The position vector of a particle related to time $t$ is given by

$\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m$

The direction of net force experienced by the particle is :

To find the direction of the net force experienced by the particle, we need to find the acceleration vector of the particle and then use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration vector. <br/><br/> The position vector of the particle is given by: <br/><br/> $\vec{r} = (10t\hat{i} + 15t^2\hat{j} + 7\hat{k})\,\text{m}$ <br/><br/> Differentiating $\vec{r}$ twice with respect to time $t$, we get the acceleration vector: <br/><br/> $$ \vec{a} = \frac{d^2\vec{r}}{dt^2} = \frac{d}{dt}(10\hat{i} + 30t\hat{j}) = 30\hat{j}\,\text{m/s}^2 $$ <br/><br/> Therefore, the acceleration vector is $\vec{a} = 30\hat{j}\,\text{m/s}^2$. <br/><br/> Using Newton's second law, the net force on the particle is given by: <br/><br/> $\vec{F}_{net} = m\vec{a}$ <br/><br/> where $m$ is the mass of the particle. <br/><br/> Since we are only interested in the direction of the net force, we can ignore the magnitude of the acceleration and focus on its direction, which is along the positive $y$-axis.