Answer:
The detailed calculations are shown below;
Step-by-step explanation:
a)The maximum acceleration of the particle:
It is seen that the maximum change in velocity is at the time between 8s to 10s.
Maximum acceleration:
![(V)/(t)](https://img.qammunity.org/2021/formulas/physics/middle-school/wa4y4u0vz27rro5bd6kcmi958xfav67v0v.png)
=
![(20)/(2)](https://img.qammunity.org/2021/formulas/chemistry/middle-school/j9tho7i4jqbx4958yjor2utb1lwd2u3sh4.png)
= 10 m/
![s^(2)](https://img.qammunity.org/2021/formulas/physics/high-school/gc1nu4waym469je1mfqno1uqpq0jqmefce.png)
b) The deceleration of the particle
The velocity of particle is decreased after 10s so,
deceleration = -
![(40)/(6)](https://img.qammunity.org/2021/formulas/physics/middle-school/xupqmuqgg59pl5n9pgmu59c46eveztaxh9.png)
= - 6.67 m/
![s^(2)](https://img.qammunity.org/2021/formulas/physics/high-school/gc1nu4waym469je1mfqno1uqpq0jqmefce.png)
c)The total distance traveled by the particle = Area under the curve
=
* 4*20 + 4*20 +
* 2*20+ 2*20+
* 40*16
= 290 m
d)The average velocity of the particle =
![(Area under the curve)/(Total time)](https://img.qammunity.org/2021/formulas/physics/middle-school/76bqnjajzg1teerzr40hi7yipmad9qhl3d.png)
=
![(290)/(16)](https://img.qammunity.org/2021/formulas/physics/middle-school/lokn3fsgorb62oasuwpma5r5sv4ojne5x1.png)
= 18.12 m/s