Q42P - University Physics with Modern Physics

Question

The position of a dragonfly that is flying parallel to the ground is given as a function of time by $\overrightarrow{\mathbf{r}}\mathbf{=}\left[2.90m+\left(0.900m/s^2\right)t^2\right]\hat{\mathbf{i}}\mathbf{-}\left(0.0150 m/s^2\right){\mathit{t}}^{\mathbf{3}}\hat{\mathbf{j}}$ . (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?

Step-by-Step Solution

Answer

a) The time at which velocity vector makes angle with x-axis is 2.31 s .

b) The magnitude and direction of acceleration at time 2.31 s is $0.275m/s^2$ and $310.87°$ respectively.

Step 1: Identification of given data

The given data can be listed below,

• The position vector given is, $\overrightarrow{r}=\left[2.90m+\left(0.900m/s^2\right)t^2\right]\hat{i}-\left(0.0150 m/s^2\right)t^3\hat{j}$

Step 2: Concept/Significance of acceleration vector

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.

Step 3: (a) Determination of value of t does the velocity vector of the dragonfly make an angle of 30 ° clockwise from the +x-axis

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,

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

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,

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

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

Step 4: (b) Determination of the magnitude and direction of the dragonfly’s acceleration vector

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,

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

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 $0.275 m/s^2$ and $310.87°$ respectively.