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Ch. 05 - Using Newton's Laws: Friction, Circular Motion, Drag Forces
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 5, Problem 62b

The position of a particle moving in the xy plane is given by r\(\overrightarrow{r}\) = (2.0m) cos [(3.0 rad/s)t ] i^\(\hat{i}\) +(2.0m) sin [(3.0 rad/s)t ] j^\(\hat{j}\), where r is in meters and t is in seconds. Calculate the velocity and acceleration vectors as functions of time.

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Start by understanding the given position vector: \( \vec{r}(t) = (2.0 \text{ m}) \cos[(3.0 \text{ rad/s})t] \hat{i} + (2.0 \text{ m}) \sin[(3.0 \text{ rad/s})t] \hat{j} \). This describes the motion of the particle in the xy-plane as a function of time.
To find the velocity vector \( \vec{v}(t) \), take the time derivative of the position vector \( \vec{r}(t) \). Use the derivative rules for trigonometric functions: \( \frac{d}{dt}[\cos(\omega t)] = -\omega \sin(\omega t) \) and \( \frac{d}{dt}[\sin(\omega t)] = \omega \cos(\omega t) \).
Apply the derivative to each component of \( \vec{r}(t) \): \( \frac{d}{dt}[(2.0 \text{ m}) \cos(3.0t)] = -(2.0 \text{ m})(3.0 \text{ rad/s}) \sin(3.0t) \) for the \( \hat{i} \)-component, and \( \frac{d}{dt}[(2.0 \text{ m}) \sin(3.0t)] = (2.0 \text{ m})(3.0 \text{ rad/s}) \cos(3.0t) \) for the \( \hat{j} \)-component.
Combine the results to write the velocity vector: \( \vec{v}(t) = -(6.0 \text{ m/s}) \sin(3.0t) \hat{i} + (6.0 \text{ m/s}) \cos(3.0t) \hat{j} \).
To find the acceleration vector \( \vec{a}(t) \), take the time derivative of the velocity vector \( \vec{v}(t) \). Use the same derivative rules for trigonometric functions. For the \( \hat{i} \)-component, \( \frac{d}{dt}[-(6.0 \text{ m/s}) \sin(3.0t)] = -(6.0 \text{ m/s})(3.0 \text{ rad/s}) \cos(3.0t) \), and for the \( \hat{j} \)-component, \( \frac{d}{dt}[(6.0 \text{ m/s}) \cos(3.0t)] = -(6.0 \text{ m/s})(3.0 \text{ rad/s}) \sin(3.0t) \). Combine these to write \( \vec{a}(t) = -(18.0 \text{ m/s}^2) \cos(3.0t) \hat{i} - (18.0 \text{ m/s}^2) \sin(3.0t) \hat{j} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Position Vector

The position vector describes the location of a particle in space relative to a reference point. In this case, the position vector r→ is expressed in terms of its components along the x and y axes, using trigonometric functions to indicate circular motion. Understanding the position vector is crucial for deriving other motion-related quantities such as velocity and acceleration.
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Velocity Vector

The velocity vector represents the rate of change of the position vector with respect to time. It is calculated by taking the derivative of the position vector r→ with respect to time t. In this scenario, the velocity will also exhibit a periodic nature due to the trigonometric functions involved, reflecting the particle's circular motion in the xy plane.
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Acceleration Vector

The acceleration vector indicates the rate of change of the velocity vector with respect to time. It is obtained by differentiating the velocity vector. For a particle in circular motion, the acceleration can be both tangential and centripetal, and understanding its components is essential for analyzing the dynamics of the particle's motion in the xy plane.
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