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 1.6 - Equations of Motion from Velocity-Time Graph

In the previous section, we determined the acceleration from a velocity-time graph. In this section we will determine the distance travelled. 
• Consider our original velocity-time graph (fig.1.19 in the previous section). 
• Suppose we want to know the distance travelled by the car in a duration of 8 seconds starting from 12. 
• Then our t₁ is 8 and t₂ is 20. 
• To find that distance, we use the following procedure:
1. Erect two perpendiculars from t₁ and t₂. Extend them upwards until they meet the velocity-time graph. This is shown in the fig.1.24 below:

Fig.1.24

2. The distance travelled by the car in the duration of 12 seconds (from t₁ = 8 to t₂ = 20) is the area of the quadrilateral ABCD. Let us see the proof:
3. It is easy to prove this if, the velocity-time graph is horizontal. We saw it earlier in fig.1.18 of the previous section. The quadrilateral we saw there is a rectangle. But here, the quadrilateral is a trapezium.
4. Consider a very small duration of time dt, after say 12 seconds (In 'dt', the 'd' stands for 'delta', the Greek symbol used to denote very small quantities'). It is shown shaded in the fig.1.25 below:

Fig.1.25

5. The shaded area is a small trapezium. Let us see it's details:
• The height of it's left vertical side gives the initial velocity (velocity just before the beginning of dt)
• The height of it's right vertical side gives the final velocity after the duration dt
• But dt is so small that, the increase in velocity during that duration can be ignored.
• So the height on the right side is same as that on the left side. That means, the car travelled at a constant velocity for a duration of dt.
• So, it is not a trapezium but a rectangle
6. We know that, if it is a rectangle, the distance travelled during dt will be equal to the area of the rectangle
7. So the distance travelled by the car for the duration of dt after 12 seconds, is the area of the rectangle, whose width is dt, shown shaded in the fig.1.25 above
8. We can divide the total area of the trapezium ABCD into so many such rectangles. All rectangles will have the same width of dt. But the heights will be different. See fig.1.26(a) below:

Fig.1.26

• Consider a rectangle of width dt near the left side in fig.1.26(a) 
    ♦ It's height will be less. 
    ♦ So it's area will be less 
    ♦ So the distance travelled during that dt will be less
• Consider a rectangle of width dt near the right side in fig.1.26(a) 
    ♦ It's height will be more.
    ♦ So it's area will be more
    ♦ So the distance travelled during that dt will be more
9. We divide the total duration between A and B into a number of dt's. This is shown in fig.1.26(b). (In the fig.b, only a few rectangles are shown. In reality, the space between A and B is completely filled with rectangles. And the width of all those rectangles is dt) 
10. The distance travelled in each of these dt's = Area of the rectangle in that dt
11. Sum of the distances travelled in all the dt's = Total distance travelled in the duration from t₁ to t₂
12. But left side of (11) is the 'sum of areas of all the rectangles'. So this can be included in (11)
13. So (11) becomes:
Sum of the distances travelled in all the dt's 
= Total distance travelled in the duration from t₁ to t₂ 
= Sum of areas of all the rectangles 
14. Consider the last part in the above: 'Sum of areas of all the rectangles'. This is the 'total area of the trapezium ABCD'.
15. So (13) becomes:
Sum of the distances travelled in all the dt's 
= Total distance travelled in the duration from t₁ to t₂ 
= Sum of areas of all the rectangles
= Total area of the trapezium ABCD
16. Take the second and fourth parts from (15). We get:
■ Total distance travelled in the duration from t₁ to t₂ = Total area of the trapezium ABCD 
This is the required proof.

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Now we will try to write a general equation for the area. So that, we will be able to calculate the distance  's' quickly. Consider the velocity-time graph given below:

Fig.1.27

1. We want the area of the trapezium ABCD in the fig.1.27
2. From the fig. it is clear that the height AD = u and height BC = v. Also AB = t = (t₂ - t₁)
3. We have: Area of a trapezium = ^h⁄₂ × (a+b)
Where a is the top parallel side, b is the bottom parallel side and h is the height Details here.
4. To use this formula in our present case, we must consider:
• AD as the top parallel side
• BC as the bottom parallel side
• AB as the height
5. So we get:
Area of trapezium ABCD = Ditance travelled s = ^t⁄₂ × (u+v)
6. But in the previous section we obtained Eq.1.6:
v = u + at 
7. Substituting this v in (5) we get:
s =  ^t⁄₂ × (u + u + at )
⇒ s =   ^t⁄₂ × (2u + at ) 
From this we get the equation below:
Eq.1.7:
s = ut + ¹⁄₂ at²
This is the second equation of motion.

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We have to derive one more equation. But it does not require the help of a graph. We can derive it from the first two.
1. While we were deriving the second equation of motion, we had reached step (5) above. In that step, we got: s =  ^t⁄₂ × (u+v)
2. But we can obtain the value of t from the first equation of motion: v = u + at
⇒ t = ^(v-u)⁄_a.
3. Substituting this in (1) we get:
s =  ^(v-u)⁄_2a × (u+v). 
But [(v+u)(v-u)] is an identity. We have:
[(v+u)(v-u)] = (v² - u²)     
4. So (3) becomes:
s = ^(v2-u2)⁄_2a.
5. From this we get the third equation of motion:
Eq.1.8:
v² = u² + 2as

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Let us write the three equations together:
■ First equation of motion: v = u + at 
■ Second equation of motion: s = ut + ¹⁄₂ at²
■ Third equation of motion: v² = u² + 2as

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Now we will see some solved examples:
Solved example 1.6
The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table below:

Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.
Solution:
The required distance time graph is shown in fig.1.28 below:

Fig.1.28

Interpretation:
1. At Station A, there are two red dots in the graph. 
• These red dots have the same y coordinates '0' because, the train remains at the same point (Station A) from 8 hours to 8:15 hours. 
• There is no change in distance during those 15 minutes. So the y coordinates does not change. 
• But the x coordinates change because the time passes from 8:00 hours to 8:15 hours  (Note that 0:15 hours is 15 minutes, which is 0.25 hours)
• At 8:15 hours it begins it travel to Station B. So the distance from Station A increases. This is the 'yellow sloping line' from (8.25,0) to (11.25,120)

2. At Station B, there are two red dots in the graph. 
• These red dots have the same y coordinates '120' because, the train remains at the same point (Station B) from 11:15 hours to 11:30 hours. 
• There is no change in distance during those 15 minutes. So the y coordinates does not change. 
• But the x coordinates change because the time passes from 11:15 hours to 11:30 hours  (Note that 0:30 hours is 30 minutes, which is 0.50 hours)
• At 11:30 hours it begins it travel to Station C. So the distance from Station A (The zero distance point) increases. This is the 'yellow sloping line' from (11.5,120) to (13,180)

3. At Station C, there are two red dots in the graph. 
• These red dots have the same y coordinates '180' because, the train remains at the same point (Station C) from 13:00 hours to 13:15 hours. 
• There is no change in distance during those 15 minutes. So the y coordinates does not change. 
• But the x coordinates change because the time passes from 13:00 hours to 13:15 hours  (Note that 0:15 hours is 15 minutes, which is 0.25 hours)
• After station C there no further travel is given in the table

Solved example 1.7
Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table below shows the distance travelled by them in different times.

Plot the distance-time graph for their motions on the same scale and interpret.
Solution:
The required distance time graph is shown in fig.1.29 below:

Fig.1.29

• The Distance-time graph of the travel made by Feroz is shown in yellow colour
• The Distance-time graph of the travel made by Sania is shown in cyan colour
• In the x-axis, only minutes are taken. If we take hours also, we will have to convert into decimal form as follows:
8:05 = 8 + 5/60 = 8 + 0.08333 = 8.08333
8:10 = 8 + 10/60 = 8 + 0.16667 = 8.16667
8:15 = 8 + 15/60 = 8 + 025 = 8.25
So on...
• So it is convenient to take minutes alone
• Once the scales on x and y axes are fixed, we can plot the graphs easily. The coordinates can be taken directly from the table. No calculations are required.

Interpretation:
• Feroz was travelling at a greater speed because he travelled 3.6 Kms in 20 minutes, while Sania took 25 minutes to travel the same distance
• After the plotting, if we examine carefully, we can find that, the points do not fall on straight lines. 
• This will be clear if we join the first and last points separately for the two graphs. It is shown in the fig.1.30 below:

Fig.1.30

• Note that, dashed lines are used to join the first and last points. This is to distinguish from the main graph.
• Since they are not straight lines, we can infer the following:
    ♦ Feroz did not travel at an uniform speed
    ♦ Sania did not travel at an uniform speed
Reason:
■ If the travel is at an uniform speed, we must get similar triangles for any random pairs of points that we take in a Distance-time graph. So that the ratio ^altitude⁄_base is always a constant. If it is not a straight line, we will not get similar triangles at any random pair we take. This we saw in fig.1.15
■ Also, from the readings in the given table, it is evident that the speeds were not uniform:
Feroz travelled 1 Km in the first 5 minutes
He travelled (1.9 - 1) = 0.9 Km in the next 5 minutes
He travelled (2.8 - 1.9) = 0.9 Km in the next 5 minutes
He travelled (3.6 - 2.8) = 0.8 Km in the next 5 minutes
■ Sania travelled 0.8 Km in the first 5 minutes
She travelled (1.6 - 0.8) = 0.8 Km in the next 5 minutes
She travelled (2.3 - 1.6) = 0.7 Km in the next 5 minutes
She travelled (3.0 - 2.3) = 0.7 Km in the next 5 minutes
She travelled (3.6 - 3.0) = 0.6 Km in the next 5 minutes

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In the next section, we will see a few more solved examples. 

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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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