In the previous section we saw some applications of the position-time graph. In this section we will see average speed and average velocity.
• In the previous section we saw an object moving with a uniform velocity.
• But in day to day life, we often see objects moving with non-uniform velocity.
■ Consider the object that we saw in the previous section.
• We turned the stop watch on only when the object reached P.
♦ From P onwards, it travelled with a uniform velocity of 12 ms-1.
• But the object started it's journey from O. It started from rest at O.
■ If started from rest at O, how did it attain a velocity of 12 ms-1?
(ii) When it is at P, the velocity is 12 ms-1
(iii) So the velocity increased from zero to 12 ms-1
(iv) How did this increase happen?
• To find the answer, we will do another experiment. This time, we observe the motion of a car
• Also this time, will turn on the stop watch at the very instant when the car starts it's motion from O
• We will need the help of the odometer or trip meter of the car
(Trip meter is more preferable because, it can be reset to zero)
Let us begin:
(i) At the instant when the car begins it's motion, t = 0, x = 0
(ii) When t = 1 s, the trip meter reading is 1.8 m
• That means, in the first second, the car moved 1.8 m.
• Let this new position be A.
♦ Then OA = 1.8 m
This is shown in fig.3.29 below:
(iii) When t = 2 s, the trip meter reading is 7.2 m
• That means, in the second second, the car moved (7.2 - 1.8) = 5.4 m.
• Let this new position be B. Then:
♦ AB = 5.4 m
♦ OB = 7.2 m
This is shown in fig.3.23 above
(iv) When t = 3 s, the trip meter reading is 16.2 m
• That means, in the third second, the car moved (16.2 - 7.2) = 9 m.
• Let this new position be C. Then:
♦ BC = 9 m
♦ OC = 16.2 m
This is shown in fig.3.23 above
(v) When t = 4 s, the trip meter reading is 28.8 m
• That means, in the fourth second, the car moved (28.8 - 16.2) = 12.6 m.
• Let this new position be D. Then:
♦ CD = 12.6 m
♦ OD = 28.8 m
This is shown in fig.3.23 above
(vi) When t = 5 s, the trip meter reading is 45 m
• That means, in the fifth second, the car moved (45 - 28.8) = 16.2 m.
• Let this new position be E. Then:
♦ DE = 16.2 m
♦ OE = 45 m
This is shown in fig.3.23 above
(vii) When t = 6 s, the trip meter reading is 64.8 m
• That means, in the sixth second, the car moved (64.8 - 45) = 19.8 m.
• Let this new position be E. Then:
♦ EF = 19.8 m
♦ OF = 64.8 m
This is shown in fig.3.23 above
(viii) When t = 7 s, the odometer reading is 88.2 m
• That means, in the seventh second, the car moved (45 - 28.8) = 23.4 m.
• Let this new position be E. Then:
♦ EG = 23.4 m
♦ OG = 88.2 m
From the above values, the following two points become very clear:
(i) In the previous experiment,
♦ The time durations are all same: 1 s
♦ The distance travelled in each of these durations are the same: 12 m
• That means the object travelled with a uniform velocity
(ii) In the present experiment
♦ The time durations are all same: 1 s
♦ But the distance travelled in each of these durations are not the same
• That means, the object travelled with non-uniform velocity
Let us plot the position-time graph. It is shown in fig.3.30 below:
• To avoid congestion, coordinates of only alternate points are written.
■ We see that, the graph is a curve
• We have seen that, slope of the position-time graph will give the velocity
• If the object moves with uniform velocity, the position-time graph will be a straight line
• Then we can easily calculate the slope. That slope will be same at all points
• But here, the graph is a curve. The slope will be different at different points.
• That means, the velocity is different at different points
• In such a situation, we use average velocity.
We can explain this using an example:
1. Consider the points D and F
• The car passed through both D and F
2. Suppose someone wants to know the 'velocity with which the car travelled between D and F'
• We cannot give a definite answer. Because the velocity continuously changed
3. In such a situation we take the ratio: Displacement from D to F⁄Time required for the travel from D to F
• We know that:
♦ Displacement from D to F = Δx = (x2-x1) = (y coordinate of F - y coordinate of D)
♦ = (64.8 - 28.8) = 36
Time required to travel from D to F = Δt = (t2-t1) = (x coordinate of F - x coordinate of D)
♦ = (6 - 4) = 2
• So the ratio is 36⁄2 = 18 ms-1.
4. This velocity of 18 ms-1 is the average velocity with which the object travelled from D to F.
We can write the definition:
■ Average velocity is defined as the displacement (Δx) divided by the time duration (Δt) in which the displacement occurs
• The formula is:
♦ The bar over v is a standard notation used to indicate average quantities
5. We can check whether this formula is dimensionally correct:
• On the left side we have velocity [LT-1]
• On the right side,
♦ in numerator we have distance [L]
♦ in denominator we have distance [T]
• Thus we get [LT-1] on both sides
We have seen that, if the velocity is uniform, we can easily calculate the velocity as the slope
■ Can we relate average velocity to slope?
Let us try:
• The portion between D and F is enlarged and shown in fig.3.31 below:
• We know that DF is a curve. But now, a straight line is draw between them in cyan colour
• Also a triangle DFF' is formed
• Now take the ratio: altitude⁄base
• We get: altitude⁄base = FF'⁄DF' = (64.8-28.8)⁄(6-4) = 36⁄2 = 18 ms-1
So we can write:
■ Average velocity is the slope of the line joining the initial and final positions of the object in the position-time graph
• We have seen the details about average velocity.
• Average speed can also be calculated in a similar way.
• But in the numerator, instead of 'displacement', we use 'path length'. Thus we write:
Average speed =
Total path length between two points⁄Time required for the travel between the two points
■ Average velocity is a vector quantity. It has both magnitude and direction
■ Average speed is a scalar quantity. It has magnitude but no direction
The difference between the two can be understood using an example. We will write it in steps:
1. Fig.3.32 below shows the positions of an object at various times shown by a stop watch
From the fig., the following information can be obtained:
• In the previous section we saw an object moving with a uniform velocity.
• But in day to day life, we often see objects moving with non-uniform velocity.
■ Consider the object that we saw in the previous section.
• We turned the stop watch on only when the object reached P.
♦ From P onwards, it travelled with a uniform velocity of 12 ms-1.
• But the object started it's journey from O. It started from rest at O.
■ If started from rest at O, how did it attain a velocity of 12 ms-1?
We can put this question in another way using the following 4 statements:
(i) At the instant when the object started from rest, the velocity will be zero(ii) When it is at P, the velocity is 12 ms-1
(iii) So the velocity increased from zero to 12 ms-1
(iv) How did this increase happen?
• To find the answer, we will do another experiment. This time, we observe the motion of a car
• Also this time, will turn on the stop watch at the very instant when the car starts it's motion from O
• We will need the help of the odometer or trip meter of the car
(Trip meter is more preferable because, it can be reset to zero)
Let us begin:
(i) At the instant when the car begins it's motion, t = 0, x = 0
(ii) When t = 1 s, the trip meter reading is 1.8 m
• That means, in the first second, the car moved 1.8 m.
• Let this new position be A.
♦ Then OA = 1.8 m
This is shown in fig.3.29 below:
 |
| Fig.3.29 |
• That means, in the second second, the car moved (7.2 - 1.8) = 5.4 m.
• Let this new position be B. Then:
♦ AB = 5.4 m
♦ OB = 7.2 m
This is shown in fig.3.23 above
(iv) When t = 3 s, the trip meter reading is 16.2 m
• That means, in the third second, the car moved (16.2 - 7.2) = 9 m.
• Let this new position be C. Then:
♦ BC = 9 m
♦ OC = 16.2 m
This is shown in fig.3.23 above
(v) When t = 4 s, the trip meter reading is 28.8 m
• That means, in the fourth second, the car moved (28.8 - 16.2) = 12.6 m.
• Let this new position be D. Then:
♦ CD = 12.6 m
♦ OD = 28.8 m
This is shown in fig.3.23 above
(vi) When t = 5 s, the trip meter reading is 45 m
• That means, in the fifth second, the car moved (45 - 28.8) = 16.2 m.
• Let this new position be E. Then:
♦ DE = 16.2 m
♦ OE = 45 m
This is shown in fig.3.23 above
(vii) When t = 6 s, the trip meter reading is 64.8 m
• That means, in the sixth second, the car moved (64.8 - 45) = 19.8 m.
• Let this new position be E. Then:
♦ EF = 19.8 m
♦ OF = 64.8 m
This is shown in fig.3.23 above
(viii) When t = 7 s, the odometer reading is 88.2 m
• That means, in the seventh second, the car moved (45 - 28.8) = 23.4 m.
• Let this new position be E. Then:
♦ EG = 23.4 m
♦ OG = 88.2 m
From the above values, the following two points become very clear:
(i) In the previous experiment,
♦ The time durations are all same: 1 s
♦ The distance travelled in each of these durations are the same: 12 m
• That means the object travelled with a uniform velocity
(ii) In the present experiment
♦ The time durations are all same: 1 s
♦ But the distance travelled in each of these durations are not the same
• That means, the object travelled with non-uniform velocity
Let us plot the position-time graph. It is shown in fig.3.30 below:
 |
| Fig.3.30 |
■ We see that, the graph is a curve
• We have seen that, slope of the position-time graph will give the velocity
• If the object moves with uniform velocity, the position-time graph will be a straight line
• Then we can easily calculate the slope. That slope will be same at all points
• But here, the graph is a curve. The slope will be different at different points.
• That means, the velocity is different at different points
• In such a situation, we use average velocity.
We can explain this using an example:
1. Consider the points D and F
• The car passed through both D and F
2. Suppose someone wants to know the 'velocity with which the car travelled between D and F'
• We cannot give a definite answer. Because the velocity continuously changed
3. In such a situation we take the ratio: Displacement from D to F⁄Time required for the travel from D to F
• We know that:
♦ Displacement from D to F = Δx = (x2-x1) = (y coordinate of F - y coordinate of D)
♦ = (64.8 - 28.8) = 36
Time required to travel from D to F = Δt = (t2-t1) = (x coordinate of F - x coordinate of D)
♦ = (6 - 4) = 2
• So the ratio is 36⁄2 = 18 ms-1.
4. This velocity of 18 ms-1 is the average velocity with which the object travelled from D to F.
We can write the definition:
■ Average velocity is defined as the displacement (Δx) divided by the time duration (Δt) in which the displacement occurs
• The formula is:
♦ The bar over v is a standard notation used to indicate average quantities
5. We can check whether this formula is dimensionally correct:
• On the left side we have velocity [LT-1]
• On the right side,
♦ in numerator we have distance [L]
♦ in denominator we have distance [T]
• Thus we get [LT-1] on both sides
We have seen that, if the velocity is uniform, we can easily calculate the velocity as the slope
■ Can we relate average velocity to slope?
Let us try:
• The portion between D and F is enlarged and shown in fig.3.31 below:
 |
| Fig.3.31 |
• Also a triangle DFF' is formed
• Now take the ratio: altitude⁄base
• We get: altitude⁄base = FF'⁄DF' = (64.8-28.8)⁄(6-4) = 36⁄2 = 18 ms-1
So we can write:
■ Average velocity is the slope of the line joining the initial and final positions of the object in the position-time graph
• We have seen the details about average velocity.
• Average speed can also be calculated in a similar way.
• But in the numerator, instead of 'displacement', we use 'path length'. Thus we write:
Average speed =
Total path length between two points⁄Time required for the travel between the two points
■ Average velocity is a vector quantity. It has both magnitude and direction
■ Average speed is a scalar quantity. It has magnitude but no direction
The difference between the two can be understood using an example. We will write it in steps:
1. Fig.3.32 below shows the positions of an object at various times shown by a stop watch
 |
| Fig.3.32 |
(i) The object moves along the x axis
(ii) It started it's journey from the origin
♦ At that instant the stop watch was turned on
(iii) It travelled in the +x direction
(iv) After 18 seconds it reached P
♦ P is 360 m away from O
(v) Then it began to travel in the -x direction
(vi) After 6 more seconds, it reached Q
♦ Q is 240 m away from O
2. We are asked to find the following two items:
(i) The average velocity with which the object travelled from O to Q
(ii) The average speed with which the object travelled from O to Q
Solution:
Part 1: To find average velocity:
1. To get a better understanding of the problem, we will see the position-time graph of the motion of an object:
• The graphs are curves. So it is clear that, the object travelled with non-uniform velocity
• But that does not affect this problem because we are asked to find average velocity and speed
2. Average velocity depends only on initial and final positions and also the time.
• It does not matter whether it is uniform or non-uniform velocity
• We have:
x2 = position of the final point Q = 240
x1 = position of the initial point O = 0
• So displacement = Δx = (x2-x1) = (240-0) = 240
t2 = time at which the final point Q is reached = 24 s
t1 = time at which the object is at the initial position = 0 s
• So duration required for the displacement = Δt = (t2-t1) = (24-0) = 24 s
3. Thus average velocity = 240⁄24 = 10 ms-1
Another method:
1. We have seen that, average velocity is the slope of the line joining initial and final points.
• This line is shown in fig.3.33 below:
2. Clearly, altitude = QQ' = 240
• Base = OQ' = 24
• Thus we get: Slope = altitude⁄base = QQ'⁄OQ' = 240⁄24 = 10 ms-1
Part 2:
To find average speed:
1. For this, we have to take the total path length
• Total path length = (Distance from O to P + Distance from P to Q) = (360+120) = 480
2. Total time - 24 s
3. We have: Average speed =
Total path length between two points⁄Time required for the travel between the two points = 480⁄24 = 20 ms-1s
■ So we find that average velocity and average speed need not be the same
Now we will see a solved example. The link is given below:
Solved example 3.1
 |
| Fig.3.32 |
• But that does not affect this problem because we are asked to find average velocity and speed
2. Average velocity depends only on initial and final positions and also the time.
• It does not matter whether it is uniform or non-uniform velocity
• We have:
x2 = position of the final point Q = 240
x1 = position of the initial point O = 0
• So displacement = Δx = (x2-x1) = (240-0) = 240
t2 = time at which the final point Q is reached = 24 s
t1 = time at which the object is at the initial position = 0 s
• So duration required for the displacement = Δt = (t2-t1) = (24-0) = 24 s
3. Thus average velocity = 240⁄24 = 10 ms-1
Another method:
1. We have seen that, average velocity is the slope of the line joining initial and final points.
• This line is shown in fig.3.33 below:
 |
| Fig.3.33 |
• Base = OQ' = 24
• Thus we get: Slope = altitude⁄base = QQ'⁄OQ' = 240⁄24 = 10 ms-1
Part 2:
To find average speed:
1. For this, we have to take the total path length
• Total path length = (Distance from O to P + Distance from P to Q) = (360+120) = 480
2. Total time - 24 s
3. We have: Average speed =
Total path length between two points⁄Time required for the travel between the two points = 480⁄24 = 20 ms-1s
■ So we find that average velocity and average speed need not be the same
Now we will see a solved example. The link is given below:
Solved example 3.1
In the next section, we will see instantaneous velocity.
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