Mandy's on the run!

What path would take our book enthusiast the farthest from the library? Well, any path going away in a straight line from the library (still assuming it's all happening on flat land). That's obvious but it allows us to reduce the problem to a 1-dimensional situation.

There is enough information to then construct $v(t)$ , Mandy's velocity as a function of time. Her routine is drive 3h, stop 30min, drive 3h...

We know that she drives away from the library. That means accelerates ( $a= 1 m/s²$ ) from 0 to 108km/h (or 30m/s) in t=\frac { v }{ a } =\frac { 30 }{ 1 } =30s , then drives at 108km/h for 2h59 (10740s) and decelerates ( $a= -1m/s²$ ) from 30m/s to 0.

That's 3 hours of driving, how far does that take her?

So what we're looking for is the area under $v(t)$ (remember $dr=vdt$ ).

The easiest way to do so is by looking at $v(t)$ from t=0 to t=10800s and noticing it's a very simple 'curve':

So, basically a 30x30 square and a 10740x30 rectangle.

We found the distance traveled during the driving pattern. This pattern occurs 7 times in 24 hours.

Therefore,

$\large \frac { \tilde { R } }{ 7 }$ $= { 30 }^{ 2 }+30\times 10740 =$ \int _{ 0 }^{ 30 }{ tdt } +\int _{ 0 }^{ 10740 }{ 30dt } +\int _{ 0 }^{ 30 }{ -t+30 } dt = 323100

and

$\tilde R = 2261700m = \boxed{2262km}$

There is no chance to spot her further than 2262km from the library after she's been gone for exactly 24h.