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Chapter 3: Problem 42

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

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

Step by step solution

01

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, \[\vec{v} = \frac{d\vec{r}}{dt} = \left( \frac{d}{dt}[2.90 + 0.0900t^2] \right) \hat{\imath} - \left( \frac{d}{dt}[0.0150t^3] \right) \hat{\jmath} \]. \[\vec{v} = [0 + 0.1800 t] \hat{\imath} - [0.0450t^2] \hat{\jmath} = [0.1800 t] \hat{\imath} - [0.0450 t^2] \hat{\jmath}\].
02

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, \[\tan(-30.0\degree) = \frac{-0.0450 t^2}{0.1800 t}\].
03

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 \(t\): \[t = 4 \times \frac{1}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \approx 2.31 \text{ sec}\].
04

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, \[\vec{a} = [0.1800] \hat{\imath} - [0.0900 t] \hat{\jmath}\].
05

Calculate Acceleration at \(t = 2.31\) sec

Substitute \(t = 2.31\) sec into the acceleration expression: \[\vec{a} = [0.1800] \hat{\imath} - [0.0900 \times 2.31] \hat{\jmath} = [0.1800] \hat{\imath} - [0.2079] \hat{\jmath}\].
06

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): \(\theta = \arctan\left(\frac{-0.2079}{0.1800}\right) \approx -49.4\degree\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A rock is thrown with a velocity \(v_0\), at an angle of \(\alpha_0\) from the horizontal, from the roof of a building of height \(h\). Ignore air resistance. Calculate the speed of the rock just before it strikes the ground, and show that this speed is independent of \(\alpha_0\).

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle \(\textbf{(Fig. P3.63). }\) The takeoff ramp was inclined at 53.0\(^{\circ}\), the river was 40.0 m wide, and the far bank was 15.0 m lower than the top of the ramp. The river itself was 100 m below the ramp. Ignore air resistance. (a) What should his speed have been at the top of the ramp to have just made it to the edge of the far bank? (b) If his speed was only half the value found in part (a), where did he land?

When a train's velocity is 12.0 m/s eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined 30.0\(^\circ\) to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth? With respect to the train? (b) What is the magnitude of the velocity of the raindrop with respect to the earth? With respect to the train?

A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center (\(\textbf{Fig. E3.27}\)). The linear speed of a passenger on the rim is constant and equal to 6.00 m/s. What are the magnitude and direction of the passenger's acceleration as she passes through (a) the lowest point in her circular motion and (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

CALC The coordinates of a bird flying in the \(xy\)-plane are given by \(x(t)\) = \(at\) and \(y(t)\) = 3.0 m - \(\beta t^2\), where \(\alpha\) = 2.4 m/s and \(\beta\) = 1.2 m/s\(^2\). (a) Sketch the path of the bird between \(t\) = 0 and \(t\) = 2.0 s. (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t\) = 2.0 s. (d) Sketch the velocity and acceleration vectors at \(t\) = 2.0 s. At this instant, is the bird's speed increasing, decreasing, or not changing? Is the bird turning? If so, in what direction?
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