SI unit of speed is meter per second. The average speed of the body is given by
\({ v }_{ av }\quad =\frac { { v }_{ 1 }{ t }_{ 1 }+{ v }_{ 2 }{ t }_{ 2 } }{ { t }_{ 1 }+{ t }_{ 2 } } \)
Here, \({ v }_{ 1 }\)= 6 km/h, \({ t }_{ 1 }\)=2 h,\({ v }_{ 2 }\) = 4 km/h,\({ t }_{ 2 }\) = 3 h
\({ v }_{ av }\)       = \(\frac { 6\times 2+4\times 3 }{ 2+3 } \)= 4.8 km/h
\(\therefore \)Total distance = \({ v }_{ 1 }{ t }_{ 1 }+{ v }_{ 2 }{ t }_{ 2 }\) = \(6\times 2+4\times 3\) = 24km

The uses of a velocity-time graph are as follows:
(i) To determine the speed of a body at any instant of time
(ii) To determine the acceleration of a body 
(iii) To detemine the total distance travelled by a body in a given time-intrerval.

The first equation of motion. Let a body start with initial velocity u and after time t, its velocity becomes v due to uniform acceleration a from the definition of acceleration
Acceleration = \(\frac { Change\quad in\quad velocity }{ Time\quad taken } =\frac { Final\quad velocity-Intial\quad velocity }{ Time\quad taken } \)
\(a=\frac { v-u }{ t } \quad \) or at = v - u
or  v = u + at.

Here, v = 90 km/h = \(\frac { 90X1000m }{ 3600s } =\) 25 m/s
s = 100 m
Minimum time, \(t=\frac { s }{ v } =\frac { 100 }{ 25 } =4s\)

(a) Average speed = \(\frac { s_{ 1 }+s_{ 2 } }{ t_{ 1 }t_{ 2 } } =\frac { 30+30 }{ 4-2 } =10\quad ms^{ -1 }\)
(b) No, the direction of motion of the object is also needed for finding its average velocity.

a = 8 \(ms^{ -2 }\),v = 32 \(ms^{ -1 }\) 
(ii) t = \(2\sqrt { 2\quad s } \)

\(v_{ av }=\frac { s_{ 1 }+s_{ 2 } }{ t_{ 1 }+t_{ 2 } } =\frac { s }{ \frac { s_{ 1 } }{ v_{ 2 } } +\frac { s_{ 2 } }{ v_{ 2 } } } \)
\(40=\frac { 100 }{ \frac { 30 }{ 30 } +\frac { 70 }{ v_{ 2 } } } \)
\(=1+\frac { 70 }{ v_{ 2 } } =\frac { 100 }{ 40 } =\frac { 5 }{ 2 } \)
\(\frac { 7 }{ v_{ 2 } } =\frac { 5 }{ 2 } -1=\frac { 3 }{ 2 } \)
\(v_{ 2 }=\frac { 7X2 }{ 3 } =\frac { 14 }{ 3 } =4.67\quad km\quad h^{ -1 }\)

(a) Speed is defined as the distance travelled by a body per unit time.
(b) The SI unit of speed is m s^-1
(c) Average speed is the total distance travelled by a body, divided by the total time taken to cover that distance.
(d) No, the average speed of an object can't be negative.

Let one-way distance = x km
Time is taken in forwarding trip at a speed of 20 km/h = \(\frac { Distance }{ Speed } =\frac { X }{ 20 } h\)
Time is taken in return trip at a speed of 30 km/h = \(\frac { X }{ 30 } h\)
Total time for the whole trip = \(\frac { Distance }{ Speed } =\frac { x }{ 20 } +\frac { x }{ 30 } =\frac { 3x+2x }{ 60 } =\frac { 5x }{ 60 } h\)
Total distance covered = x + x = 2 x  km
Average speed =\(\frac { Total\quad Distance }{ Total\quad time } =\frac { 2x }{ 5x/60 } +\frac { 2xX60 }{ 5x/60 } \) 
 = 24 km \(h^{ -1 }\) 

Time taken = 2 min 20 s = 2 x 60 + 20 = 140 s
Radius,r = 100 m
In 140 s, the athlete will complete three and a half round
Distance covered = \(2\pi \) x 3.5
= 2 x \(\frac { 22 }{ 7 } \) x 100 x 3.5 = 2200 m.
At the end of his motion, the athlete will be in the diametrically opposite position
\(\therefore \)  Displacement = diameter = 200 m

The second equation of motion. Suppose a body starts with initial velocity u and due to uniform acceleration a,its final velocity becomes v after t. Then
Average velocity=\(\frac { Intial\quad velocity+Final\quad velocity }{ 2 } =\frac { u+v }{ 2 } \)
So, the distance covered by the body in time t is
s = Average velocity x Time
= \(\frac { u+v }{ 2 } \) x t = \(=\frac { u+(u+at) }{ 2 } Xt\)   \([\therefore v=u+at]\)
\(=\frac { 2ut+at^{ 2 } }{ 2 } \)
or \(s=ut+\frac { 1 }{ 2 } at^{ 2 }\) 

Acceleration. In non-uniform motion, the velocity of a body changes with time It has different velocities at different instants of time and at different points of its path In such a situation we define a physical quantity called acceleration which is a measure of the change in the velocity of a body per unit time.
Acceleration is defined as the rate of change of velocity.If the velocity of a body changes from u to v in time t, then
Acceleration=\(\frac { Change\quad in\quad velocity }{ Time\quad taken } =\frac { Final\quad velocity-Initial\quad velocity }{ Time\quad taken } \)
or \(a=\frac { v-u }{ t } \)
As acceleration has both magnitude and direction, it is a vector quantity.

Graphical representation of motion. The nature of motion of bodies can be easily studied from its graphical representation here ,time is plotted along X-axis while distance or velocity is plotted along y-axis Graphs give detailed information about the nature of motion than when motion is expected in a tabular form.
Uses of graphs
(i) From distance -time graph the postion of the body at any instant of time can be determined
(ii) Distance covered by the body during particular interval of time can be seen from the graph.
(iii) By simply looking at the graph one can tell wheteher motion is uniform or not
(iii) The velocity of the body at any instant of time can be determined
(iv) By simply looking at the graph ,one can tell whether motion is uniform or not
(v) Slope of velocity -time graph gives the acceleration of the body
(vi) Graphs are very useful for comparing the motions of two moving bodies Distance time graphs easily tell when and where one body crosses