Car A and car B start moving simultaneously in the same direction along the line joining them. Car A moves with a constant acceleration $a=4m{s}^{-2}$, while car B moves with a constant velocity  $v=1m{s}^{-1}$.At time $t=0$ , car A is 10 m behind car B. Find the time in $\frac{5}{t}$sec when car A overtakes car B.

Given  $u_A=0,u_B=1m{s}^{-1},a_A=4m{s}^{-1}and{a}_B=0$

Assuming car B to be at rest, we have
$$\begin{gathered} u_{AB}=u_r=u_A-u_B=0-1=-1m{s}^{-1} \\ {u}_{AB}=a_r=a_A-a_B=4-0=4m{s}^{-1} \end{gathered}$$

Now, the problem can be assumed in simplified form as follows:

Substituting the proper values in equation
$$\begin{gathered} s_r=u_{r}t+\frac{1}{2}{a}_r{t}^2 \\ \Rightarrow 10=-t+\frac{1}{2}\left(4\right){\left(t\right)}^2 \\ \Rightarrow 2t^2-t-10=0 \\ \Rightarrow t=\frac{1\pm \sqrt{1+80}}{4}=\frac{1+\sqrt{81}}{4} \\ \Rightarrow t=\frac{1+9}{4} \\ \Rightarrow t=2.5sand-2s \end{gathered}$$
Ignoring the negative value, the desired time is 2.5s