A bus starts from rest, moves with a uniform acceleration ‘a’. Simultaneously a passenger at a distance X from the bus starts running to catch the bus. The minimum velocity of the passenger to catch the bus is                                        

Given that

A bus starts from rest

i.e. u = 0

Moving with uniform acceleration $=a$

Distance between man & bus $=x$

Man started running to catch the bus with minimum speed = v

Let he can catch the bus after t sec

Then we get distance travelled by man

= distance travelled by bus + distance between man & bus

$\text{s}={\text{s}}_{\text{1}}+x$

u = 0 a = 0 s = ut

${\text{s}}_1=\frac{1}{2}{\text{at}}^2$

$\text{s}={\text{s}}_1+x$

$\text{ut}=\frac{1}{2}{\text{at}}^2+x$

$\text{2ut}={\text{at}}^2+2x$

${\text{at}}^2-\text{2ut}+2x=0$

$t=\frac{\text{2u}\pm \sqrt{{\left(\text{2u}\right)}^2-4\text{a}\times \text{2}x}}{2\text{a}}$

$t=\frac{\text{2u}\pm 2\sqrt{u^2-2\text{a}x}}{2\text{a}}$              

$=\frac{\text{u}\pm \sqrt{u^2-2\text{a}x}}{\text{a}}$

$u>\sqrt{2\text{a}x}$

$u^2\ge 2\text{a}x$