In the previous section, we learned about Average Velocity. In this section we will see Rate of change of Velocity.
Consider the situation when we are travelling in a bus along a straight line.
1. Let the bus be moving at a uniform velocity speed of 40 kmph.
2. When the bus approaches the destination, the driver has to stop it. That means, the velocity of the bus has to become zero.
3. He cannot apply the brakes all of a sudden. Then the bus will stop abruptly, and all of us will fall forward, causing injuries.
4. So, the driver gradually reduces the speed. He continues to reduce the speed until it becomes zero.
5. Remember that, the bus is travelling in a straight line. So there is no change in direction. But when the speed is reduced and thus changed to zero, there is a change in velocity.
Another example:
1. Consider the situation when we are travelling in a bus along a straight line. Let the bus be moving at a uniform velocity speed of 40 kmph.
2. After some time, the driver sees a road side sign board. It says: “Speed limit 30 kmph”.
3. So the driver reduces the speed and continues the travel at 30 kmph. Here also there is a change in velocity.
One more example:
1. Consider the situation when we are sitting in a bus which is at rest. The bus will start soon. But at present, the velocity is zero.
2. After some time the bus starts and begins it's journey. The bus is now moving at 40 kmph along a straight line.
3. But how did the zero increase to 40?
Ans: The driver gradually increased the speed from zero. He kept on increasing it until it became 40.
4. Here also there is a change in velocity
• There are several ways in which this change from 40 to 60 can be achieved. Let us see a step by step procedure:
1. While travelling at 40, the driver presses down gently on the 'accelerator pedal'. The speed gradually increases. It becomes 41, 42, 43 . . . and reaches 45.
2. Let the bus travel at this 45 for a while. We will analyse the increase from 40 to 45.
3. The change in velocity from 40 to 45 is (45 - 40) = 5 kmph
4. This change is achieved gradually. That means, some time is required to achieve this increase from 40 to 45.
5. Let the time required be 8 seconds. It is shown by the first yellow line in the graph below:
6. Time is plotted along the x-axis and velocity is plotted along the y-axis. So the coordinates of the ends of the first yellow line are (0,40) and (8,45).
• x coordinates which give time, increase from zero to 8
• y coordinates which give velocity, increase from 40 to 45
7. The change in velocity is achieved in 8 seconds.
• So the change in velocity achieved in 1 s = 5 Kmph⁄8 s
• But 5 kmph = 5000 m⁄3600 s = 1.39 m s-1
• So 'change in velocity achieved in 1 s' = 1.39 ms-1 ⁄8 s = 0.174 m s-2.
8. The 'Change in velocity in 1 s' is the 'rate of change of velocity', and it is called acceleration.
We can write: Acceleration = Change in velocity⁄time.
We can write it in the form of a formula:
Eq.1.3:
a = v-u⁄t.
Where a is the acceleration, v is the final velocity and u is the initial velocity
• The coordinates of the ends of the purple line are (8,45) and (13,45)
• Note that, there is no change in the y-coordinates
10. After 5 seconds of steadiness, the velocity increases. It reaches 55 kmph. So the increase is (55-45) = 10 kmph
11. This increase of 10 is achieved in 2 seconds. So rate of change of velocity = acceleration
= 10 Kmph⁄2 s =
• 10 kmph = 10000 m⁄3600 s = 2.78 m s-1
• So acceleration = 2.78 ms-1 ⁄2 s = 1.39 m s-2.
[We can also use Eq.1.3 above. But for that, we must convert the numerator and denominator into same units before applying the equation.
• We have: v-u = 55 - 45 = 10 kmph
• t = 2 s
• 10 kmph = 10000 m⁄3600 s = 2.78 m s-1.
• So acceleration = a = v-u⁄t = 2.78 ms-1 ⁄2 s = 1.39 m s-2.]
12. This acceleration of 1.39 m s-2 is greater than the previous acceleration of 0.174 m s-2.
• We can 'feel' it in the graph. Because, the second yellow line is steeper than the first yellow line.
• Steeper yellow line indicates a sudden increase in velocity.
• Higher velocity is achieved in a smaller time. That means, acceleration is high. The passengers will get a backward jolt.
13. The velocity of 55 kmph is not enough. We want 60. So let us continue the journey:
After achieving 55, the speed is maintained steady for 4 seconds. That means, during those 4 seconds, there is no change in velocity. This is indicated by the second purple coloured horizontal line in the graph.
• The coordinates of the ends of this purple line are (15,55) and (19,55)
• Note that, there is no change in the y-coordinates
14. After these 4 seconds of steadiness, the velocity increases. It reaches 60 kmph. So the increase is (60-55) = 5 kmph
15. This increase of 5 is achieved in 10 seconds. So rate of change of velocity = acceleration
= 5 Kmph⁄10 s =
• 5 kmph = 5000 m⁄3600 s = 1.39 m s-1
• So acceleration = 1.39 ms-1 ⁄10 s = 0.139 m s-2.
16. This acceleration of 0.139 m s-2 is less than the previous acceleration of 1.39m s-2.
• We can 'feel' it in the graph. Because, the third yellow line is more flat than the second yellow line.
• Flat yellow line indicates a gradual increase in velocity.
• Higher velocity is achieved in a greater time. This does not cause any discomfort for the passengers.
Another method to increase the velocity from 40 to 60 kmph:
Now we will see some solved examples
Solved example 1.4
A bus decreases its speed from 80 kmph to 60 kmph in 5 s. Find the acceleration of the bus.
Solution:
• Initial velocity u = 80 kmph
• Final velocity v = 60 kmph
Here the final velocity is lesser. So it is deceleration.
• Time t = 5 s
• v-u = 60 -80 = -20 kmph = -20000 m⁄3600 s = -5.56 m s-1
• So a = v-u⁄t = -5.56⁄5 = -1.11 m s-2
Solved example 1.5
A train starting from a railway station and moving with uniform acceleration attains a speed 40 kmph in 10 minutes. Find its acceleration.
Solution:
• Initial velocity u = 0 kmph
• Final velocity v = 40 kmph
• Time t = 10 minutes = 10 × 60 = 600 s
• v-u = 40 -0 = 40 kmph = 40000 m⁄3600 s = 11.11 m s-1
• So a = v-u⁄t = 11.11⁄60 = 0.185 m s-2
In the next section, we will see the Graphical representation of Motion.
Consider the situation when we are travelling in a bus along a straight line.
1. Let the bus be moving at a uniform velocity speed of 40 kmph.
2. When the bus approaches the destination, the driver has to stop it. That means, the velocity of the bus has to become zero.
3. He cannot apply the brakes all of a sudden. Then the bus will stop abruptly, and all of us will fall forward, causing injuries.
4. So, the driver gradually reduces the speed. He continues to reduce the speed until it becomes zero.
5. Remember that, the bus is travelling in a straight line. So there is no change in direction. But when the speed is reduced and thus changed to zero, there is a change in velocity.
Another example:
1. Consider the situation when we are travelling in a bus along a straight line. Let the bus be moving at a uniform velocity speed of 40 kmph.
2. After some time, the driver sees a road side sign board. It says: “Speed limit 30 kmph”.
3. So the driver reduces the speed and continues the travel at 30 kmph. Here also there is a change in velocity.
One more example:
1. Consider the situation when we are sitting in a bus which is at rest. The bus will start soon. But at present, the velocity is zero.
2. After some time the bus starts and begins it's journey. The bus is now moving at 40 kmph along a straight line.
3. But how did the zero increase to 40?
Ans: The driver gradually increased the speed from zero. He kept on increasing it until it became 40.
4. Here also there is a change in velocity
In this way, we encounter 'change in velocity' on many occasions in our day to day life. We have to learn about it in more detail.
• Consider a bus travelling at 40 kmph. When it reaches a freeway, the velocity is increased to 60 kmph. So there is a change in velocity. • There are several ways in which this change from 40 to 60 can be achieved. Let us see a step by step procedure:
1. While travelling at 40, the driver presses down gently on the 'accelerator pedal'. The speed gradually increases. It becomes 41, 42, 43 . . . and reaches 45.
2. Let the bus travel at this 45 for a while. We will analyse the increase from 40 to 45.
3. The change in velocity from 40 to 45 is (45 - 40) = 5 kmph
4. This change is achieved gradually. That means, some time is required to achieve this increase from 40 to 45.
5. Let the time required be 8 seconds. It is shown by the first yellow line in the graph below:
 |
| Fig.1.7 |
• x coordinates which give time, increase from zero to 8
• y coordinates which give velocity, increase from 40 to 45
7. The change in velocity is achieved in 8 seconds.
• So the change in velocity achieved in 1 s = 5 Kmph⁄8 s
• But 5 kmph = 5000 m⁄3600 s = 1.39 m s-1
• So 'change in velocity achieved in 1 s' = 1.39 ms-1 ⁄8 s = 0.174 m s-2.
8. The 'Change in velocity in 1 s' is the 'rate of change of velocity', and it is called acceleration.
We can write it in the form of a formula:
Eq.1.3:
a = v-u⁄t.
Where a is the acceleration, v is the final velocity and u is the initial velocity
9. The velocity of 45 kmph is not enough. We want 60. So let us continue the journey:
After achieving 45, the speed is maintained steady for 5 seconds. That means, during those 5 seconds, there is no change in velocity. This is indicated by the purple coloured horizontal line in the graph.• The coordinates of the ends of the purple line are (8,45) and (13,45)
• Note that, there is no change in the y-coordinates
10. After 5 seconds of steadiness, the velocity increases. It reaches 55 kmph. So the increase is (55-45) = 10 kmph
11. This increase of 10 is achieved in 2 seconds. So rate of change of velocity = acceleration
= 10 Kmph⁄2 s =
• 10 kmph = 10000 m⁄3600 s = 2.78 m s-1
• So acceleration = 2.78 ms-1 ⁄2 s = 1.39 m s-2.
[We can also use Eq.1.3 above. But for that, we must convert the numerator and denominator into same units before applying the equation.
• We have: v-u = 55 - 45 = 10 kmph
• t = 2 s
• 10 kmph = 10000 m⁄3600 s = 2.78 m s-1.
• So acceleration = a = v-u⁄t = 2.78 ms-1 ⁄2 s = 1.39 m s-2.]
12. This acceleration of 1.39 m s-2 is greater than the previous acceleration of 0.174 m s-2.
• We can 'feel' it in the graph. Because, the second yellow line is steeper than the first yellow line.
• Steeper yellow line indicates a sudden increase in velocity.
• Higher velocity is achieved in a smaller time. That means, acceleration is high. The passengers will get a backward jolt.
13. The velocity of 55 kmph is not enough. We want 60. So let us continue the journey:
After achieving 55, the speed is maintained steady for 4 seconds. That means, during those 4 seconds, there is no change in velocity. This is indicated by the second purple coloured horizontal line in the graph.
• The coordinates of the ends of this purple line are (15,55) and (19,55)
• Note that, there is no change in the y-coordinates
14. After these 4 seconds of steadiness, the velocity increases. It reaches 60 kmph. So the increase is (60-55) = 5 kmph
15. This increase of 5 is achieved in 10 seconds. So rate of change of velocity = acceleration
= 5 Kmph⁄10 s =
• 5 kmph = 5000 m⁄3600 s = 1.39 m s-1
• So acceleration = 1.39 ms-1 ⁄10 s = 0.139 m s-2.
16. This acceleration of 0.139 m s-2 is less than the previous acceleration of 1.39m s-2.
• We can 'feel' it in the graph. Because, the third yellow line is more flat than the second yellow line.
• Flat yellow line indicates a gradual increase in velocity.
• Higher velocity is achieved in a greater time. This does not cause any discomfort for the passengers.
In this method, the intervals of steadiness are not given. So we see no horizontal lines. The graph is shown below:
 |
| Fig.1.8 |
The acceleration is in three stages.
One more method:
1. In this method, the acceleration from 40 to 60 kmph is done in a single stage. The graph will be a single line as shown below:
 |
| Fig.1.9 |
2. The coordinates of the ends of the line are: (0,40) and (32,60)
3. So we can say this: The increase in velocity from 40 to 60 is achieved in 32 seconds
4. Using Eq.1.3:
• a = v-u⁄t.
• v-u = 60 - 40 = 20 kmph = 20000 m⁄3600 s = 5.55 m s-1
• t = 32 s
• So a = 5.55⁄32 = 0.173 m s-2.
What is the difference between the second and third methods?
1. Let us consider the second method. See fig.1.8 above. The acceleration is in three stages.
• In the first stage, there is an acceleration of 0.174 m s-2.
• In the second stage, there is an acceleration of 1.39 m s-2.
• In the third stage, there is an acceleration of 0.139 m s-2.
This acceleration is not uniform. In other words, the increase from 40 to 60 kmph was achieved in a non-uniform manner. It is a case of non-uniform acceleration
2. Let us consider the third method. See fig.1.9 above.
• The increase from 40 to 60 kmph was achieved in a time interval of 32 seconds.
• We can take any random '1 second time interval' from among the 32 seconds. Which ever we take, the acceleration in that 1 second will be the same 0.173 m s-2.
• The increase from 40 to 60 kmph was achieved in a uniform manner. It is a case of uniform acceleration.
Now we will see some solved examples
Solved example 1.3
Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s-1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.
Solution:
Case 1:
• Initial velocity u = 0 m s-1
• Final velocity v = 6 m s-1
• Time t = 30 s
• So acceleration = a = v-u⁄t = 6-0⁄30 = 6⁄30 = 1⁄5 = 0.2 m s-2
Case 2:
• Initial velocity u = 6 m s-1
• Final velocity v = 4 m s-1
• Time t = 5 s
• So acceleration = a = v-u⁄t = 4 -6⁄5 = -2⁄5 = -0.4 m s-2
■ Note that, in case 2, we get a negative value for the acceleration. This means that, the velocity is being reduced. When velocity is reduced, we call the acceleration as retardation or deceleration
Solution:
Case 1:
• Initial velocity u = 0 m s-1
• Final velocity v = 6 m s-1
• Time t = 30 s
• So acceleration = a = v-u⁄t = 6-0⁄30 = 6⁄30 = 1⁄5 = 0.2 m s-2
Case 2:
• Initial velocity u = 6 m s-1
• Final velocity v = 4 m s-1
• Time t = 5 s
• So acceleration = a = v-u⁄t = 4 -6⁄5 = -2⁄5 = -0.4 m s-2
■ Note that, in case 2, we get a negative value for the acceleration. This means that, the velocity is being reduced. When velocity is reduced, we call the acceleration as retardation or deceleration
Solved example 1.4
A bus decreases its speed from 80 kmph to 60 kmph in 5 s. Find the acceleration of the bus.
Solution:
• Initial velocity u = 80 kmph
• Final velocity v = 60 kmph
Here the final velocity is lesser. So it is deceleration.
• Time t = 5 s
• v-u = 60 -80 = -20 kmph = -20000 m⁄3600 s = -5.56 m s-1
• So a = v-u⁄t = -5.56⁄5 = -1.11 m s-2
Solved example 1.5
A train starting from a railway station and moving with uniform acceleration attains a speed 40 kmph in 10 minutes. Find its acceleration.
Solution:
• Initial velocity u = 0 kmph
• Final velocity v = 40 kmph
• Time t = 10 minutes = 10 × 60 = 600 s
• v-u = 40 -0 = 40 kmph = 40000 m⁄3600 s = 11.11 m s-1
• So a = v-u⁄t = 11.11⁄60 = 0.185 m s-2
In the next section, we will see the Graphical representation of Motion.
No comments:
Post a Comment