Help with vector question

The forces,
F(1)= 4i+8j
F(2)= -6i+2j
F(3)= 3i-14j
are acting on a particle that is on a smooth surface. The vectors i and j are perpendicular and lie on the horizontal surface.

a) Find the resultant force
b)the mass of the particle is 5kg. Find the acceleration of the particle
c)At time,t=0, the particle has a velocity of 5j and the particle is at the origin. Find the position vector of the particle relative to the origin when it is moving parallel to the vector i+j

What i got;
a) just add up all the force, i got (i-4j)
b) a=0.2i-0.8j
c) this is the part I am stuck on, the correct answer is 2.5i+15j. but i have no idea how to get that.

Please help :)

c)
First, use the formula v = v_0 + at
to find an equation for v.
v is the velocity, v_0 is the velocity at start, a is acceleration and t is time.

Try to do it yourself before looking at hint 1 below the line. (You already know v_0 and a.)
The answer is:
v = 0.2ti + (5–0.8t)j

Now, find the correct value for t such that v is parallel to the vector i+j.
v is parallel to i+j if v can be written as n(i+j) where n is a number.

Try to do it yourself before looking at hint 2 below the line.
The answer is:
t = 5

And finally, use the formula r = v_0 t + ½at²
to find r, where r is the position vector.

Try to do it yourself before looking at hint 3 below the line.
You already know what the answer is.
---------------------------------------…
Hint 1: at = (0.2i – 0.8j) t = 0.2ti – 0.8tj

Hint 2: v = n(i+j) = ni + nj = 0.2ti + (5–0.8t)j
Since n = n, then:
0.2t = 5–0.8t

Hint 3: r = v_0 t + ½at² = 5j · 5 + ½(0.2i–0.8j) · 5²