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(A) The position vector is given by $\overrightarrow{r} = 10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}$.The velocity vector $\overrightarrow{v}$ is the der

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Positive $y$-axis

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(A) The position vector is given by $\overrightarrow{r} = 10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}$.The velocity vector $\overrightarrow{v}$ is the derivative of position with respect to time: $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 10 \hat{i} + 30t \hat{j}$.The acceleration vector $\overrightarrow{a}$ is the derivative of velocity with respect to time: $\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = 30 \hat{j}$.According to Newton's Second Law,$\overrightarrow{F} = m\overrightarrow{a}$. Since mass $m$ is a positive scalar,the direction of the net force $\overrightarrow{F}$ is the same as the direction of acceleration $\overrightarrow{a}$.Since $\overrightarrow{a} = 30 \hat{j}$,the net force is directed along the positive $y$-axis.

The position vector of a particle related to time $t$ is given by $\overrightarrow{r} = (10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}) \text{ m}$. The direction of the net force experienced by the particle is:

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