Question

A bus starts from a bus stop with a constant acceleration of 0.4 $m/s^2$. A passenger arrives at the stop 6 sec after bus leaves the stop. Minimum constant speed (in m/s) of passenger so that he can catch bus is (Write upto two digits after the decimal point.)

Answer 1

$x_B=\frac{1}{2}\times 0.4t^2$ and $x_p=v(t-6)$

$x_B=x_p\Rightarrow 0.2t^2=vt-6v$

$0.2t^2-vt+6v=0\Rightarrow t=\frac{v\pm \sqrt{v^2-4\times 0.2\times 6v}}{2\times 0.2}$

For 't' to have a real solution,
$v^2-4\times 0.2\times 6v\ge 0\Rightarrow v=4.8m/s$