Two cars A and B are moving in the same direction with velocities $v_A=30\text{ km/h}$ and $v_B=20\text{ km/h}$. When car A is $10\text{ km}$ behind car B, car A accelerates at $a_A=4{\text{ km/h}}^2$ and car B decelerates at $a_B=-6{\text{ km/h}}^2$. How long will it take for car A to overtake car B?

Solution

1. Define the relative initial velocity as $$\begin{array}{rl}{u}_{rel}& =v_A-v_B\\ & =30-20\\ & =10\text{ km/h}\end{array}$$ and the relative acceleration as $$\begin{array}{rl}{a}_{rel}& =a_A-a_B\\ & =4-(-6)\\ & =10{\text{ km/h}}^2\end{array}$$.
2. Use the kinematic equation $s=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$ with the initial separation $s=10\text{ km}$ to get $10=10t+5t^2$.
3. Simplify the equation to $t^2+2t-2=0$ and solve for the positive root, which yields $t=1\text{ hour}$.
4. Hence the answer is (B).