Answer
If the magnitude of the acceleration $a_B<2.5ms^{-2}$ , then both of the cars will be side-by-side for exactly $2$ times,
Work Step by Step
Step-1: From the previous question, we derived the equation for the displacement of both of the cars.
For $A$:$$x=2t+20$$For $B$:$$x=12t−12(|a_B|)t^2$$
Step-2: Now, solving both of the equations,$$2t+20=12t−12(|a_B|)t^2$$$$\implies 0.5(|a_B|)t^2−10t+20=0$$
Step-3: The discriminant of the above quadratic equation is, $$\Delta=(−10)2−4×(0.5(|a_B|))×20$$$$\Delta=100−40|a_B|$$
Step-4: When $|a_B| < 2.5$ , $\Delta>0$ , which implies that there are two solutions for $t$ for which $x$ would be same; that is, there are two values of $t$ for which the cars will be side-by-side.
A graph is shown below for $a_B=−2.3ms^{−2}$. You can see that both the curves intersect at two points, which means that the cars A and B are side-by-side exactly $2$ times.
