Distance travelled by the particle $=$ Area under the given graph

$=\frac{1}{2} \times(10-0) \times(12-0)=60 m$

$\text {Average speed}= \frac{\text {Distance}}{\text {Time}}=\frac{60}{10}=6 m / s$

Let $s_{1}$ and $s_{2}$ 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 $s=s_{1}+s_{2} \ldots (i)$

For distance si:

Let $u^{\prime}$ be the velocity of the particle after 2 s and $a^{\prime}$ 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 first equation of motion, acceleration can be obtained as:

$v=u+a t$

Where,

$v=$ Final velocity of the particle

$12=0+a^{\prime} \times 5$

$a^{\prime}=\frac{12}{5}=2.4 m / s ^{2}$

Again, from first equation of motion, we have

$v=u+a t$

$=0+2.4 \times 2=4.8 m / s$

Distance travelled by the particle between time $2 s$ and $5 s$ i.e., in $3 s$ $s_{1}=u^{\prime} t+\frac{1}{2} a^{\prime} t^{2}$

$=4.8 \times 3+\frac{1}{2} \times 2.4 \times(3)^{2}$

$=25.2 \dots (ii)$

For distance $s_{2}$

Let $a^{\prime \prime}$ be the acceleration of the particle between time $t=5 s$ and $t=10 s$ From first equation of motion, $v=u+a t$ (where $v=0$ as the particle finally comes to rest) $0=12+a^{\prime \prime} \times 5$

$a''=\frac{-12}{5}$

$=-2.4 m / s ^{2}$

Distance travelled by the particle in 1 s (i.e., between $t=5$ s and $t=6$ s)

$s_{2}=u^{\prime \prime} t+\frac{1}{2} a t^{2}$

$=12 \times a+\frac{1}{2}(-2.4) \times(1)^{2}$

$=12-1.2=10.8 \;m\dots (iii)$

From equations $(i), (ii)$, and $(iii)$, we get

$s=25.2+10.8=36 m$