High school kinematic physics problem. Solving for displacement

Question

I recently got asked the problem that follows:

The leader of a bicycle race is traveling with a constant velocity of +14.20 m/s and is 16.6 m ahead of the second-place cyclist. The second-place cyclist has a velocity of +8.70 m/s and an acceleration of +1.15 m/s2. How much time elapses before he catches the leader?

I got the answer 12.89 seconds, although I know this is incorrect. I have no clue how to even begin solving this. Here's what I did as a guess though:

First, I set up my matrix:

Displacement: 16.6

Vi: 8.7

Vf: ?

Acceleration: 1.15

Time: ?

Of the four kinematic equations, the only one that could be used is d = vi*t + 1/2*a*t^2

The problem is I am unable to rearrange this equation to get the desired product, which is time.

Answer 1

What if you set up two equations $$x_1 = 14.2t + 16.6$$ and $$x_2 = \frac{1.15}{2}t^2+8.7t$$ and set them equal to each other. Each expressions represents the displacement value of the respective rider and when they equal each other, the 2nd rider has caught the 1st.

Answer 2

Simply consider it in terms of a relative frame $s_1-s_2=(s_{0,1}-s_{0,2})+(v_1-v_2)\cdot t+1/2 (a_1-a_2)\cdot t^2)=16.6+5.5 \cdot t-1/2\cdot1.15 \cdot t^2=0$.

The kinematic equation to use actually is $$ s(t)= s(0)+v(0)t+1/2 at^2$$ for constant $a(t)=a(0) = a$.

addendum as per your comment

Sorry for going too much ahead. Suppose you know the equation for $s(t)$ that I put last, which comes from $$ \ddot s(t) = a\quad \Rightarrow \quad \dot s(t) = v(0) + at\quad \Rightarrow \quad s(t) = s(0) + v(0)t + {1 \over 2}at^{\,2} $$ Then just express it for the leader and for the 2nd cyclist, where the starting positions are $s_1(0)=s_2(0)+d$, and find the $t$ value for which $s_1(t)=s_2(t)$