How to go about solving a basic mechanics problem? The bus leaves the bus stop with an acceleration of 1.5 m.s^-2 which is mantained until it reaches the speed of 12 m.s^-1. At the same instant a girl who is 5 m away from the bus stop starts to run after the bus at a constant speed of 7 m.s^-1. Will the girl catch the bus? How should I link the two situations together mathematically? What would be the way?
The answer is that the girl will catch the bus and time would be 0.76 seconds.
The girl is about 5 m away from the bus stop so would it be s=s-5? Key s=distance
 A: Time reference the moment the bus leaves and space reference the position of the bus when it leaves. Let $x_b$ the bus abscissa and $x_g$ the girl abscissa.
$x_g(t)=7t-5$ and $x_b(t)$ is more complicated to find. The bus has two type of motion; an accelerated one and an uniform one. Let’s find the starting of the uniforme phase.
$v=\gamma t\Longrightarrow t=\frac{v}{\gamma}=\frac{12}{1,5}=8s.$ The bus has therefore these two equations: $$x_b(t)=0,75t^2\quad    
t\le 8$$ and $$x_b(t)=12(t-8)+0,75\times8^2=12t-48\quad t>8$$. Now solve $x_b(t)=x_g(t)$ with the time constraint i.e for $t\le8$ and $t>8$

For the first case i.e $t\le 8$ we solve $7t-5=0.75t^2\iff 0.75t^2-7t+5=0\iff t=8.55 \operatorname{ or } t=0.779$ since $t<8$ we choose $t=0.779$. The second case we solve $12t-48=7t-5$ which has one solution $t=\frac{43}{5}=8.6s$
Comment:

The first solution is the time that the girl catch the bus. If Amelia(the girl) decide to not jump in the bus and keep running she will exceed the bus but at $t=8.6s$  the bus will catch up Amelia.

