In the previous section, we learned how to obtain the 'distance travelled by an object during a particular duration of time' from the Distance-Time Graph. In this section we will see another application.
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 (12,360) will have the y coordinate 360
7. G lies on the vertical through D. Any point on the vertical through D (18,540) will have the x coordinate 18
8. The point G lies on both the above horizontal and vertical. So the coordinates of G are (18,360)
9. Consider the two points C and G.
• C is at a horizontal distance of 12 from the y-axis
• G is at a horizontal distance of 18 from the y-axis
• So the horizontal distance between C and G
= The base of the ⊿CGD
= 18 -12 = 6 minutes
10. Consider the two points G and D.
• D is at a vertical distance of 540 from the x-axis
• G is at a vertical distance of 360 from the x-axis
• So the vertical distance between G and D
= The altitude of the ⊿CGD
= 540 - 360 = 180 metres
11. Now take the ratio altitude⁄base
We get 180 m⁄6 min = 30 m⁄1 min = 30 m⁄60 s = 0.5 m s-1
12. This was the speed at which the object travelled from C to D
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 'distances of C and D from O' (We took them from table 1.1)
• So the difference is actually the distance between C and D
(iii) Now consider the base.
• We calculated the 'difference between the x coordinates' to find the base
• But the x coordinates are the 'times for reaching C and D, after passing O' (We took them from table 1.1)
• So the difference is actually the time of travel between C and D
■ So, the altitude is the distance between C and D
■ Base is the time required to travel from C to D
(iv) Their ratio is the speed at which the object travelled from C to D (∵ speed = distance⁄time)
We get 369 m⁄12.3 min = 30 m⁄1 min = 30 m⁄60 s = 0.5 m s-1
12. This was the speed at which the object travelled from U to W
The explanation was given in the previous example.
• In fig.1.14, we find that, the object travelled from U to W at a speed of 0.5 m s-1
• In fig.1.13, we find that, the object travelled from C to D at a speed of 0.5 m s-1
• 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 0.5 m s-1
• The reader is advised to select random pairs, and do the calculations to confirm it.
■ Why do we get the same speed in all cases?
Ans:
• The graph from O to F 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.
• In our present case, this ratio is 30 m⁄1 min which works out to 0.5 m s-1
2. Speed was not recorded
3. When the distance-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 speed
6. And the uniform speed was calculated as 0.5 m s-1
■ Whenever we get a straight line for the distance-time graph, we can confirm that the object moved at uniform speed
■ And this speed can be calculated using the formula:
Eq.1.4:
This formula is based on the fig.1.16 below:
• t1 and t2 are the times
• s1 and s2 are the distances
• base of the triangle is t2 - t1
• Altitude of the triangle is s2 - s1
• Speed = altitude⁄base . Hence we get the Eq.1.4 above.
■ We have completed the discussion on Distance-Time graph. Note that, in the experiment, if the object was travelling in a straight line, we can use 'velocity' in place of 'speed.
In the next section, we will see Velocity-Time graph.
Speed from Distance-Time Graph
We are discussing about the distance-time graph. In this graph, no information seems to be available about the speed at which the object travelled. But in fact, the speed is 'hiding' inside the graph. If we want to know the speed, we will have to bring it out. Let us see how:
1. Take any two convenient points on the original graph in fig.1.11. Let us take C (12,360) and D (18, 540). This is shown in fig.1.13 below
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| Fig.1.13 |
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 (12,360) will have the y coordinate 360
7. G lies on the vertical through D. Any point on the vertical through D (18,540) will have the x coordinate 18
8. The point G lies on both the above horizontal and vertical. So the coordinates of G are (18,360)
9. Consider the two points C and G.
• C is at a horizontal distance of 12 from the y-axis
• G is at a horizontal distance of 18 from the y-axis
• So the horizontal distance between C and G
= The base of the ⊿CGD
= 18 -12 = 6 minutes
10. Consider the two points G and D.
• D is at a vertical distance of 540 from the x-axis
• G is at a vertical distance of 360 from the x-axis
• So the vertical distance between G and D
= The altitude of the ⊿CGD
= 540 - 360 = 180 metres
11. Now take the ratio altitude⁄base
We get 180 m⁄6 min = 30 m⁄1 min = 30 m⁄60 s = 0.5 m s-1
12. This was the speed at which the object travelled from C to D
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 'distances of C and D from O' (We took them from table 1.1)
• So the difference is actually the distance between C and D
(iii) Now consider the base.
• We calculated the 'difference between the x coordinates' to find the base
• But the x coordinates are the 'times for reaching C and D, after passing O' (We took them from table 1.1)
• So the difference is actually the time of travel between C and D
■ So, the altitude is the distance between C and D
■ Base is the time required to travel from C to D
(iv) Their ratio is the speed at which the object travelled from C to D (∵ speed = distance⁄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 O, A, B, C, D, E and F. In such a case, we will have to find the coordinates ourselves
Consider fig.1.14 below.
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| Fig.1.14 |
1. Two random points U and W are marked on the graph. To find the coordinates of these points we do the following:
• 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 306 and 675
• The vertical lines meet the x-axis at 10.2 and 22.5
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 = 22.5-10.2 = 12.3 min
• The altitude of ⊿UVW = 675-306 = 369 m
6. Take the ratio altitude⁄baseWe get 369 m⁄12.3 min = 30 m⁄1 min = 30 m⁄60 s = 0.5 m s-1
12. This was the speed at which the object travelled from U to W
The explanation was given in the previous example.
• In fig.1.13, we find that, the object travelled from C to D at a speed of 0.5 m s-1
• 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 0.5 m s-1
• The reader is advised to select random pairs, and do the calculations to confirm it.
■ Why do we get the same speed in all cases?
Ans:
• The graph from O to F 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.
• In our present case, this ratio is 30 m⁄1 min which works out to 0.5 m s-1
If the graph had any bends or curves, we will not get the same speed always. An example is given in fig.1.15 below:
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| Fig.1.15 |
• We can see that the triangles are not similar. Details about simialar triangles can be seen here.
• So the ratio altitude⁄base will be different
Let us write a summary of the above discussion:
1. In the experiment, only time and distance were recorded in the field2. Speed was not recorded
3. When the distance-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 speed
6. And the uniform speed was calculated as 0.5 m s-1
■ Whenever we get a straight line for the distance-time graph, we can confirm that the object moved at uniform speed
■ And this speed can be calculated using the formula:
Eq.1.4:
This formula is based on the fig.1.16 below:
 |
| Fig.1.16 |
• s1 and s2 are the distances
• base of the triangle is t2 - t1
• Altitude of the triangle is s2 - s1
• Speed = altitude⁄base . Hence we get the Eq.1.4 above.
■ We have completed the discussion on Distance-Time graph. Note that, in the experiment, if the object was travelling in a straight line, we can use 'velocity' in place of 'speed.
In the next section, we will see Velocity-Time graph.
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