The displacement and the increase in the velocity of a moving particle in the time interval of t to ( t + 1…

Considering acceleration is constant, we can write

$v=u+at$

As increase in velocity in $1 \mathrm{s}$ is given as $50 \mathrm{m} {\mathrm{s}}^{-1}$, therefore

$\Rightarrow a=50 \mathrm{m} {\mathrm{s}}^{-2}$

Now, displacement covered between given time interval is, 

$125=ut+\frac{1}{2}a{t}^2$

$\Rightarrow 125=u+\frac{a}{2}$

$\Rightarrow u=100 \mathrm{m} {\mathrm{s}}^{-1}$

Velocity after $1 \mathrm{s}$ will be, $v=100+50=150 \mathrm{m} {\mathrm{s}}^{-1}$

Therefore, displacement covered in $\left(t+1\right) \mathrm{s} \text{to} \left(t+2\right) \mathrm{s}$ will 

$\therefore S_{n^{th}}=v{t}^{\text{'}}+\frac{1}{2}a{\left(t^{\text{'}}\right)}^2, \text{where} t^{\text{'}}=1 \mathrm{s}$

$=175 \mathrm{m}$