Question

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

Answer 1

Given $u_A=0,u_B=1m/s,a_A=4m/s^{2}and{a}_B=0$
Assuming car B to be at rest, we have
$u_{AB}=u_A-u_B=0-1=-1m/s$
$a_{AB}=a_A-a_B=4-0=4m/s^2$

Now, the problem can be assumed in simplified form as follows:
Substituting the proper values in equation
$s=ut+\frac{1}{2}a{t}^2$
we get $10=-t+\frac{1}{2}\left(4\right)\left(t^2\right)$
or $2t^2-t-10=0$
or $t=\frac{1\pm \sqrt{1+80}}{4}=\frac{1\pm \sqrt{81}}{4}$
$\frac{1\pm 9}{4}ort=2.5sand-2s$
Ignoring the negative value, the desired time is 2.5 s.