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:
• 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:
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:
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:
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
• 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:
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:
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)
(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
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:
• 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:
• 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:
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:
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:
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:
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:
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
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:
■ In general, the position time graph of an object at rest will be a line parallel to the 'time axis'
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 |
♦ 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 |
• 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 |
• 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 |
• 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
• 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 |
(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 |
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 |
• 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
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 |
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 |
♦ 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 |
• 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 |
• 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 |
• 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 |
(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 |
• 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
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 the next section, we will see average velocity and average speed.
Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved
No comments:
Post a Comment