Question

The position of a dragonfly that is flying parallel to the ground is given as a function of time by \(\vec{r} = [2.90 m + (0.0900 m/s^2)t^2] \hat{\imath} - (0.0150 m/s^3)t^3\hat{\jmath}\). (a) At what value of \(t\) does the velocity vector of the dragonfly make an angle of 30.0\(^\circ\) 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?

Short answer

(a) \(t \approx 2.31\) sec. (b) Magnitude: \(0.276\text{ m/s}^2\), Direction: \(-49.4\degree\).

Step-by-step solution

Differentiate the Position Vector to Find Velocity

The velocity vector, \(\vec{v}\), is the derivative of the position vector \(\vec{r}\) with respect to time \(t\). Thus, \[\vec{v} = \frac{d\vec{r}}{dt} = \left( \frac{d}{dt}[2.90 + 0.0900t^2] \right) \hat{\imath} - \left( \frac{d}{dt}[0.0150t^3] \right) \hat{\jmath} \]. \[\vec{v} = [0 + 0.1800 t] \hat{\imath} - [0.0450t^2] \hat{\jmath} = [0.1800 t] \hat{\imath} - [0.0450 t^2] \hat{\jmath}\].

Set Up the Velocity Angle Equation

The angle \(\theta\) of the velocity vector with the positive x-axis is given by \(\tan(\theta) = \frac{v_y}{v_x}\). Given that the angle is 30.0\(^\circ\) clockwise, it is equivalent to \(-30.0\degree\) in the coordinate plane. Therefore, \[\tan(-30.0\degree) = \frac{-0.0450 t^2}{0.1800 t}\].

Solve for Time \(t\)

Simplify and solve the equation \(\tan(-30.0\degree) = -\frac{1}{\sqrt{3}}\). Set up the equation: \(-\frac{1}{\sqrt{3}} = \frac{-0.0450 t}{0.1800}\). Simplifying gives \(\frac{1}{\sqrt{3}} = \frac{t}{4}\). Solve for \(t\): \[t = 4 \times \frac{1}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.31 \text{ sec}\].

Differentiate Velocity to Find Acceleration

Find the acceleration vector \(\vec{a}\) by differentiating the velocity vector. \[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt} [0.1800 t] \hat{\imath} - \frac{d}{dt} [0.0450 t^2] \hat{\jmath} \]. Thus, \[\vec{a} = [0.1800] \hat{\imath} - [0.0900 t] \hat{\jmath}\].

Calculate Acceleration at \(t = 2.31\) sec

Substitute \(t = 2.31\) sec into the acceleration expression: \[\vec{a} = [0.1800] \hat{\imath} - [0.0900 \times 2.31] \hat{\jmath} = [0.1800] \hat{\imath} - [0.2079] \hat{\jmath}\].

Find Magnitude and Direction of Acceleration

Find the magnitude: \(|\vec{a}| = \sqrt{(0.1800)^2 + (0.2079)^2} \approx 0.276\text{ m/s}^2\). Find direction (angle with +x-axis): \(\theta = \arctan\left(\frac{-0.2079}{0.1800}\right) \approx -49.4\degree\).

Key concepts

Velocity Vector

Understanding the velocity vector is crucial in kinematics as it represents the rate of change of position with respect to time. In this exercise, the velocity vector, \(\vec{v}\), is derived from the position vector \(\vec{r}\). By differentiating the position vector with respect to time, we obtain the velocity vector:
\[\vec{v} = \frac{d\vec{r}}{dt} = [0.1800 t] \hat{\imath} - [0.0450 t^2] \hat{\jmath}\]Understanding its components:

• The \(\hat{\imath}\) component \([0.1800 t]\) represents the velocity along the x-axis, which increases linearly with time.
• The \(\hat{\jmath}\) component \([-0.0450 t^2]\) represents the velocity along the y-axis, which follows a quadratic relationship, indicating a more complex motion as time progresses.

Together, these components illustrate how the dragonfly moves through space over time. To explore further, we calculate the angle it makes with the x-axis, crucial for determining the specific orientation of the velocity vector.

Acceleration Vector

The acceleration vector represents the change in velocity over time, key to understanding how quickly an object speeds up or slows down. It's derived from the velocity vector:
\[\vec{a} = \frac{d\vec{v}}{dt} = [0.1800] \hat{\imath} - [0.0900 t] \hat{\jmath}\]Breaking it down:

• The consistent \([0.1800] \hat{\imath}\) component suggests a constant acceleration along the x-axis.
• The \([-0.0900 t] \hat{\jmath}\) component indicates an increasing deceleration along the y-axis, reflecting a complex dynamic as time advances.

To gain further insight, we calculate the magnitude and direction at a specific time (e.g., \(t = 2.31\) seconds). The acceleration's magnitude gives us the intensity of change in velocity, while its direction (angle with the x-axis) shows the path orientation. These calculations help us comprehend the complete effect of forces acting on the dragonfly.

Differentiation

Differentiation is a vital mathematical tool used to find the rate of change of functions, crucial in physics for analyzing motion like that of the dragonfly. In this context, we used differentiation to derive both the velocity and acceleration vectors from the position function:
\[\vec{v} = \frac{d\vec{r}}{dt}\]and for acceleration:
\[\vec{a} = \frac{d\vec{v}}{dt}\]These operations require a step-by-step approach:

• First, apply the power rule to each term, essentially multiplying by the exponent and subtracting one from the exponent.
• The time factor "\(t\)" plays a critical role in how each component changes.
• The \(\hat{\imath}\) and \(\hat{\jmath}\) components derived show us changes along perpendicular axes.

Differentiation pinpoints precise instances where changes in motion are significant, like identifying when the velocity changes direction or acceleration intensity spikes.

Tangent Angle

The tangent angle is an essential concept in kinematics, as it represents the angle a velocity vector makes with a reference direction, normally the positive x-axis. It is calculated using the tangent function:
\[\tan(\theta) = \frac{v_y}{v_x}\]In this exercise, we seek the time when \(\theta\) equals \(-30.0^\circ\). Steps for finding this angle involve:

• Calculate \(\theta\) using the known components of the velocity vector. Here, this involves solving \(-30.0^\circ = \tan^{-1}\left(\frac{-0.0450 t^2}{0.1800 t}\right)\).
• Simplify and solve for time \(t\), giving insight into when the velocity vector aligns at the specified angle.

This angle informs exactly how the dragonfly's motion is oriented at a particular point in time, bridging the change in position to its path.