A particle is moving in a straight line with velocity v m.s⁻¹

Question

Question

My Attempt

I have done part A and got it right, but am struggling with part B. I have attached my attempt on the question as well as the full question.

The correct answer for part B is: t1 = 4.07 s or t2 = 7.93 s (5 marks)

But I'm not sure where I went wrong and what I have to do.

Answer 1

Your formula 1) for the position is fine, but it is valid only up to $t=4$. Hence your solution $t=4.065$ must be discarded.

Your formula 2) isn't correct, because position $s(t)$ at $t=4$ is $8/15$: a constant must then be added to account for that. The correct formula is as follows: $$ s(t)=\int_0^t v(\tau)\,d\tau= \int_0^4 (\tau^2-0.4\tau-4.4)\,d\tau+ \int_4^t (30-5\tau)\,d\tau=30t-{5\over2}t^2-80+{8\over15}. $$