Chapter 1.9 - Solved examples on Motion of Objects

In the previous section, we completed the discussion on the motion of objects. In this section we will see some solved examples.

Solved example 1.16
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
1. Diameter of the circular track = d = 200 m. So radius = r = 100 m
2. Perimeter of the track = 2πr = πd = 200π m
3. Time for completing one round = t = 40 s
4. Speed of the athlete = ^distance⁄_time  = ^πd⁄_t  = ^200π⁄₄₀ = 5π m s^-1.
5. So the athlete covers a distance of 5π m in 1 s. We have to find the distance that the athlete covers in 2 mins and 20 s
6. 2 mins 20 s = 140 s
7. So the distance covered in this time = 140 × 5π = 700π m
8. Now we have to find the displacement at the end of 140 s
9. When the athlete covers one perimeter 200π, from the starting point, he reaches back to his starting point. At that time, his displacement is zero
10. In this way, when he travels 700π, he reaches back the starting point 3 times. After the third time he travels a 'certain distance' to make the 700π. This 'certain distance is the displacement'. It can be calculated as follows:
11. 700π = 3 × 200π + 100π.
So the 'certain distance' = displacement = 100π.

Solved example 1.17
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution:
Case 1. From A to B:
1. Distance = 300 m
2. Time = 2 mins 30 s = 150 s
3. Speed = ^distance⁄_time = ³⁰⁰⁄₁₅₀ = 2 m s^-1.
4. Velocity = ^displacement⁄_time 
• Here displacement = 300 m
• So velocity = ^displacement⁄_time = ³⁰⁰⁄₁₅₀ = 2 m s^-1.
Case 2. From A to C
1. Distance = (300 m + 100 m) = 400 m
2. Time = (2 mins 30 s + 1 min) = 3 mins 30 s = 210 s
3. Speed = ^distance⁄_time = ⁴⁰⁰⁄₂₁₀ = 1.904 m s^-1.
4. Velocity = ^displacement⁄_time 
• Here displacement = 200 m
• So velocity = ^displacement⁄_time = ²⁰⁰⁄₂₁₀ = 0.952 m s^-1.

Solved example 1.18
Abdul, while driving to school, computes the average speed for his trip to be 20 kmph. On his return trip along the same route, there is less traffic and the average speed is 30 kmph. What is the average speed for Abdul’s trip?  
Solution:
1. Let the distance from home to school be 's'
2. Let the time required in the morning be t1
3. Then Abdul calculated his average speed in the morning to be 20 kmph in the following way:
Speed = ^distance⁄_time = ^s⁄_t1 = 20 kmph
4. From this we get: t1 = ^s⁄₂₀ .
5. In the evening he travels the same distance 's'
6. Let the time required in the evening be t2
7. Then he calculated his average speed in the evening to be 30 kmph in the following way:
Speed = ^distance⁄_time = ^s⁄_t2 = 30 kmph
8. From this we get: t2 = ^s⁄₃₀ .
9. So total time required for the travel = (^s⁄₂₀ + ^s⁄₃₀) = ^5s⁄₆₀.
10. Total distance = s + s = 2s
11. So average speed = ^{total distance}⁄_{total time} = 2s ÷ ^5s⁄₆₀ = 2s × ⁶⁰⁄_5s = 24 kmph

Solved example 1.19
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^-2 for 8.0 s. How far does the boat travel during this time? 
Solution:
1. The motorboat starts from rest. So initial velocity u = 0
• Time of travel = 8 s
• Acceleration = 3 m s^-2.
2. We want the distance s
The second equation of motion connects the above quantities. So we will use it:
3. s = ut + ¹⁄₂ at² ⇒ s = 0 × 300 + ¹⁄₂ × 3 × 8² ⇒ s = 96 m

Solved example 1.20
A driver of a car travelling at 52 kmph applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 30 kmph in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Solution:
The fig. shows the speed-time graph for the two cars:

1. The yellow line AB is the speed-time graph of the first car. The purple line PQ is the speed-time graph of the second car.
Note that, the speeds are converted from kmph to m s^-1. 
2. The area enclosed by a speed-time graph and the time axis will give the distance travelled by the object. We saw the details here. The areas are shown shaded in the above fig.
3. First car:
Area = Area of the triangle OAB = ¹⁄₂ × 5 × 14.44 = 36.1 m
[ ∵ Base = 5 units and altitude = 14.44 units]
4. Second car : 
Area = Area of the triangle OPQ = ¹⁄₂ × 10 × 8.33 = 41.65 m
[ ∵ Base = 10 units and altitude = 8.33 units]

Solved example 1.21
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Solution:
1. Height of fall = distance of travel = s = 20 m
2. Acceleration = a = 10 m s^-2.
3. The ball falls from rest. So it has an initial velocity u = 0
4. We have to find the final velocity v. We can use the third equation of motion. Because it connects all these quantities
5. So we can write:
v² = u² + 2as ⇒ v² = 0² + 2 × 10 × 20 ⇒ v² = 400 ⇒ v  = 20 m s^-1.
6. To find the time, we can use the first equation of motion:
v = u + at ⇒ 20 = 0 + 10t ⇒ 20 = 10 t ⇒ t = 2 s

Solved example 1.22
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Solution:
1. Radius of the orbit = r = 42250 km
2. Perimeter of the orbit = 2πr = 2 × π × 42250 = 84500π km = 84500000 m
3. Time for completing one round revolution = t = 24 hours = 24 × 60 × 60 = 86400 s
4. Speed of the satellite = ^distance⁄_time  =  ^84500000π⁄₈₆₄₀₀ = 3070.95 m s^-1

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So we have completed the discussion on the 'Motion of objects'. In the next chapter, we will see 'Forces acting on objects'. 

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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 1.7 - Equations of Motion - Solved examples

In the previous section, we derived the three equations of motion. They are:
■ First equation of motion: v = u + at 
■ Second equation of motion: s = ut + ¹⁄₂ at²
■ Third equation of motion: v² = u² + 2as
In this section we will see some solved examples.

Solved example 1.8
A train starting from rest attains a velocity of 72 kmph in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.
Solution:
Part (i):
1. The train starts from rest. So u = 0
2. The final velocity attained in 5 minutes is 72 kmph. So v = 72 kmph. 
3. And t = 5 minutes
• Note the units: Speed is given in kmph and time is given in minutes
• But 'minutes' is not generally used when we specify velocity. 'Seconds' or 'hours' are used
    ♦ In kmph we use hours
    ♦ In m s^-1 we use seconds 
• So we have to change the '5 minutes' to either hours or seconds. In this problem, it is convenient to use seconds. So we get:
t = 5 minutes = 5 × 60 = 300 Seconds
4. Since time is in seconds we must convert velocity into m s^-1. We get:
72 kmph = ^{(72 × 1000)}⁄_{(1 × 3600)} = 20 m s^-1
5. So we have u, v and t. We have to find the acceleration a
6. Out of the three equations of motion, the one which connects the four quantities in (5), is the first equation. So we will use it.
7. v = u + at ⇒ a = ^(v-u)⁄_t ⇒ a = ^(20-0)⁄₃₀₀ = ¹⁄₁₅ m s^-2.
Part (ii):
1. We have u, v, t and a. We have to find s.
2. Out of the three equations of motion, there is none which connects the five quantities in (1). But we do not need all the five quantities to be connected. The second equation will help us to get the result. So we will use it.
3. s = ut + ¹⁄₂ at² ⇒ s = 0 × 300 + ¹⁄₂ ×(¹⁄₁₅)×300² ⇒ s = 3000 m = 3 Km
■ So the train was able to increase it's velocity from zero to 20 m s^-1 within a distance of 3 Km because, it was subjected to an acceleration of ¹⁄₁₅ m s^-2 . And the train took 300 seconds (5 minutes) to attain the velocity of 20 m s^-1.

Solved example 1.9
A car accelerates uniformly from 18 kmph to 36 kmph in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.
Solution:
Part (i):
1. u = 18 kmph = ^{(18 × 1000)}⁄_{(1 × 3600)} = 5 m s^-1
2. v = 36 kmph = ^{(36 × 1000)}⁄_{(1 × 3600)} = 10 m s^-1
3. t = 5 s
4. We have to find acceleration a. Out of the three equations of motion, the one which connects the above three quantities and a is, the first equation. So we will use it.
5. v = u + at ⇒ a = ^(v-u)⁄_t ⇒ a = ^(10-5)⁄₅ = ⁵⁄₅ = 1 m s^-2.
Part (ii):
1. We have u, v, t and a. We have to find s.
2. Out of the three equations of motion, there is none which connects the five quantities in (1). But we do not need all the five quantities to be connected. The second equation will help us to get the result. So we will use it.
3. s = ut + ¹⁄₂ at² ⇒ s = 5 × 5 + ¹⁄₂ × 1 × 5² ⇒ s = 37.5 m
■ So the car was able to increase it's velocity from 5 m s^-1 to 10 m s^-1 within a distance of 37.5 m because, it was subjected to an acceleration of 1 m s^-2 . And the car took 5 seconds to attain the velocity of 10 m s^-1.

Solved example 1.10
The brakes applied to a car produce an acceleration of 6 m s^-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Solution:
1. a = -6 m s^-2. The negative sign is applied because it is deceleration. It is opposing the motion.
2. t = 2 s
3. We have to find the distance s. 
• Consider the three equations of motion. Only the second one connects s, a and t. 
• So we will have to use it. But to use it, we need to find the initial velocity u. It is the velocity at which the car was travelling just before brakes began to slow it down. 
• So we will use the first equation to find that u:
4. v = u + at ⇒ u = v - at ⇒ u = 0 - (-6) × 2 ⇒ u = 12 m s^-1. (∵ final velocity is zero as the car comes to a stop)
5. Now we can use the second equation:
s = ut + ¹⁄₂ at² ⇒ s = 12 × 2 + ¹⁄₂ × (-6) × 2² ⇒ s = 24 + (-12) = 12 m
6. Note that, once we find u, we can use the third equation also:
v² = u² + 2as ⇒ 0² = 12² + 2 × (-6) × s ⇒ 0 = 144  -12s ⇒ 12s = 144 ⇒ s = 12 m
■ So the car was able to decrease it's velocity from 12 m s^-1 to 0 m s^-1 because, it was subjected to a deceleration of (-6) m s^-2 . And the car took 2 seconds to come to a stop. Also note that the car travelled 12 m during this 2 seconds. This is why drivers are cautioned to maintain some distance between vehicles while travelling on the road.

Solved example 1.11
A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Solution:
Part (i):
1. u = 0, a = 0.1 m s^-2, t = 120 s
2. We have to find the final speed v. 
3. We can use the first equation of motion. It connects all the above four quantities.
4. v = u + at ⇒v = 0 + 0.1 × 120 = 12 m s^-1.
Part (ii):
1. We have to find s. We can use the second equation:
s = ut + ¹⁄₂ at² ⇒ s = 0 × 120 + ¹⁄₂ × 0.1 × 120² ⇒ s = 0 + 120 × 6 = 720 m

Solved example 1.12
A train is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of (-0.5) m s^-2. Find how far the train will go before it is brought to rest.
Solution:
1. a = -0.5 m s^-2. The negative sign is applied because it is deceleration. It is opposing the motion.
2. t = 2 s
3. u = 90 kmph = ^{(90 × 1000)}⁄_{(1 × 3600)} = 25 m s^-1
4. We have to find the distance s. We can use the third equation of motion. 
5. v² = u² + 2as ⇒ 0² = 25² + 2 × (-0.5) × s ⇒ 0 = 625  - s ⇒ s = 625 m

Solved example 1.13
A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?
Solution:
1. u = 0, a = 2 cm s^-2, t = 3 s
2. We have to find the final speed v. 
3. We can use the first equation of motion. It connects all the above four quantities.
4. But before that, we have to convert the acceleration from cm s^-2 to m s^-2
We have: 2 cm = ²⁄₁₀₀ m = 0.02 m. So 2 cm s^-2 = 0.02 m s^-2.
5. v = u + at ⇒v = 0 + 0.02 × 3 = 0.06 m s^-1.

Solved example 1.14
A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?
Solution:
1. u = 0, a = 4 m s^-2, t = 10 s
2. We have to find the distance s. We can use the second equation of motion: 
3. s = ut + ¹⁄₂ at² ⇒s = 0 + ¹⁄₂ × 4 × 10² ⇒ s = 200 m

Solved example 1.15
A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Solution:
1. u = 5 m s^-1, a = -10 m s^-2.
2. We have to find s and t. We can not use the second equation s = ut + ¹⁄₂ at² because it has two unknowns s and t. Let us try the third equation:
3. v² = u² + 2as. In this also there seems to be two unknowns v and s. But in reality, v is not an unknown. Let us analyse: 
• The maximum height reached by the stone is the s that we want. 
• When the stone is thrown upwards, it's velocity continuously decreases because of the deceleration of 10 ms^-2. 
• The velocity continuously decreases and finally becomes zero. So v = 0. After that, it falls downwards.
4. So we can use the third equation: 
v² = u² + 2as ⇒ 0² = 5² + 2 × (-10) × s ⇒ 0 = 25  - 20s ⇒ s = 1.25 m 
5. To find the time, we can use the first equation:
v = u + at ⇒0 = 5 + (-10) × t ⇒0 = 5 -10t ⇒5 = 10t ⇒t = 0.5 s

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So we have completed the discussion on the 'Motion along a straight line'. In the next section, we will see 'Motion along a circular path'. 

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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)

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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 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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