Two particles A and B are moving in the same direction on a straight line with velocities $v_A=10\text{ m/s}$ and $v_B=5\text{ m/s}$ respectively. When A is $20\text{ m}$ behind B, A starts accelerating at $a_A=2{\text{ m/s}}^2$. How long will it take for A to overtake B?

Solution

1. Define the relative initial velocity $$\begin{array}{rl}{u}_{rel}& =v_A-v_B\\ & =10-5\\ & =5\text{ m/s}\end{array}$$, relative acceleration $$\begin{array}{rl}{a}_{rel}& =a_A-a_B\\ & =2-0\\ & =2{\text{ m/s}}^2\end{array}$$, and relative displacement $s_{rel}=20\text{ m}$.
2. Apply the kinematic equation $s_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$ to get $20=5t+\frac{1}{2}(2)t^2$.
3. Simplify the equation to the quadratic form $t^2+5t-20=0$, which factors to $(t-4)(t+5)=0$.
4. Solving for the positive time value, we find $t=4\text{ s}$.
5. Hence the answer is (B).