^{(a) Distance travelled by the particle = Area under the given graph
= (1/2) × (10 - 0) × (12 - 0) = 60 m
Average speed = Distance / Time = 60 / 10 = 6 m/s
(b)Let s₁ and s₂ be the distances covered by the particle between time
t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
s = s₁ + s₂
For distance s₁
Let u₁ be the velocity of the particle after 2 s and a₁ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from the first equation of motion, acceleration can be obtained as:
v = u + at
Where,
v = Final velocity of the particle
12 = 0 +a₁ × 5
$a_1=\frac{12}{5}=2.4 m/s^2$Again, from the first equation of motion, we have velocity at t=2sec
v = u + at}

^{u₁= 0 + 2.4 × 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s
$s_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$
$=4.8\times 3+\frac{1}{2}\times 2.4\times {\left(3\right)}^2$= 25.2 m      ........(i)
For distance s₂:
Let a₂ be the acceleration of the particle between time t = 5 s and t = 10 s.
From first equation of motion,
v = u + at (where v = 0 as the particle finally comes to rest)
0 = 12 + a₂ × 5
a₂ = -12 / 5 = - 2.4 m/s2
Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)
$s_2=u_{2}t+\left(\frac{1}{2}\right)a_2{t}^2$
$=12\times 1+\left(\frac{1}{2}\right)\left(-2.4\right)\times {\left(1\right)}^2$= 12 - 1.2 = 10.8 m    .........(ii)
From equations (i) and (ii) we get
s = 25.2 + 10.8 = 36 m
Therefore,
 Average speed = Total distance/time taken = 36 / 4 = 9 m/s.}