A2A:
For the first car, x1=v0t+x0.
v0=20 is its constant velocity, and let x0 be the position of its nose when passing begins.
For the second car, x3=at2+v1t+x2.
Its constant acceleration is a=2. Let x2=x0+5 be the position of its nose when passing begins. (These are big cars!) Passing ends when x3=x1−5. Now we can write
x1=20t+x0,
x1−5=2t2+v1t+x0+5.
We have 2 equations in 4 unknowns. We need two more pieces of information in order to be able to solve the system. I have not used the knowledge that the second car is initially at rest.
If I assume that “initially” means “when passing begins”, then I can let v1=0. We then have
x1=20t+x0,
x1−5=2t2+x0+5.
The answer we seek is the value of x1−x0, which allows us to treat two unknowns as one. From the first equation,
(1)x1−x0=20t.
From the second equation,
(2)x1−x0=2t2+10.
We can equate the right sides of the equations to get
20t=2t2+10
or t2−10t+5=0,
which factors to
(t−1)(t−5)=0.
This has two solutions: t=1 and t=5. This is expected, because the second car should later overtake the first, some time after having been overtaken, because the second car is accelerating and the first is not. Equation (1) yields x1−x0=20 and 100 meters and (2) yields 12 and 60 meters, so I have made a mistake!
It's morning and I haven't had my coffee yet (that's my excuse), so I shall take a break for breakfast and then look at this again.
Edit: I had factored it wrong. Using the quadratic formula
