Chapter 2 - Motion Along a Straight Line - Problems - Page 34: 39c

The graph is shown below. The blue line represents the equation of car $A$ and the green line represents the equation of car $B$.

Step-1: The equation for the movement of car B can be derived after knowing that $a_B=−2.5ms^{−2}$. The equation is, $$x=ut+12\times aB\times t^2$$$$x=12t−12\times 2.5\times t^2=12t−1.25t^2$$ Step-2: Plot the equation.