Two points $$A$$ and $$B$$ move from rest along a straight line with constant acceleration $$f$$ and $$f'$$ respectively. If $$A$$ takes $$m$$ sec. more than $$B$$ and describes '$$n$$' units more than $$B$$ in acquiring the same speed then

Let the common speed finally acquired by both points be $$v$$.

For point $$A$$ (acceleration $$f$$, starting from rest):
  Final speed  $$v = f\,t_A$$  $$\Rightarrow \; t_A = \frac{v}{f}$$

For point $$B$$ (acceleration $$f'$$, starting from rest):
  Final speed  $$v = f'\,t_B$$  $$\Rightarrow \; t_B = \frac{v}{f'}$$

Given that point $$A$$ takes $$m$$ seconds more than point $$B$$ to reach this speed,

$$t_A - t_B = m$$ $$\frac{v}{f} - \frac{v}{f'} = m$$ $$v\left(\frac{1}{f} - \frac{1}{f'}\right) = m$$  $$-(1)$$

Distances travelled while accelerating from rest:

For $$A$$: $$s_A = \frac12 f\,t_A^2 = \frac12 f\left(\frac{v}{f}\right)^2 = \frac{v^2}{2f}$$
For $$B$$: $$s_B = \frac12 f'\,t_B^2 = \frac12 f'\left(\frac{v}{f'}\right)^2 = \frac{v^2}{2f'}$$

Given that $$A$$ covers $$n$$ units more than $$B$$,

$$s_A - s_B = n$$ $$\frac{v^2}{2}\left(\frac{1}{f} - \frac{1}{f'}\right) = n$$  $$-(2)$$

Let $$\Delta = \frac{1}{f} - \frac{1}{f'}$$. From $$(1):\; v\Delta = m \;\Rightarrow\; v = \frac{m}{\Delta}$$

Substituting $$v$$ in $$(2):$$

$$\frac{1}{2}\left(\frac{m}{\Delta}\right)^2 \Delta = n$$ $$\frac{m^2}{2\Delta} = n$$ $$\Delta = \frac{m^2}{2n}$$  $$-(3)$$

But $$\Delta = \frac{1}{f} - \frac{1}{f'} = \frac{f' - f}{ff'}$$. Equating this with $$(3):$$

$$\frac{f' - f}{ff'} = \frac{m^2}{2n}$$ $$\bigl(f' - f\bigr)\,n = \frac12\,f\,f'\,m^2$$

Thus, the required relation is
$$\boxed{(f' - f)\,n = \dfrac12\,f\,f'\,m^2}$$

Option D which is: $$(f' - f)n = \frac{1}{2}ff'm^2$$