Time = $\frac{Distance}{Speed}$ , right??

Even though not stated, $O$ can be taken to be the origin without any loss of generality. The initial position of $P$ can be taken as $(0, 16)$ . (It can also be taken as $(0, -16)$ , both of which will give the same answer.)

Let the position of the point at any time $t$ be $\overrightarrow{r} = x \hat{i} + y \hat{j}$ . This gives that $\overrightarrow{PO} = - x \hat{i} - y \hat{j}$ and the unit vector will be $-\dfrac{x \hat{i}}{\sqrt{x^2+y^2}} - \dfrac{y \hat{j}}{\sqrt{x^2+y^2}}$ .

$\therefore \overrightarrow{v} = 5 \Bigg(-\dfrac{x \hat{i}}{\sqrt{x^2+y^2}} - \dfrac{y \hat{j}}{\sqrt{x^2+y^2}}\Bigg) + 3\hat{i}$

$\implies \overrightarrow{v} = \Bigg(\dfrac{-5x}{\sqrt{x^2+y^2}}\ + 3\Bigg)\hat{i} + \Bigg(\dfrac{-5y}{\sqrt{x^2+y^2}}\Bigg)\hat{j}$

Using the definition of velocity, $\overrightarrow{v} = \dfrac{dx}{dt}\hat{i} + \dfrac{dy}{dt}\hat{j}$

$\dfrac{dx}{dt} = \dfrac{-5x}{\sqrt{x^2+y^2}}\ + 3 \hspace{3 cm} (1)$

$\dfrac{dy}{dt} = \dfrac{-5y}{\sqrt{x^2+y^2}} \hspace{3.7 cm} (2)$

Diving $(1)$ by $(2)$ :

$\dfrac{dx}{dy} = \dfrac{5x - 3\sqrt{x^2 + y^2}}{5y} = \dfrac{x}{y} - \dfrac{3}{5}\sqrt{1+\Bigg(\dfrac{x}{y}\Bigg)^2}$

Using the substitution: $x = uy \implies \frac{dx}{dy} = u + y\frac{du}{dy}$ :

$u + y\dfrac{du}{dy} = u - \dfrac{3}{5}\sqrt{1 + u^2}$

$\implies \dfrac{du}{\sqrt{1+u^2}} = -\dfrac{3}{5} \dfrac{dy}{y}$

Integrating both sides:

$\ln(\sqrt{1+u^2} + u) = -\dfrac{3}{5}\ln(y) + C$

Using initial conditition of $x = 0$ when $y=16$ , $C$ can be found to be equal to $\dfrac{3}{5}\ln(16)$

$\therefore \ln(\sqrt{1+u^2} + u) = -\dfrac{3}{5}\ln\bigg(\dfrac{y}{16}\bigg)$

Taking anti-log:

$\sqrt{1+u^2} + u = \bigg(\dfrac{y}{16}\bigg)^{-\dfrac{3}{5}} \hspace{3 cm} (3)$

$\sqrt{1+u^2} - u = \bigg(\dfrac{y}{16}\bigg)^{\dfrac{3}{5}} \hspace{3.1 cm} (4) \hspace{3 cm} (\text{Using the fact that }(\sqrt{1+u^2} + u)(\sqrt{1+u^2} - u) = 1$

Adding the two:

$\sqrt{1+u^2} = \dfrac{1}{2} \Bigg[\bigg(\dfrac{y}{16}\bigg)^{-\dfrac{3}{5}}+ \bigg(\dfrac{y}{16}\bigg)^{\dfrac{3}{5}} \Bigg]$

$\implies \dfrac{\sqrt{x^2+y^2}}{y} = \dfrac{1}{2} \Bigg[\bigg(\dfrac{y}{16}\bigg)^{-\dfrac{3}{5}}+ \bigg(\dfrac{y}{16}\bigg)^{\dfrac{3}{5}} \Bigg]$

Subsituting the above in $(2)$ :

$\dfrac{1}{2} \Bigg[\bigg(\dfrac{y}{16}\bigg)^{-\dfrac{3}{5}}+ \bigg(\dfrac{y}{16}\bigg)^{\dfrac{3}{5}} \Bigg] = -5\dfrac{dt}{dy}$

$\implies\dfrac{1}{2} \displaystyle\int_{16}^{0} \Bigg[\bigg(\dfrac{y}{16}\bigg)^{-\dfrac{3}{5}}+ \bigg(\dfrac{y}{16}\bigg)^{\dfrac{3}{5}} \Bigg] dy = -5 \displaystyle\int_{0}^{T} dt$

$\implies\dfrac{1}{2} \Bigg[16\dfrac{1}{\frac{2}{5}}\bigg(\dfrac{y}{16}\bigg)^{\dfrac{2}{5}}+ 16\dfrac{1}{\frac{8}{5}}\bigg(\dfrac{y}{16}\bigg)^{\dfrac{8}{5}} \Bigg]_{16}^{0}= -5 T$

$\implies\dfrac{5}{2} \Bigg[16\dfrac{1}{2}(-1)+ 16\dfrac{1}{8}(-1) \Bigg]= -5 T$

$\implies T = \boxed{5}$