2 dimensional motion - calculate particles position

[new^2^physics]

Homework Statement

A particle starts from rest x=0 and moves for 10 s with an acceleration component of +2.0 cm/s². for the next 20 s, the acceleration component of the particle is -1.0 cm/s². What is the position of the particle at the end of this motion?

Possible answers:

a. zero
b. + 3.0 m
c. - 1.0 m
d. +2.0 m
e. -3.0 m

2. The attempt at a solution

a. deltaX = Vit + 1/2 At²
= 0 (10) + 1/2 (0.02)(10²)
= 1m

b. velocity
vf = vi +at
= 0 + (0.02)(10)
= 0.2 m/s

c. deltaX = Vit + 1/2 At²
= (0.2)(20) = 1/2 (-0.01)(20²)
= 2m

d. total distance traveled is 1m + 2m = + 3.0m

Therefore I would argue that "b" would be the correct answer. Is this correct?

 

[G01]

Looks correct to me.