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:
• 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:
• 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
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
• 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
• 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
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 15th 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
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
■ 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 |
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 |
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
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 |
• 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
• 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
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 15th 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
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
• 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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