Chapter 3.6 - Types of Velocity-time graphs

In the previous section we saw acceleration. We also saw that, velocity-time graph of an object moving with constant acceleration will be a straight line. In this section we will see the different orientations possible for such straight line graphs.

1. Consider the velocity-time graph shown below:

Fig.3.44

• The inclined pink line is the velocity-time graph
2. The coordinates of some random points are marked on the graph.
• At the instant when the experiment was begun (t= 0), the object was moving with a velocity of 80 ms^-1.
• After 1 s (t= 1), the velocity became 71 ms^-1
• After 1 more second (t = 2), the velocity became 62 ms^-1. 
• So, with the passage of time, the velocity is decreasing 
3. We know that the body was moving with uniform acceleration (∵ the graph is a straight line)
The value of this acceleration can be calculated using the triangle AA'B
• We have: Slope of the pink line = ^altitude⁄_base 
• Altitude in numerator = AA' = Δv = (v₂ - v₁) = (39.5 - 62) = -22.5
• Base in denominator = A'B =  Δt = (t₂ - t₁) = (4.5 - 2) = 2.5     
• Thus slope = ^AA'⁄_A'B = ^Δv⁄_Δt = ^-22.5⁄_2.5 = -9 ms^-2.
4. We get a -ve value for acceleration. Why is that so?
Ans: With the passage of time, velocity is decreasing. 
• That means, the acceleration is acting in the direction opposite to that of the velocity. 
• So indeed acceleration will be negative in this case  
■ We can write the general case:
If the object moves with a 'uniform negative acceleration', the velocity-time graph will be a falling line
■ The converse is also applicable:
If the given velocity-time graph is a falling line, then the object was moving with a 'uniform negative acceleration'. That is., the object as slowing down at a uniform rate 
5. We started this experiment at the instant when the object passed a convenient point
• The stop watch was turned on at that instant
• The velocity of the object was 80 ms^-1 at that instant 
5. Let us write the relation to coordinate geometry:
• Equation of the pink line is:
v = at + 80
'a' has a negative value: -9
So the equation is: v = -9t + 80
6. Using this equation, we can find the time at which the body will come to rest.
• For that, we put v = 0
• Thus we get: 0 = -9t + 80
⟹ 9t = 80 ⟹ t = 8.9 s
• If we extend the pink line in fig.3.44 downwards, it will meet the time axis at (8.9,0)

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Now we will see another orientation. 
1. Consider the velocity-time graph shown below:

Fig.3.45

• The pink line is below the time axis  
• All the y coordinates are negative
• That means, the object moved with a negative velocity
• 'Negative velocity' indicates that, the object was moving towards the negative side of the x axis
2. Note that, even though it was moving towards the negative side of the x axis, it was having positive acceleration. That is., the velocity of the body was increasing every instant.
• This is clear because, the graph is a rising line

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1. Consider a velocity-time graph shown below:

Fig.3.46

• The pink line is below the time axis  
• All the y coordinates are negative
• That means, the object moved with a negative velocity
• 'Negative velocity' indicates that, the object was moving towards the negative side of the x axis
2. Note that,it was moving towards the negative side of the x axis and at the same time, it was having negative acceleration. That is., the velocity of the body was decreasing every instant.
• This is clear because, the graph is a falling line

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We will see one more orientation.
1. Consider the velocity-time graph shown below:

Fig.3.47

• Part of the pink line is above the time axis  The rest is below the time axis
• Above the time axis, all the y coordinates are positive. 
    ♦ Below the time axis, all the y coordinates are negative
• That means, initially, the object moved with a positive velocity.
    ♦ After some it moved with negative velocity
2. How is that possible?
Let us analyse:
(i) Consider the point of intersection of the pink line with the x axis. The coordinates are: (3.67,0)
(ii) The object was moving with positive velocity until the stop watch showed 3.67 s
(iii) After that it moved with negative velocity
3. So it is clear:
• The object was moving towards the positive side of the x axis until the stop watch showed 3.67 s
• After that, it changed direction. It began to move in the exact opposite direction. That is., towards the negative side of the x axis
3. Note that, during the entire journey, it was having negative acceleration.
• This is clear because, the graph is a falling line

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Now we will see an application of the velocity-time graph
1. Consider the velocity time graph shown below:

Fig.3.48

• It is clear that the object moved with uniform velocity of 'u' ms^-1. There was no acceleration
2. Consider the following two instants:
• Instant at which the stop watch showed t₁ seconds. It is marked as P
• Instant at which the stop watch showed t₂ seconds. It is marked as Q
3. Draw vertical lines at P and Q 
• These vertical lines intersect the velocity-time graph at P' and Q' 
4. Now calculate the area of the rectangle PQQ'P'. We get:
• Area = (PP' × PQ) = u(t₂ - t₁)
5. In the right side of the above equation, we are multiplying two quantities:
(i) velocity 
(ii) a time duration.
• We know that, when we multiply those two quantities, we get the 'displacement in that duration'
■ So the 'area below the above velocity-time graph' is the 'displacement of the object'

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• The above velocity-time graph is horizontal. So it was easy to calculate the area
• Is it true for all types of velocity-time graphs? Let us try:
1. Consider the velocity time graph shown below:

Fig.3.49

• It is clear that the object moved with uniform acceleration
2. Consider the following two instants:
• Instant at which the stop watch showed t1 seconds. It is marked as P
• Instant at which the stop watch showed t2 seconds. It is marked as Q
3. Draw vertical lines at P and Q 
• These vertical lines intersect the velocity-time graph at P' and Q' 
• We have a trapezium PQQ'P'. 
• It's area can be calculated by splitting it into a rectangle and a triangle
4. Using the principles of calculus, we can prove that, this area is equal to:
The displacement of the object in the duration (t₂ - t₁)
• We saw some basics about this proof here.
■ So we can write the general form:
The area enclosed between a velocity-time graph and the x axis will give the displacement of the object

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In the next section, we will see the equations of motion.

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Chapter 3.5 - Acceleration

In the previous section we saw instantaneous velocity. In this section we will see acceleration.
■ To study about velocity, we used position-time graph
■ Now, to study about acceleration, we can use velocity-time graph
We will write the steps:
1. Consider the velocity-time graph shown below:

Fig.3.41

• The horizontal pink line is the velocity-time graph
2. The coordinates of some random points are marked on the graph. 
• We can see that all the points have the same y coordinate '40'
3. So, whichever 'instant of time' we consider, the velocity of the object at that instant will be 40 ms^-1.
• We can say: The object was moving with a uniform velocity
4. Now consider another velocity-time graph shown below:

Fig.3.42

• The inclined pink line is the velocity-time graph
5. The coordinates of some random points are marked on the graph. 
• We can see that, when the x coordinate changes, the y coordinate also changes.
■ So the object was moving with a non-uniform velocity. We have to study the details about such a velocity
6. In the above graph, 
    ♦ the instantaneous velocity at t = 1 s is 32 ms^-1
    ♦ the instantaneous velocity at t = 2.5 s is 50 ms^-1
• So 'change in velocity' = (50-32) = 18 ms^-1
• Time duration in which the change in velocity occured = (2.5-1) = 1.5 s
7. The object attained a 'change in velocity' of '18 ms^-1' during a time gap of 1.5 s
■ How much 'change in velocity' did it attain during a time gap of 1 s?
• To find the answer, we simply divide. We can write:
'Change in velocity' attained in a duration of 1 s
= ^{Change in velocity}⁄_{Duration required for that change in velocity} = ¹⁸⁄_1.5 = 12
8. So, in the 'duration of 1.5 s' that we considered above, the object was attaining a 'change in velocity' of 12 ms^-1 in every second
• We can write:
(i) The object was moving at 32 ms^-1 at the beginning of those 1.5 seconds
(ii) Because of the attainment of 'change in velocity', the object was able to move at 50 ms^-1 after those 1.5 seconds
9. What we calculated above is:
• Change in velocity per unit time
Another name for it is:
• Rate of change of velocity with time
Yet another name is:
• Acceleration
10. So the '12' that we calculated is 'acceleration'
■ Now we want a unit to measure this acceleration 
(i) Consider the ratio '^{Change in velocity}⁄_{Duration required for that change in velocity}'
• In the numerator, we have: Change in velocity
    ♦ It's unit is ms^-1
• In the denominator, we have: Duration
    ♦ It's unit is s
(ii) So the unit of acceleration is: 'ms^-1 per s' 
• This is same as: ms^-2
■ So the unit for measuring acceleration is ms^-2

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In the above example, we can write:
• The object moved with an acceleration of 12 ms^-2 during those 1.5 seconds
Can we obtain this '12 ms^-2' graphically?
Let us try:
1. In fig.3.43 below, the triangle PQQ' is completed

Fig.3.43

• We get: Slope of the pink line = ^altitude⁄_base = ^QR'⁄_PQ' = ^Δv⁄_Δt = ^(50-32)⁄_(2.5-1) = ¹⁸⁄_1.5 = 12 ms^-2.
• So it is clear that, we can find acceleration graphically also
2. The pink line is a 'single straight line'. It's slope will be the same every where
So we can write:
• The object was moving with the 'same acceleration of 12 ms^-2' during it's entire journey
Another way of saying this is:
• The object was moving with 'uniform acceleration of 12 ms^-2' during it's entire journey 

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• So we can write a general rule:
The velocity-time graph of an object moving with uniform acceleration will be a straight line
• We can write the converse also:
If the given velocity-time graph is a straight line, then the object was moving with uniform acceleration

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Since it is a straight line, let us connect it to our knowledge of '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.3.43 above, we have the same situation.
    ♦ The slope of the yellow line is '12', which is the acceleration 'a'
    ♦ 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, velocity 'v' is plotted
    ♦ so 'y' in y = mx + c should be replaced by v
• Thus the equation of the pink line is:
v = at + 20
4. Using this equation, we can calculate the velocity of the object at any given time 't'.
For example:
• when t = 15 s, we have:
v = [(12 × 15)] + 20 = [180] + 20 = 200 ms^-1 
•  So, we can write:
(i) The object started it's journey from some unknown point
• This starting point is not significant for our present discussion
(ii) The object was moving with a uniform acceleration
• As a result, it's velocity was changing continuously
(ii) We began the experiment at the instant when the object passed a convenient point
• The stop watch was turned on at that instant
• At that instant, the object was moving with a velocity of 20 ms^-1.
    ♦ This '20' is the y-intercept
• The position of this 'convenient point' is also not significant for our present discussion
(iii) In each second, the object attained a velocity of 12 ms^-1.
So, at the end of the 15^th second, it would attain a velocity of (12 × 15) = 180 ms^-1.
(iii) But before the experiment began (at t= 0), it had already attained a velocity of 20 ms^-1
(iv) So the total velocity at t = 15 = (180+20) = 200 ms^-1

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From the above discussion, we can write a general equation also. We will write the steps:
1. We saw that, equation of the pink line is: v = at + 20
2. '20' is the 'y intercept'. 
• It is the velocity with which the object was travelling, when the stop watch was just turned on. 
• That is., t = 0
3. If we denote this velocity as v₀, the equation in (1) can be written as:
Eq.3.1
v = v₀ + at
• This is a general form. It can be applied to any object which is moving with an uniform acceleration 'a'
    ♦ v₀ is the velocity at the instant when t = 0
    ♦ v is the velocity after a duration of 't' seconds

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• We now know that, the velocity-time graph of an object moving with uniform acceleration is a straight line. 
• But such straight lines can come in all types of orientations. We must be able to identify them. 
• In the next section, we will see those different orientations.

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Chapter 3.4 - Instantaneous Velocity

In the previous section we saw average velocity and average speed. In this section we will see instantaneous velocity. We will write it in steps:
1. Fig.3.34 below shows a position-time graph of an object:

Fig.3.34

• In the above graph a point Q is marked. The following two items are clear:
(i) Q is at a distance of 5.12 m from the origin
(ii) The object passed Q when the stop watch showed 4 s
[Note: It is difficult to obtain the value 5.12 from the graph. So we can use the equation of the curve.
The equation is x = 0.08t³.
Substituting t = 4, we get:
s = 0.08 × (4)³= 0.08 × 64 = 5.12]  
2. Consider the instant at which the object passed Q
• We want to find the exact velocity at that instant
Let us try:
(i) The point of interest is Q. It is the position of the object when it travelled for 4 seconds
(ii) We need two neighbouring points of Q. 
• One before Q and 
• The other after Q
(iii) These new points should satisfy a condition:
• They must be at equal time intervals from Q
(iv)Let us choose a point P before Q such that:
    ♦ The object reached P, 1 s before Q. That is., when the stop watch showed 3 s  
• Let us choose a point R after Q such that:  
    ♦ The object reached R, 1 s after Q. That is., when the stop watch showed 5 s  
■ So the duration of travel from P to R = Δt = (t₂ - t₁) = (5 - 3) = 2 s
This is shown in fig.3.35 below:

Fig.3.35

(v) In the above graph, the distances corresponding to t = 3 and t = 5 are calculated using the same procedure that we used for t = 4:
s(3) = 0.08 × (3)³= 0.08 × 27 = 2.16 m
s(5) = 0.08 × (5)³= 0.08 × 125 = 10 m
(vi) Now we join PR. This is shown in fig.3.36 below:

Fig.3.36

• Slope of PR can be calculated by completing the triangle PRR'
• We get: Slope of PR = ^altitude⁄_base = ^RR'⁄_PR' = ^Δx⁄_Δt = ^(10-2.16)⁄_(5-3) = ^7.84⁄₂ = 3.92 ms^-1.
(vii) But we know that, this slope gives us only the 'average velocity' between P and R
• Though Q falls between P and R, 3.92 ms^-1 is not the instantaneous velocity at Q
• The duration of travel from P to R is 2 s. This '2 s' is not an 'instant'
3. So let us reduce the time interval:
• We need two new neighbouring points of Q. One before Q and the other after Q
• Like before, these new points should also satisfy a condition:
They must be at equal time intervals from Q
(i) Let us choose a point P₁ before Q such that:
    ♦ The object reached P₁, 0.5 s before Q. That is., when the stop watch showed 3.5 s  
Let us choose a point R₁ after Q such that:  
    ♦ The object reached R₁, 0.5 s after Q. That is., when the stop watch showed 4.5 s  
■ So the duration of travel from P₁ to R₁ = Δt = (t₂ - t₁) = (4.5 - 3.5) = 1 s
(ii) After marking P and R, we join them. This is shown in fig.3.37 below:

Fig.3.37

• In the above graph, the distances corresponding to t = 3.5 and t = 4.5 are calculated using the same procedure that we used for t = 4:
s(3) = 0.08 × (3.5)³= 0.08 × 27 = 3.43    
s(5) = 0.08 × (4.5)³= 0.08 × 125 = 7.29
(iii) Slope of P₁R₁ can be calculated by completing the triangle P₁R₁R₁'
• To avoid congestion, we will zoom in on the required area. This is shown in fig.3.38 below:

Fig.3.38

• We get: Slope of DF = ^altitude⁄_base = ^R₁R'₁ ⁄_P1R'1= ^Δx⁄_Δt = ^(7.29-3.43)⁄_(4.5-3.5) = ^3.86⁄₁ = 3.86 ms^-1
(iv) But we know that, this slope gives us only the 'average velocity' between P₁ and R₁
• Though Q falls between P₁ and R₁, 3.86 ms^-1 is not the instantaneous velocity at Q
• The duration of travel from P1 to R1 is 1 s. This '1 s' is not an 'instant'
4. So let us reduce the time interval:
• We need two new neighbouring points of Q. One before Q and the other after Q
• Like before, these new points should also satisfy a condition:
They must be at equal time intervals from Q
(i) Let us choose a point P₂ before Q such that:
    ♦ The object reached P₁, 0.25 s before Q. That is., when the stop watch showed 3.75 s  
Let us choose a point R₂ after Q such that:  
    ♦ The object reached R₂, 0.25 s after Q. That is., when the stop watch showed 4.25 s  
■ So the duration of travel from P₂ to R₂ = Δt = (t₂ - t₁) = (4.25 - 3.75) = 0.5 s
(ii) After marking P₂ and R₂, we join them. This is shown in fig.3.39 below. The triangle is also shown:

Fig.3.39

• We get: Slope of DF = ^altitude⁄_base = ^R₂R'₂ ⁄_P2R'2
= ^Δx⁄_Δt = ^{(6.1413-4.2188)}⁄_(4.25-3.75) = ^1.9225⁄_0.5 = 3.845 ms^-1.
(iv) But we know that, this slope gives us only the 'average velocity' between P₂ and R₂
• Though Q falls between P₂ and R₂, 3.845 ms^-1 is not the instantaneous velocity at Q
• The duration of travel from P₂ to R₂ is 0.5 s. This '0.5 s' is not an 'instant'

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Let us write a summary of what we have done so far:
Step 1: We took points P and R which are centred at 4 s
• The duration of travel (Δt) from P to R was 2 s
• The average velocity with which the object travelled from P to R was 3.92 ms^-1
• But the '2 s' is not an instant. So '3.92 ms^-1' was discarded
Step 2: We took points P₁ and R₁ which are centred at 4 s
• The duration of travel (Δt) from P₁ to R₁ was 1 s
• The average velocity with which the object travelled from P₁ to R₁ was 3.86 ms^-1
• But the '1 s' is not an instant. So '3.86 ms^-1' was discarded
Step 3: We took points P₂ and R₂ which are centred at 4 s
• The duration of travel (Δt) from P₂ to R₂ was 0.5 s
• The average velocity with which the object travelled from P₂ to R₂ was 3.845 ms^-1
• But the '0.5 s' is not an instant. So '3.845 ms^-1' was discarded

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• From the above 3 steps, it is clear that, with each step, the following two items are decreasing:
    ♦ Distance between P and R
    ♦ Time duration required for the travel between P and R
This can be shown as in fig.3.40 below:

Fig.3.40

• We can see that, the lines are becoming more and more aligned with the curve
• We must continue the steps until the time of travel between P and R is very small
• But it is not convenient to use graphs for further steps. because, the points will be very close to each other. We can use a table to do the calculations. It is shown below:

• The three steps that we already did are given in the first 3 rows of the table.
• Two more steps are done in the table. In the fifth step, Δt is very small. It is 0.01
■ But even this is not sufficient. Let us see the reason:
• Consider the ratio ^Δx⁄_Δt . We know that this ratio gives the average velocity
• To get the instantaneous velocity, the denominator Δt must be very small (like 0.00000001 s)
• It must be very small. We call it: 'infinitesimal time' . The meaning of 'infinitesimal' can be seen here.
    ♦ And at the same time, it must not be equal to zero. Because, division by zero is undefined
• Mathematically, such a small denominator is indicated as: '^Δx⁄_Δt when Δt → 0'
    ♦ That is.,we need the ratio when 'Δt tends to zero'
• Such a ratio can be easily calculated using the principles of calculus which we will study in maths classes.
• The reader may try to write the answer after learning calculus. The answer will be obtained as 3.84 ms^-1.
We know that, '^Δx⁄_Δt' is a slope.

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• When we use calculus, we get a value of ^Δx⁄_Δt
    ♦ (The value of the ratio when the denominator Δt is very small)
• This value of the ratio, 'when Δt is very small', is also a slope
• It is the slope of the tangent to the curve at the point where we seek the instantaneous velocity
• The lines PR, P₁R₁, P₂R₂ etc., touches the curve at two points
• But the tangent will touch the curve only at one point, which is Q

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Another example:
In the above example, the equation of motion was x = 0.08t³. We used this equation to find the distances at various times. Now consider an object moving according to the equation: 
x = 8.5 + 2.5t².
• What is the instantaneous velocity at t = 0 s ?
• What is the instantaneous velocity at t = 2 s ?
• What is the average velocity with which it travels between t = 2 and t = 4 s
Solution:
A. To find instantaneous velocities:
1. Based on the principles of calculus, the ratio '^Δx⁄_Δt when Δt → 0' is: 5t. 
2. So we get:
• At t = 0, the instantaneous velocity = 5 × 0 = 0 ms^-1 
• At t = 2, the instantaneous velocity = 5 × 2 = 10 ms^-1
B. To find average velocity:
1. Distance travelled when t = 2 s
= x(2) = (8.5 + 2.5×(2)²) = 18.5 m 
• Distance travelled when t = 4 s
= x(4) = (8.5 + 2.5×(4)²) = 48.5 m
2. So Δx = (48.5 - 18.5) = 30
• Δt = (t₂ - t₁) = (4 - 2) = 2 s 
3. Average velocity = ^Δx⁄_Δt = ³⁰⁄₂ = 15 ms^-1.

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In the next section, we will see acceleration.

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Chapter 3.3 - Average velocity and Average speed

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?

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

(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

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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

• To avoid congestion, coordinates of only alternate points are written.
■ We see that, the graph is a curve

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• 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

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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

• 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

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• 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}

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■ 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

From the fig., the following information can be obtained:

(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: 

Fig.3.32

• 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:

Fig.3.33

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

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Now we will see a solved example. The link is given below:
Solved example 3.1

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In the next section, we will see instantaneous velocity.

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