The position of a dragonfly that is flying parallel to the ground is given as a function of time by$\overrightarrow{\mathbf{r}}\mathbf{=}\left[2.90m+\left(0.900m/s^2\right)t^2\right]\hat{\mathbf{i}}\mathbf{-}\left(0.0150m/s^2\right){\mathit{t}}^{\mathbf{3}}\hat{\mathbf{j}}$ . (a) At what value of t does the velocity vector of the dragonfly make an angle of $\mathbf{30}\mathbf{°}$clockwise from the +x-axis? (b) At the time calculated in part (a), what are the magnitude and direction of the dragonfly’s acceleration vector?

The given data can be listed below,

The acceleration vector is the rate at which velocity changes, which is a vector quantity as well. It has magnitude and acts in one direction.

The velocity of the dragonfly is given by,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Here,$\overrightarrow{r}$ is the position vector of dragonfly.

Substitute all the values in the above,

$$\begin{gathered} \overrightarrow{v}=\frac{d\left(\left[2.90m+\left(0.90m/s^2\right)t^2\right]\hat{i}-\left(0.0150m/s^2\right)t^3\hat{j}\right)}{dt} \\ =\left(0.18t\right)\hat{i}-\left(0.045t^2\right)\hat{j} \end{gathered}$$

The angle of velocity is given by,

$\tan \theta =\frac{v_y}{v_x}$

Here,$v_y$ is the y-component of velocity and$v_x$ is x-component of velocity.

Substitute all the values in the above,

$$\begin{gathered} \tan \left(-30°\right)=\frac{-0.045t^2}{0.18t} \\ t=\frac{0.18\tan \left(-30°\right)}{-0.045} \\ =2.31s \end{gathered}$$

Thus, the time at which velocity vector makes angle with x-axis is 2.31 s .

The acceleration of the dragonfly is given by,

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$

Here,$\overrightarrow{v}$ is the velocity vector of the dragonfly.

Substitute all values in the above,

$$\begin{gathered} \overrightarrow{a}=\frac{d\left(\left(0.18t\right)\hat{i}-\left(0.045t^2\right)\hat{j}\right)}{dt} \\ =0.18\hat{i}-\left(0.09t\right)\hat{j} \end{gathered}$$

Acceleration at time 2.31 s is given by,

$$\begin{gathered} \overrightarrow{a}=0.18\hat{i}-\left(0.09\right)\left(2.31s\right)\hat{j} \\ =0.18\hat{i}-\left(0.208\right)\hat{j} \end{gathered}$$

The magnitude of acceleration is calculated as,

$$\begin{gathered} a=\sqrt{{\left(0.18\right)}^2+{\left(0.208\right)}^2}m/s^2 \\ =0.275m/s^2 \end{gathered}$$

The direction of the acceleration is given by,

$$\begin{gathered} \tan \theta =\left(\frac{a_y}{a_x}\right) \\ \theta ={\tan }^{-1}\left(\frac{-0.208}{0.18}\right) \\ =-49.13° \\ =310.87° \end{gathered}$$

Thus, the magnitude and direction of acceleration at time 2.31 s is$0.275m/s^2$ and$310.87°$ respectively.