Answer
The acceleration vector as a function of time is
$$
\vec{a}=\frac{d \vec{v}}{d t}=\frac{d}{d t}\left(8.00 t \hat{\mathrm{i}}+3.00 t^{2} \hat{\mathrm{j}}\right) \mathrm{m} / \mathrm{s}=(8.00 \hat{\mathrm{i}}+6.00 t \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}
$$
The magnitude of the force acting on the particle is
$$
F=m a=m|\vec{a}|=(3.00) \sqrt{(8.00)^{2}+(6.00 t)^{2}}=(3.00) \sqrt{64.0+36.0 t^{2}} \mathrm{N}
$$
Thus, $F=35.0 \mathrm{\ N}$ corresponds to $t=1.415 \mathrm{\ s}$ , and the acceleration vector at this instant is
$$
\vec{a}=[8.00 \hat{\mathrm{i}}+6.00(1.415) \hat{\mathrm{j}}] \mathrm{m} / \mathrm{s}^{2}=\left(8.00 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(8.49 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .
$$
The angle $\vec{a}$ makes with $+x$ is
$$
\theta_{a}=\tan ^{-1}\left(\frac{a_{y}}{a_{x}}\right)=\tan ^{-1}\left(\frac{8.49 \mathrm{m} / \mathrm{s}^{2}}{8.00 \mathrm{m} / \mathrm{s}^{2}}\right)=46.7^{\circ} .
$$
Work Step by Step
The acceleration vector as a function of time is
$$
\vec{a}=\frac{d \vec{v}}{d t}=\frac{d}{d t}\left(8.00 t \hat{\mathrm{i}}+3.00 t^{2} \hat{\mathrm{j}}\right) \mathrm{m} / \mathrm{s}=(8.00 \hat{\mathrm{i}}+6.00 t \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}
$$
The magnitude of the force acting on the particle is
$$
F=m a=m|\vec{a}|=(3.00) \sqrt{(8.00)^{2}+(6.00 t)^{2}}=(3.00) \sqrt{64.0+36.0 t^{2}} \mathrm{N}
$$
Thus, $F=35.0 \mathrm{\ N}$ corresponds to $t=1.415 \mathrm{\ s}$ , and the acceleration vector at this instant is
$$
\vec{a}=[8.00 \hat{\mathrm{i}}+6.00(1.415) \hat{\mathrm{j}}] \mathrm{m} / \mathrm{s}^{2}=\left(8.00 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(8.49 \mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .
$$
The angle $\vec{a}$ makes with $+x$ is
$$
\theta_{a}=\tan ^{-1}\left(\frac{a_{y}}{a_{x}}\right)=\tan ^{-1}\left(\frac{8.49 \mathrm{m} / \mathrm{s}^{2}}{8.00 \mathrm{m} / \mathrm{s}^{2}}\right)=46.7^{\circ} .
$$
