Question

Car $A$ starts with an acceleration of $2m{s}^{-2}.$ After $5s$, car $B$ starts from same point with uniform velocity $20m{s}^{-1}$ to chase $A$. After starting of car $A$, the time taken by car $B$ to catch car $A$ is $t$ seconds. The value of $t$ is .

Answer 1

Let $x_A$ be displacement of $A$ and
$x_B$ be displacement of $B$.

For $B$ to catch $A$, $x_A$ in $ts$ must be equal to $x_B\text{in}(t-5)s$
Using $s=ut+\frac{1}{2}a{t}^2$ for $A$ and putting values,
$x_A=\frac{1}{2}\left(2\right)t^2$
Using $s=ut+\frac{1}{2}a{t}^2$ for $B$ and putting values,
$x_B=20(t-5)$
Now,
$x_A=x_B$
$t^2=20(t-5)$
$\Rightarrow t^2-20t+100=0$
$\Rightarrow t=\frac{20\pm \sqrt{400-400}}{2}$
$\Rightarrow t=10s$