Answer
If the magnitude of the acceleration $a_B>2.5ms^{-2}$, then both the cars will $never$ be side-by-side.
Work Step by Step
Step-1: From the previous question, we derived the equation for the displacement of both the cars.
For $A$: $$x=2t+20$$
For $B$: $$x=12t-\frac{1}{2}(|a_B|)t^2$$
Step-2: Now solving both of the equations, $$2t+20=12t-\frac{1}{2}(|a_B|)t^2$$$$\implies 0.5(|a_B|)t^2-10t+20=0$$
Step-3: The discriminant of the above quadratic equation is, $$\Delta=(-10)^2-4\times (0.5(|a_B|))\times 20$$$$\Delta = 100-40|a_B|$$
Step-4: When $|a_B| > 2.5$, $\Delta<0$, which implies that there is no solution for $t$ for which $x$ would be same, meaning that there is no value of $t$ for which the cars will be side-by-side.
