Chapter 1.8 - Uniform Circular motion

In the previous section, we completed the discussion on the motion of objects along a straight line. In this section we will discuss the motion of an object along a circular path.

We saw that velocity has both magnitude and direction. For example, if an object is moving with a velocity 12 m s^-1, it must be along a straight line path. Because if there are any bends or curves in that path, the direction of motion will change. Then the velocity will also change.

So velocity can change because of any one of the following three reasons:
• Changing speed keeping direction the same
    ♦ An example: An object was travelling with a velocity of 20 m/s, at an angle of 25^o with the x-axis. After some time, the speed changed to 15 m/s. In this case, the velocity has changed even if the object continues to travel in the same direction. The new velocity should be specified by the new speed and the same direction  
• Changing direction keeping speed the same
    ♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. In this case, the velocity has changed even if the car continues to travel with the same speed. The new velocity should be specified by the speed 30 kmph and new direction.
• Changing both speed and direction.

We saw the above details here. Now we are going to discuss about a situation in which changes in velocity occur due to the change in direction only.

1. Consider an athlete running along a closed path ABCD shown in fig.1.30(a). 

Fig.1.30

2. The path is square in shape. He is running at constant speed. Let us check if he is running at constant velocity:
3. Consider his motion along AB. When he reaches B, he will have to change his direction. Otherwise, he will not be able to keep himself within the track. 
4. Because of this change in direction, his velocity will change at B. This will happen at C, D and A also. So when he completes one lap, his direction will have changed 4 times.
5. Consider a path of another shape. This time, a hexagonal path shown in fig.b. Here also the there are changes in direction. The change happen 6 times. They are: at A, B, C, D, E and F
6. What if the path is octagonal as shown in the fig.c? 
Ans: The change happens 8 times. They are: at A, B, C, D, E, F, G and H
7. So we see that, when the 'number of sides' n of the path increase, the 'number of times that the athlete has to change his direction' will also increase. This is shown in fig.1.31 below:

Fig.1.31

8. What if the number of sides of the path is very large?
Ans: Then the length of the sides will become very less. So less that the sides will be just ‘points’. And the path will appear as a circle. In maths we use this principle to obtain the general formula for the perimeter of a circle. Details here.
9. So, if the athlete is moving with constant speed along a circular path, his velocity is changing at every point of the track. That means, his velocity is changing continuously. 
10. We saw such a 'continuous change in velocity' earlier when we discussed ‘motion along a straight line’. There, the velocity change was due to the change in speed. There was no change in direction. The change in speed was due to an application of acceleration. So we can write this:
• In straight line motion, if there is a change in magnitude of the velocity, that velocity has changed. 
    ♦ The change occurred due to the acceleration. This acceleration causes a continuous change in velocity.
• In circular motion, even if the magnitude of the velocity remains the same, the direction continuously changes. So the velocity continuously changes. Thus 'motion along a circle' is also an accelerated motion.
11. We know that the perimeter of a circle is 2πr. Where r is the radius of the circle. 
So, if the athlete requires t seconds to complete one path, then: 
Speed of the athlete = ^distance⁄_time  = ^2πr⁄_t .

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1. Consider a small stone tied to a string. Hold one end of the string and move the stone in a circular path. This is shown in fig.1.32 (a) below:

Fig.1.32

2. Let the stone move at a constant speed. After some time, let the stone go by releasing the string. In what direction will the stone move?
Ans: When the string is released the stone will move in a straight line path. This straight line path is tangential to the circle. The point of tangency is the position of the stone at the instant when the string is released.
3. Let us analyse the above answer:
■ In the fig.1.32 (b), the string is released when the stone reaches B
(i) Then, instead of the circular path, the stone will begin to travel in a straight line path.
(ii) We want to know this path. For that, draw a tangent line at B. In the fig., this tangent is named as AC
(iii) The movement of the stone after the release will be along the line AC. The direction is shown by the arrow at A
■ Another point of release is also shown in the fig.1.32 (b). It is at Q. If it is released at Q:
(i) Then, instead of the circular path, the stone will begin to travel in a straight line path.
(ii) We want to know this path. For that, draw a tangent line at Q. In the fig., this tangent is named as PQ
(iii) The movement of the stone after the release will be along the line PQ. The direction is shown by the arrow at P
• In higher classes we will learn to draw tangent at any given point on a circle.

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So we have completed the discussion on the 'Motion of objects'. In the next chapter, we will see  some more solved examples. 

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Chapter 1.5 - Velocity-Time Graph

In the previous section, we completed the discussion on Distance-Time graph. In this section we will see another type of graph.

Velocity-Time Graph

Let us see another experiment. A car is being driven along a straight road. After some time, a person sitting next to the driver watches the speedometer of the car and notes down the velocity every 5 seconds. Here is what he got:

Table.1.2

Let us look at the table in some detail:
1. The driver and the assistant starts the engine and begin the travel along a straight line
2. They continue their travel. The experiment has not begun yet
3. After some time the assistant starts his stop watch. At that instant, the experiment has begun
4. At that instant. the time is zero
5. So we see that, when the experiment began, time is zero. But velocity is not zero
• When the stop watch shows 5 s, the assistant notes down the reading of the speedometer. It is 40 kmph
• When the stop watch shows 10 s, the assistant notes down the reading of the speedometer. It is 40 kmph
• When the stop watch shows 15 s, the assistant notes down the reading of the speedometer. It is 40 kmph
- - -
- - -
• The readings are taken upto 30 s. Thus we get Table 1.2 above
• We must not be discouraged on seeing 40 kmph every time. In science and engineering, all readings are important. The constant reading simply shows that the car was travelling at a constant speed of 40 kmph since the beginning of the experiment.

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Let us make a graph. The table 1.2 above shows the recordings made in the field. As the time is in seconds, we will convert the velocity from kmph to m s^-1. The modified table is given below: 

Table.1.3

The fig. below shows the graph based on table 1.3
• Time is taken along the x-axis and velocity is taken along the y-axis

Fig.1.17

• The reader may plot the above graph on a fresh graph paper. Any convenient scale can be used. The following scale is also adequate:
    ♦ X axis: 1 cm represents 2 s
    ♦ Y axis: 1 cm represents 1 m s^-1
We see the following features:
1. All points fall on a straight line
2. This straight line is parallel to the x-axis
3. This straight line meets the y-axis at 11.1
■ So we can infer this:
When the velocity is a constant, the velocity-time graph is a straight line parallel to the x-axis.

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Now let us see an application of the above graph:
1. Draw two vertical lines shown in purple colour in the fig.1.18 below:
• One through time t₁ = 8 s  
• One through time t₂ = 22 s

Fig.1.18

2. Draw them upwards until they meet the horizontal yellow line. Now we get a rectangle which is shown shaded in the fig.1.18
• The height of the rectangle is equal to the constant velocity v, which is equal to 11.1 m s^-1
• The width of the rectangle = (t₂ – t₁) = (22 – 8) = 14 s
• So area of the rectangle = 11.1 m s^-1 × 14 s = 155.4 m
3. What is the peculiarity of this area?
Ans: 
• To find the area, we have multiplied a velocity v with a time duration of 14 seconds
• We know that when we multiply a velocity with a time duration, we get the distance 's' travelled during that time duration. That is., v×t = s
• So the shaded area in fig.1.18 is the distance travelled by the car in a time duration of 14 seconds starting from t = 8 s
■ In this manner we can find the distance during any time interval by drawing a suitable rectangle.
■ Note that, we did not take any 'reading of distances' during the experiment. But after the experiment, we are able to find any required distance that we want, from the graph.

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Let us repeat the experiment:
The procedure is the same. But this time, the velocities are different. The field recordings are shown below in table 1.4

Table.1.4

The modified table with velocities converted into m s^-1 is given below:

Table.1.5

The resulting graph is:

Fig.1.19

• In the graph, we see that, as the time increases, the velocity is also increasing. 
• Take any second from the graph. The velocity at that second will be greater than the velocity at the previous second. 
• Such an increase in velocity can not be achieved with out acceleration. 
• But no recordings related to acceleration were made at the field. 
• In the graph also, no information seems to be available about the acceleration. 
• But in fact, the acceleration is hiding inside the graph. If we want to know the acceleration, we will have to bring it out. 
• Recall how we brought out the ‘speed’ which was hiding in the 'distance-time graph'. Details here. We will be using a similar method:
1. Consider the above graph in fig.1.19. Take any two convenient points on that graph. Let us take (10,10) and (20,15). Let us name them as C and D. This is shown in fig.1.20 below:

Fig.1.20

2. Draw a horizontal line through the lower point C
3. Draw a vertical line through the higher point D
4. These two lines will meet at a point. Name this point as G
5. So we get a triangle CGD. We want the base CG and altitude GD of this triangle. Let us find them:
6. G lies on the horizontal through C. Any point on the horizontal through C (10,10) will have the y coordinate 10
7. G lies on the vertical through D. Any point on the vertical through D (20,15) will have the x coordinate 20
8. The point G lies on both the above horizontal and vertical. So the coordinates of G are (20,10)
9. Consider the two points C and G. 
• C is at a horizontal distance of 10 from the y-axis
• G is at a horizontal distance of 20 from the y-axis
• So the horizontal distance between C and G
= The base of the ⊿CGD
= 20 -10 = 10 s
10. Consider the two points G and D.  
• D is at a vertical distance of 15 from the x-axis
• G is at a vertical distance of 10 from the x-axis
• So the vertical distance between G and D
= The altitude of the ⊿CGD
= 15 - 10 = 5 m s^-1
11. Now take the ratio ^altitude⁄_base
We get ^{5 ms-1}⁄_{10 s} = 0.5 m s^-2 
12. The car was given this much acceleration for a duration of 10 seconds from t = 10 s
Explanation:
(i) We formed the triangle CGD, and took the ratio ^altitude⁄_base
(ii) Consider the altitude. 
• We calculated the 'difference between the y coordinates' to find the altitude
• But the y coordinates are the 'velocities at t = 10 and t = 20' (We took them from table 1.5 above)
• So the difference is actually the 'change in velocity'
(iii) Now consider the base. 
• We calculated the 'difference between the x coordinates' to find the base
• But the x coordinates are the 'times of travel'  (We took them from table 1.5 above)
• So the difference is actually the duration of time in which change of velocity from 10 m s^-1 to 20 m s^-1 took place.
■ So, the altitude is the change in velocity
■ Base is the duration of time in which the change in velocity took place
(iv) Their ratio is the acceleration required for the change in velocity from 10 m s^-1 to 20 m s^-1 in 10 seconds (∵ acceleration = ^{change in velocity}⁄_time)

---

• In the above calculations, we took two known points C and D. The x and y coordinates are already known to us because, they were recorded in the field. 

• Let us take two points other than those recorded in the field. In such a case, we will have to find the coordinates ourselves

Consider fig.1.21 below.

Fig.1.21

1. Two random points U and W are marked on the graph. To find the coordinates of these points we do the following:

• Draw two horizontal lines. One through U and the other through W

• Draw two vertical lines. One through U and the other through W

2. • The horizontal lines meet the y-axis at 9.25 and 15.5

• The vertical lines meet the x-axis at 8.5 and 21

3. Extend the bottom horizontal line towards the right until it intersects the vertical through W

4. Name the point of intersection as V

5. Thus we get ⊿UVW

• The base of ⊿UVW = 21- 8.5 = 12.5 s

• The altitude of ⊿UVW = 15.5 - 9.25 = 6.25 m s^-1

6. Take the ratio ^altitude⁄_base
We get ^{6.25 ms-1}⁄_{12.5 s} =  0.5 m s^-2
12. This was the acceleration at which the object travelled from U to W
The explanation was given in the previous example.

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• In fig.1.21, we find that, the object travelled from U to W at an acceleration of 0.5 m s^-2
• In fig.1.20, we find that, the object travelled from C to D at an acceleration of 0.5 m s^-2
• We can take any pair of points we like on the graph. 
    ♦ A first point and a second point. 
    ♦ In all cases we will get the same result:
    ♦ The object travelled from the first point to second point at an acceleration of 0.5 m s^-2
• The reader is advised to select random pairs, and do the calculations to confirm it. 
■ Why do we get the same acceleration in all cases?
Ans: 
• The graph is a straight line with no bends or curves. 
• For every pair of points, the triangles formed will be similar. 
• So the ratio ^altitude⁄_base is always a constant.

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If the graph had any bends or curves, we will not get the same acceleration always. An example is given below:

Fig.1.22

• We can see that the triangles are not similar.

• So the ratio ^altitude⁄_base will be different

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Let us write a summary of the above discussion:

1. In the experiment, only time and velocity were recorded in the field
2. Acceleration was not recorded
3. When the velocity-time graph was plotted, it was seen that all the points fall on a straight line
4. So the graph do not have any bends or curves
5. Because of the 'straight nature' of the curve, it was confirmed that, the object travelled at an uniform acceleration
6. And the uniform acceleration was calculated as 0.5 m s^-2
■ Whenever we get a straight line for the velocity-time graph, we can confirm that the object moved with uniform acceleration
■ And this acceleration can be calculated using the formula:
Eq.1.5:

• Where a is the acceleration
• u is the initial velocity
• v is the final velocity
• t is the time taken for the velocity to increase from u to v
■ This formula is based on the fig.1.23 below:

Fig.1.23

• t₁ and t₂ are the times. Their difference 't' will give the duration in which the initial velocity u increases to the final velocity v
• base of the triangle is t = t₂ - t₁
• Altitude of the triangle is (v - u)
• Acceleration = ^altitude⁄_base . Hence we get the Eq.1.5 above.  
■ We have obtained Eq.1.5 above. From that equation we will get: (v-u) = at
From this we get the equation below:
Eq.1.6:
v = u + at
This is the first equation of motion.
• A body was moving with a uniform acceleration 'a'
• At time t₁, it's velocity was u
• At time t₂, it's velocity increased to v
    ♦ This increase was due to the uniform acceleration 'a'.
    ♦ This increase happened in a duration 't' = (t₂-t₁)
• In such a situation, we can calculate the final velocity v of the object using the Eq.1.6

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In the next section, we will see another equation. 

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Chapter 1.4 - Applications of Distance-Time Graph

In the previous section, we learned how to obtain the 'distance travelled by an object during a particular duration of time' from the Distance-Time Graph. In this section we will see another application.

Speed from Distance-Time Graph

We are discussing about the distance-time graph. In this graph, no information seems to be available about the speed at which the object travelled. But in fact, the speed is 'hiding' inside the graph. If we want to know the speed, we will have to bring it out. Let us see how:

1. Take any two convenient points on the original graph in fig.1.11. Let us take C (12,360) and D (18, 540). This is shown in fig.1.13 below

Fig.1.13

2. Draw a horizontal line through the lower point C
3. Draw a vertical line through the higher point D
4. These two lines will meet at a point. Name this point as G
5. So we get a triangle CGD. We want the base CG and altitude GD of this triangle. Let us find them:
6. G lies on the horizontal through C. Any point on the horizontal through C (12,360) will have the y coordinate 360
7. G lies on the vertical through D. Any point on the vertical through D (18,540) will have the x coordinate 18
8. The point G lies on both the above horizontal and vertical. So the coordinates of G are (18,360)
9. Consider the two points C and G. 
• C is at a horizontal distance of 12 from the y-axis
• G is at a horizontal distance of 18 from the y-axis
• So the horizontal distance between C and G
= The base of the ⊿CGD
= 18 -12 = 6 minutes
10. Consider the two points G and D.  
• D is at a vertical distance of 540 from the x-axis
• G is at a vertical distance of 360 from the x-axis
• So the vertical distance between G and D
= The altitude of the ⊿CGD
= 540 - 360 = 180 metres
11. Now take the ratio ^altitude⁄_base
We get ^{180 m}⁄_{6 min} = ^{30 m}⁄_{1 min} = ^{30 m}⁄_{60 s} = 0.5 m s^-1
12. This was the speed at which the object travelled from C to D
Explanation:
(i) We formed the triangle CGD, and took the ratio ^altitude⁄_base
(ii) Consider the altitude. 
• We calculated the 'difference between the y coordinates' to find the altitude
• But the y coordinates are the 'distances of C and D from O' (We took them from table 1.1)
• So the difference is actually the distance between C and D
(iii) Now consider the base. 
• We calculated the 'difference between the x coordinates' to find the base
• But the x coordinates are the 'times for reaching C and D, after passing O'  (We took them from table 1.1)
• So the difference is actually the time of travel between C and D
■ So, the altitude is the distance between C and D
■ Base is the time required to travel from C to D
(iv) Their ratio is the speed at which the object travelled from C to D (∵ speed = ^distance⁄_time)

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• In the above calculations, we took two known points C and D. The x and y coordinates are already known to us because, they were recorded in the field. 

• Let us take two points other than O, A, B, C, D, E and F. In such a case, we will have to find the coordinates ourselves

Consider fig.1.14 below.

Fig.1.14

1. Two random points U and W are marked on the graph. To find the coordinates of these points we do the following:

• Draw two horizontal lines. One through U and the other through W

• Draw two vertical lines. One through U and the other through W

2. • The horizontal lines meet the y-axis at 306 and 675 

• The vertical lines meet the x-axis at 10.2 and 22.5

3. Extend the bottom horizontal line towards the right until it intersects the vertical through W

4. Name the point of intersection as V

5. Thus we get ⊿UVW

• The base of ⊿UVW = 22.5-10.2 = 12.3 min

• The altitude of ⊿UVW = 675-306 = 369 m

6. Take the ratio ^altitude⁄_base
We get ^{369 m}⁄_{12.3 min} = ^{30 m}⁄_{1 min} = ^{30 m}⁄_{60 s} = 0.5 m s^-1
12. This was the speed at which the object travelled from U to W
The explanation was given in the previous example.

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• In fig.1.14, we find that, the object travelled from U to W at a speed of 0.5 m s^-1
• In fig.1.13, we find that, the object travelled from C to D at a speed of 0.5 m s^-1
• We can take any pair of points we like on the graph. 
    ♦ A first point and a second point. 
    ♦ In all cases we will get the same result:
    ♦ The object travelled from the first point to second point at 0.5 m s^-1
• The reader is advised to select random pairs, and do the calculations to confirm it. 
■ Why do we get the same speed in all cases?
Ans: 
• The graph from O to F is a straight line with no bends or curves. 
• For every pair of points, the triangles formed will be similar. 
• So the ratio ^altitude⁄_base is always a constant. 
• In our present case, this ratio is ^{30 m}⁄_{1 min} which works out to 0.5 m s^-1

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If the graph had any bends or curves, we will not get the same speed always. An example is given in fig.1.15 below:

Fig.1.15

• We can see that the triangles are not similar. Details about simialar triangles can be seen here. 

• So the ratio ^altitude⁄_base will be different

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Let us write a summary of the above discussion:

1. In the experiment, only time and distance were recorded in the field
2. Speed was not recorded
3. When the distance-time graph was plotted, it was seen that all the points fall on a straight line
4. So the graph do not have any bends or curves
5. Because of the 'straight nature' of the curve, it was confirmed that, the object travelled at an uniform speed
6. And the uniform speed was calculated as 0.5 m s^-1
■ Whenever we get a straight line for the distance-time graph, we can confirm that the object moved at uniform speed
■ And this speed can be calculated using the formula:
Eq.1.4:

This formula is based on the fig.1.16 below:

Fig.1.16

• t₁ and t₂ are the times
• s₁ and s₂ are the distances
• base of the triangle is t₂ - t₁
• Altitude of the triangle is s₂ - s₁
• Speed = ^altitude⁄_base . Hence we get the Eq.1.4 above.  
■ We have completed the discussion on Distance-Time graph. Note that, in the experiment, if the object was travelling in a straight line, we can use 'velocity' in place of 'speed.

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In the next section, we will see Velocity-Time graph. 

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Chapter 1.3 - Distance-Time Graph

In the previous section, we learned about Acceleration. In this section we will see the Graphical representations related to Motion. We have already discussed about graphs in our maths classes. Details here.

Consider an object in motion. We are going to make a distance-time graph for that object. For making that graph, we need some readings. So let us get those readings first:

1. The line in fig.1.10(a) below shows the path on which the object moves. 

Fig.1.10

2. In that path, mark a point ‘O’ on the ground. This is the first mark. It is shown in fig.1.10(b).  
3. Make an number of markings A, B, C etc., at convenient distances. 
4. One person should be present at each of the marked points. And each of them should have a stop watch. When these arrangements are complete, the object can begin it’s journey along the path. 
5. The object begins it's journey at some distance before O. At the instant when the object passes O, all persons should start their stop watches. The experiment has begun. Now, all that we have to do is this:
6. Record the exact time instant at which the object passes each of the markings.
This is a simple process which can be done as follows:
■ Consider the person at ‘A’.
(i) He sees the object coming towards him from a distance. So he gets ready. 
(ii) At the instant when the object passes the marking at ‘A’, he stops the watch.
(iii) He records the reading of the watch. Let it be 3 minutes. He has done his work.
(iv) We can infer that the object took 3 minutes to travel from O to A. 

■ All the other stop watches are still running. Consider the person at ‘B’
(i) He sees the object coming towards him from a distance. So he gets ready. 
(ii) At the instant when the object passes the marking at ‘B’, he stops the watch.
(iii) He records the reading of the watch. Let it be 7 minutes. He has done his work.
(iv) We can infer that the object took 7 minutes to travel from O to B. 

■ All other stop watches beyond B are still running. Consider the person at ‘C’
(i) He sees the object coming towards him from a distance. So he gets ready. 
(ii) At the instant when the object passes the marking at ‘C’, he stops the watch.
(iii) He records the reading of the watch. Let it be 12 minutes. He has done his work.
(iv) We can infer that the object took 12 minutes to travel from O to C.

- - - 
- - - 

In this way we will obtain the time at all the points. 
7. Now we need distances. For that, measure OA, AB, BC, CD, . . .
From this, we can calculate the distance of each point from the origin ‘O’. This is done as follows:
OA = OA
OB = OA + AB
OC = OB + BC
OD = OC + CD
So on..
Note that all the distances are to be measured along the path. For this experiment, we do not want the straight line distances from ‘O’.
8. Now we tabulate the results. It is shown in table 1.1  below:

Table.1.1

9. Using the above table, we can plot the graph.
• Time is taken along the x-axis and distance is taken along the y-axis
• The resulting graph is shown below:

Fig.1.11

10. All the points are plotted on the graph paper. When those points are joined together, a straight line is obtained.
• In other words, all the points obtained from the experiment lie on a straight line.
• This line is shown in yellow colour. 
• The reader may plot the above graph on a fresh graph paper. Any convenient scale can be used. The following scale is also adequate:
    ♦ X axis: 1 cm represents 1 minute
    ♦ Y axis: 1 cm represents 50 m

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So we have successfully drawn the graph. We will now see some of the uses of a distance-time graph:

• We know the distances travelled by the object in 3 seconds, 7 seconds, 12 seconds etc., after passing 'O'. They were recorded in the field, and are available in table 1.1 above
• What if we want to know some other distances?
• For example, What distance did the object travel in 15 seconds after passing 'O'
The answer can be obtained as follows:
1. Make a mark on 15 in the x-axis
2. Draw a vertical line (vertical means parallel to y-axis) through this mark
3. Let this vertical line intersect the graph at 'P'. The vertical line is shown in purple colour in fig.1.12 below

Fig.1.12

4. Draw a horizontal line (horizontal means parallel to x-axis) through 'P'
5. Mark the point where this horizontal line intersects the the y-axis
6. Take this reading on the y-axis. This reading is the required distance
7. In the above graph, the horizontal line meets the y-axis at 450. So we can confirm that, the object travelled 450 m in 15 seconds after it passed 'O'
Explanation: 
• 'P' is like any other point on the yellow line. 
• It's x coordinate will be the time since the object passed 'O'
• It's y coordinate will be the distance from  'O'
• In this problem, P has coordinates (15,450). 
• So 450 m is the required answer

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In the next section, we will see another application. 

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Chapter 1.1 - Average speed and Average velocity

In the previous section, we learned about Displacement. In this section we will see Uniform and Non-uniform Motion.

Uniform motion and non uniform motion

■ Consider an object moving in a straight line. Let it travel:
• 5 m in the first second. 
• 5 more metre in the next second, 
• 5 m in the third second and 
• 5 m in the fourth second. 
So the distance covered by the object in every second is equal. It is 5 m
■ If an object covers equal distances in equal intervals of time, it is said to be in uniform motion.
• In our day to day life, such uniform motion is occurs only rarely. What we encounter most is non-uniform motion.
• For example, a car travelling in a crowded street will not be able to maintain steady speed. 
    ♦ It may be able to cover 25 m in the first 5 seconds. 
    ♦ If the crowd increases after those five seconds, the speed will have to be so decreased that, it travels only 5 m in the next 5 seconds. 
• The motion of the car is not uniform in this case.

Speed

• If an object is moving fast, it will cover a larger distance within a given time ‘t’
• If another object is moving slowly, it will cover a smaller distance within the same time ‘t’
• So we need a method to compare the motions of different objects. 
• For that comparison, we use a unit called ‘metre per second’. In short form it is written as m/s or ms^-1
• For greater speeds we use kilometres per hour. In short form it is written as km/hr or kmph
■ If some one tells us that "a car is moving at 40 km/hr", we can do the following simple analysis:
1. We see that car moving (see fig.1.4 below)
2. We make a 'mark 1' on the ground.
3. At the instant when the car crosses that mark, we start a stop watch
4. Exactly after 1 hour, we check the position of the car.
5. The position will be 40 km away from the first mark. This is shown in the fig.1.4 below:

Fig.1.4

6. That means, the car has travelled 40 km in one hour. 
• That is 40 km per hour. 
• So the speed is 40 km/hr. 
7. If after 1 more hour, we check the position of the car, it will be 80 km away from Mark 1.

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So we saw how to express the speed of an object. The car we saw was in an uniform motion. That is., it covers equal distances in equal intervals of time. What if the car was moving in a crowded street? Then it’s motion will not be uniform. In such a case, how will we express the speed?

For non uniform motion, we will have to obtain average speed. It is given by:
Eq.1.1:
Average speed = ^{Total distance}⁄_{Total time}
Let us see an example:
• An object travels 16 m in 4 seconds. Then it travels another 16 m in 2 seconds. 
• So the total distance travelled = 16 + 16 = 32 m
• Total time taken for this travel = 6 seconds
• Thus average speed = ^{Total distance}⁄_{Total time} = ^{32 m}⁄_{6 s} = 5.33 m/s

Velocity

■ Velocity is obtained when we specify direction also with speed. 
Examples: 
• An object is moving with a velocity of 20 m/s along a line which makes an angle 60^o with the y axis. See fig.1.5(a)
• A car is moving with a velocity of 35 km/hr in the North east direction. See fig.1.5(b) 

Fig.1.5

Change in velocity
We have seen that velocity has both magnitude and direction. So a velocity can be changed by any one of the following three methods:
• Changing speed keeping direction the same
    ♦ An example: An object was travelling with a velocity of 20 m/s, at an angle of 25^o with the x-axis. After some time, the speed changed to 15 m/s. In this case, the velocity has changed even if the object continues to travel in the same direction. The new velocity should be specified by the new speed and the same direction  
• Changing direction keeping speed the same
    ♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. In this case, the velocity has changed even if the car continues to travel with the same speed. The new velocity should be specified by the speed 30 kmph and new direction.
• Changing both speed and direction
♦ An example: A car was travelling with a velocity of 30 Kmph in the north east direction. After some time, the car changed course. The speed also changed to 20 kmph. In this case, the velocity has changed due to change in both speed and direction.

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• Consider an object travelling along a straight line. This line is ‘straight’, with no bends or curves. 

• So that object is travelling with absolutely no change in direction. 

• If that object is travelling with an uniform speed, it is indeed travelling with an uniform velocity.

■ In our day to day life, we are able to travel along straight roads with no bends or curves. But it is difficult to travel with a uniform speed even if the road is straight. That means, we are travelling with non uniform velocity. In such cases, how will we express the velocity?
Ans: If the velocity is not uniform, we will have to use ‘average velocity’.
It is calculated as follows:
Eq.1.2:
Average velocity = ^{Total displacement}⁄_{Total time}
This calculation can be demonstrated with an example:
1. Consider fig.1.6 below. A car is initially at rest at the origin point 'O'.

Fig.1.6

2. From O, it travels in the positive direction of the x-axis for a distance of 4 Km, and reaches 'A'. The journey from O to A takes 10 minutes.

3. Then it travels in the positive direction of the y-axis for a distance of 3 Km, and reaches 'B'. The journey from A to B takes 5 minutes.
4. B is the final position of the car. So displacement of  the car is OB = 5 Km
5. Total time for the travel = 10 + 5 = 15 minutes
6. Thus average velocity = ^{Total displacement}⁄_{Total time} = ^{5 Km}⁄_{15 mins}
• 15 mins = 0.25 = 1/4 hr 
• So average velocity = ^{5 Km}⁄_{15 mins} = ^{5 Km}⁄_{1/4 hr} = 20 kmph

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We will now see some solved examples

Solved example 1.1

The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h^-1 and m s^-1.

Solution:

• Initial reading = 2000 Km

• Final reading = 2400 Km

• So distance travelled = 2400 – 2000 = 400 Km

• Total time = 8 h

We have:

■ Average speed = ^{Total distance}⁄_{Total time} = ^{400 Km}⁄_{8 h} =  50 km h^-1

• 400 Km = 400 × 1000 = 400000 m

• 8 h = 8 × 60 × 60 = 28800 seconds
■ Average speed = ^{Total distance}⁄_{Total time} = ^{400 Km}⁄_{8 h} = ^{400000 m}⁄_{28800 s} = 13.88 m s^-1

Solved example 1.2
Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Solution:
To find average speed:

• Distance travelled = 180 m

• Total time = 1 m

We have:

■ Average speed = ^{Total distance}⁄_{Total time} = ^{180 m}⁄_{1 m}

• 1 mins = 60 s
• So average speed = ^{180 m}⁄_{1 min} = ^{180 m}⁄_{60 s} = 3 m s^-1

To find average velocity:

• Displacement  = 0 m (∵ initial position and final position are the same)

• Total time = 1 m

We have:

■ Average velocity = ^{Total displacement}⁄_{Total time} = ^{0 m}⁄_{1 m} =  0 m/min = 0 m s^-1

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In the next section, we will learn about Rate of change of velocity. 

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