Two particles, one with constant velocity $50{\text{ms}}^{-1}$ and the other with uniform acceleration $10{\text{ms}}^{-2}$, start moving simultaneously from the same place in the same direction. When will the accelerated particle be $125\text{m}$ ahead of the other particle?

The correct option is B $5(1+\sqrt{2})\text{s}$
For particle 1 moving with constant velocity, $S_1=50t$ ... (i)
For particle 2 moving with constant acceleration,
$S_2=\frac{1}{2}a{t}^2=\frac{1}{2}\times 10t^2=5t^2$ ... (ii)
As per question;
$S_2-S_1=125$, we get
From equations (i) & (ii)
$5t^2-50t=125$
$\Rightarrow t^2-10t-25=0$
On solving, we get
$t=(5\pm 5\sqrt{2})\text{s}$
$\therefore t=5(1+\sqrt{2})\text{s}$ (taking positive value)

Alternate Solution:
Relative displacement of particles 1 and 2 is
$S_{12}=-125\text{m}$
Initial velocity of 1 w.r.t. 2,
$U_{12}=50-0=50\text{m}$
Acceleration of 1 w.r.t. 2
$a_{12}=0-10=-10{\text{m/s}}^2$
Using second equation of motion,
$S_{12}=U_{12}t+\frac{1}{2}{a}_{12}{t}^2$
$-125=50t+\frac{1}{2}\times (-10)t^2$
$\Rightarrow t^2-10t-25=0$
$\Rightarrow t=5(1+\sqrt{2})$ or $t=5(1-\sqrt{2})\text{sec}$
Taking +ve value
$t=5(1+\sqrt{2})\text{s}$