(a) Let S be the distance covered by the particle between t = 0 tot = 10 s. Then,
S = area under (v-t) graph = area of the triangle, whose base is 10 (s) andheight is 12 (m s ‌–1 )
=

1

2

×10×12=60 m
Therefore, average speed =

distance‌covered

time‌taken

=

60 m

10 s

=6 ms−1
(b) If S1 and S2 are the distances covered by the particle in the time intervals t = 2 s to 5 s and t = 5 s to 6 s, then the distance covered in the time interval t = 2 s to 6 s is given by S=S1+S2
To find S1 : At t = 0, the particle is at rest (u = 0). Let v1 be velocity of the particle after 2 s and a1 be the acceleration during the time interval t1 = 0 s to 5 s. From figure, we have
a1=

12−0

5−0

=2.4 m s−2
∴‌v1=u+a1t
=0+2.4×2=4.8 m s−1
The distance S1 is covered during the time interval t = 2 s to t = 5 s i.e. in timet1 = 5 – 2 = 3 s and with the initial velocity v1.
Now, S1=v1t1+

1

2

a1t12
∴‌S1=4.8×3+

1

2

×2.4×32
=14.4+10.8=25.2 m
To find S2 : Let v2 be velocity of the particle after 5 s and a2 be theacceleration during the interval t = 5 s to 10 s.
From figure, we have
a2=

0−12

10−5

=−2.4 m s−2
Also, v2=12 m s−1
The distance S2 is covered during the time interval t = 5 s to t = 6 s i.e. in timet2 = 6 – 5 = 1 s and with initial velocity v2.
Thus, S2=v2t2+

1

2

a2t22
S2=12×1+

1

2

(−2.4)×(1)2
=12−1.2=10.8 m
Hence, the required distance,
S=25.2+10.8=36 m
Also , average speed =

distance‌covered

time‌taken

=

36

6−2

=

36

4

=9 m s−1