• So the ratio altitude⁄base is always a constant.
 |
| Fig.1.21 |
• Draw two horizontal lines. One through U and the other through W
• Draw two vertical lines. One through U and the other through W
2. • The horizontal lines meet the y-axis at 9.25 and 15.5
• The vertical lines meet the x-axis at 8.5 and 21
3. Extend the bottom horizontal line towards the right until it intersects the vertical through W
4. Name the point of intersection as V
5. Thus we get ⊿UVW
• The base of ⊿UVW = 21- 8.5 = 12.5 s
• The altitude of ⊿UVW = 15.5 - 9.25 = 6.25 m s-1
6. Take the ratio altitude⁄baseWe get 6.25 ms-1⁄12.5 s = 0.5 m s-2
12. This was the acceleration at which the object travelled from U to W
The explanation was given in the previous example.
• In fig.1.21, we find that, the object travelled from U to W at an acceleration of 0.5 m s-2
• In fig.1.20, we find that, the object travelled from C to D at an acceleration of 0.5 m s-2
• We can take any pair of points we like on the graph.
♦ A first point and a second point.
♦ In all cases we will get the same result:
♦ The object travelled from the first point to second point at an acceleration of 0.5 m s-2
• The reader is advised to select random pairs, and do the calculations to confirm it.
■ Why do we get the same acceleration in all cases?
Ans:
• The graph is a straight line with no bends or curves.
• For every pair of points, the triangles formed will be similar.
• So the ratio altitude⁄base is always a constant.
If the graph had any bends or curves, we will not get the same acceleration always. An example is given below:
• We can see that the triangles are not similar.
• So the ratio altitude⁄base will be different |
| Fig.1.22 |
Let us write a summary of the above discussion:
1. In the experiment, only time and velocity were recorded in the field2. Acceleration was not recorded
3. When the velocity-time graph was plotted, it was seen that all the points fall on a straight line
4. So the graph do not have any bends or curves
5. Because of the 'straight nature' of the curve, it was confirmed that, the object travelled at an uniform acceleration
6. And the uniform acceleration was calculated as 0.5 m s-2
■ Whenever we get a straight line for the velocity-time graph, we can confirm that the object moved with uniform acceleration
■ And this acceleration can be calculated using the formula:
Eq.1.5:
• u is the initial velocity
• v is the final velocity
• t is the time taken for the velocity to increase from u to v
■ This formula is based on the fig.1.23 below:
 |
| Fig.1.23 |
• base of the triangle is t = t2 - t1
• Altitude of the triangle is (v - u)
• Acceleration = altitude⁄base . Hence we get the Eq.1.5 above.
■ We have obtained Eq.1.5 above. From that equation we will get: (v-u) = at
From this we get the equation below:
Eq.1.6:
v = u + at
This is the first equation of motion.
• A body was moving with a uniform acceleration 'a'
• At time t1, it's velocity was u
• At time t2, it's velocity increased to v
♦ This increase was due to the uniform acceleration 'a'.
♦ This increase happened in a duration 't' = (t2-t1)
• In such a situation, we can calculate the final velocity v of the object using the Eq.1.6
In the next section, we will see another equation.
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