(i) Train-II might have gone in higher speed then the train-I as it takes less time to cover the same distance.
(ii) For train - I
Distance = 320 km
Time taken = 4 h 30 mins
= \(4\frac { 1 }{ 2 } \)
Speed = \(\frac { distance }{ Time } \)
\(\frac { 320 }{ 9/2 } km/h\)
= \(\frac { 320\times 2 }{ 9 } =\frac { 640 }{ 9 } \)
= 71.1 km.h
For train - II
Distance = 320km
Time taken = 30 mins less than the first train
=4h
Speed = \(\frac { Distance }{ Time } \)
= \(\frac { 320 }{ 4 } km/h\)
=80 km/h
(iii) The graph in figure (a) is of train-I
The graph in figure (b) is of train-II.

(a) The total distance covered in given interval of time is known as speed.
(b) Distance between Delhi & Agra = 270 km
Time taken by Train = 3 hours
Speed = Distance / Time
= 270/3 = 90 Km/hr.
(c) Distance - Time graph

The following steps may be followed:
(i) Draw two perpendicular lines to represent the two axis and mark them OX and OY. O is the intersection of the two axis
(ii) Decide the quantity to be shown along the x-axis and that to be shown along the y-axis. From the given data, we are measuring distance at given intervals of time. So time is to be shown along x-axis distance along y-axis
(iii) Choose a scale to represent the given data. Suppose we have 4 squares (or 40 divisions) along x-axis and 4 squares (or 40 divisions along y-axis. Along x-axis we can take 10 divisions equal to 15min and along y-axis we can take 10 divisions equal to 20 km.
(iv) Mark the values of time and distance on the respective axis.
(v) Now mark points on the graph paper to represent each set of values for distance and time.
(vi) Join all points on the graph. Graph obtained is shown below:

(i) Time in minutes on x-axis and speed in m/min on y-axis.
(ii) On x-axis, 10 divisions = 1 min and on y-axis 10 divisions = 25 m/min
(iii) At 3.30 minutes, the speed of car is 30 m/min.
(iv) Distance at 4 min = Speed x Time = 52.5 x 4 = 210.0 m
(v) After 6.30 minutes the car has assumed constant speed.

(i) On the x-axis, one division is equal to one day and on the y-axis, one division is equal to one cm.
(ii) Height on 5th day = 1.2cm
Height on 12th day = 9.2 cm
(iii) Initially the growth is slow, then between 9th and 11th day, the increase is quite fast, then again slows down and after 12th day the growth stops.
(iv) It appears period now shown, i.e., after 20th day the rate of growth is nil

We can check the motion of the objects C and 0 by plotting the distance-time graph for the two objects .

Graph for object C on x-axis
Scale 1 cm = 15 min and on y-axis 1 cm = 10 m
Since, the distance-time graph of the object C is a straight line. Therefore, its motion is uniform.
Graph of object D
Scale On x-axis, 1cm = 15min, On y-axis 1cm =5 m

Since, the distance-time graph of the object 0 is not a straight line, therefore, its motion is non-uniform.

From the question, Bholu covers 3 km distance in 15 min
speed of Bholu \(=\frac{3}{15}\times60=12\ km/h\)    \((1\ min=\frac{1}{60}h)\)
Now, average speed is given by
\(=\frac{Total\ distance\ travelled}{Total\ time\ taken}=\frac{9}{1}=9\ km/h\)

While choosing the most suitable scale for drawing a graph, we should keep the following points in mind
(i) The difference between the highest and lowest values of each quantity.
(ii) The intermediate values of each quantity so that with the scale chosen, it is easy to mark the values on the graph.
(iii) To utilise the maximum part of the paper on which the graph is to be drawn, assume that we have a graph paper of size 25 cm \(\times\) 25 cm and we have to accommodate following data:

Since, one of the scales will be
Distance 5 km = 1 cm and Time 6 min = 1 cm

(a) Since, the graph between B and C is parallel to time-axis, so it indicates that Boojho is at rest.
(b) Since, the graph is not straight line, so it is a non-uniform motion.
\(Speed=\frac{Distance}{Time}=\frac{225-150}{12-8}=\frac{75}{4}=18.75\ m/min\)

(a) From the graph, it is clear that the distance at 20 s is 8 m.
(b) Distance travelled by the body object in 12 s is 6 m.
(c) \(Average\ speed=\frac{Total\ distance}{Total\ time}=\frac{8}{20}=0.4m/s\)