$\langle p\rangle=\frac{(2 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}+6 \hat{j})}{4}=6$
$\vec{a}=\left(\frac{\vec{F}}{m}=\frac{1}{2} \hat{i}+\frac{3}{4} \hat{j}\right)$
$\overrightarrow{\mathrm{v}}$ at $\mathrm{t}=4 \sec =\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})$
$P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13$
$\frac{\langle P\rangle}{P_{\text {ins }}}=\frac{6}{13}$
Note: Given data is not matching.
$\mathrm{S}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}$
$\mathrm{S}=0+\frac{1}{2} \frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})}{4}(4)^{2}=4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}$
If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$
But Final position given in the question is $(6,10)$.