$\frac{}{PS}$ = $\frac{}{OS}$ - $\frac{}{OP}$ = (${}_3^{-2}$) - (${}_4^{-1}$) = (${}_{-1}^{-1}$)
Similarly $\frac{}{QR}$ = $\frac{}{OR}$ - $\frac{}{OQ}$ = (${}_y^x$) - (${}_3^2$) = (${}_{y-3}^{x-2}$)
Since $\frac{}{PS}$ = $\frac{}{QR}$, it flows that (${}_{y-3}^{x-2}$) = (${}_{-1}^{-1}$). Therefore, x - 2 = -1 which solved will yield x = 1.
Also, y - 3 = -1 which implies that y = 2

(b)(i), Since the particle starts from rest, the initial velocity (u) = 0 and substituting for final velocity (v) and distance (s) into the formula v${}^2$ = u${}^2$ + 2as to obtain 20${}^2$ = 0 + 2a(8) and when simplifies and solved for acceleration (a), to get a = 25ms${}^{-2}$
(b)(ii), S = ut + $\frac{1}{2}$at${}^2$, where s = 40, u = 0 and a = 25
Substituting these values into the equation we obtain 40 = 0 + $\frac{1}{2}$ x 25t${}^2$ and solving for t gave t = 1.79 seconds.