Problem 3 - University Physics with Modern Physics

Question

CALC A web page designer creates an animation in which a dot on a computer screen has position $$ \vec{r} =[34.0 cm +(2.5 cm/s^2)t^2] \hat{i} +(5.0 cm/s)t \hat{\jmath}.$$ (a) Find the magnitude and direction of the dot's average velocity between \(t\) = 0 and \(t\) = 2.0 s.(b) Find the magnitude and direction of the instantaneous velocity at \(t\) = 0, \(t\) = 1.0 s, and \(t\) = 2.0 s. (c) Sketch the dot's trajectory from \(t\) = 0 to \(t\) = 2.0 s, and show the velocities calculated in part (b).

Step-by-Step Solution

Answer

The dot's average velocity from 0 to 2.0 s is 7.1 cm/s at 45°. Instantaneous velocities are 5.0 cm/s at 90° (t=0 s), 7.1 cm/s at 45° (t=1.0 s), and 11.2 cm/s at 26.6° (t=2.0 s). The trajectory is an arc moving primarily horizontally.

Step 1: Define Initial Information

The position vector \( \vec{r}(t) \) of the dot is given by: \[ \vec{r} = [34.0 \text{ cm } +(2.5 \text{ cm/s}^2)t^2] \hat{i} +(5.0 \text{ cm/s})t \hat{\jmath}. \] We need to find the average velocity for \( t = 0 \text{ s} \) to \( t = 2.0 \text{ s} \), and the instantaneous velocities at specific moments \( t = 0, 1.0, 2.0 \text{ s}. \)

Step 2: Calculate the Average Velocity

The average velocity \( \vec{v}_{\text{avg}} \) is the change in position divided by the change in time: \[ \vec{v}_{\text{avg}} = \frac{\vec{r}(2.0) - \vec{r}(0)}{2.0 - 0}. \] Substitute \( t = 2.0 \text{ s} \) and \( t = 0 \text{ s} \) into the position function to find \( \vec{r}(2.0) \) and \( \vec{r}(0) \). For \( t = 0: \) \[ \vec{r}(0) = [34.0] \hat{i} + [0] \hat{\jmath} = 34.0 \hat{i} \text{ cm}. \] For \( t = 2.0: \)\[ \vec{r}(2.0) = [34.0 + 2.5 \times (2.0)^2] \hat{i} + [5.0 \times 2.0] \hat{\jmath}. \] \[ = [44.0] \hat{i} + [10.0] \hat{\jmath} \text{ cm}. \] Therefore, \[ \vec{v}_{\text{avg}} = \frac{(44.0 - 34.0) \hat{i} + (10.0 - 0) \hat{\jmath}}{2.0}. \] \[ = 5.0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \] The magnitude of \( \vec{v}_{\text{avg}}\) is found by: \[ \|\vec{v}_{\text{avg}}\| = \sqrt{(5.0)^2 + (5.0)^2} = \sqrt{50} \approx 7.1 \text{ cm/s}. \] The direction is given by: \[ \theta = \tan^{-1}\left(\frac{5.0}{5.0}\right) = 45^\circ \text{ with respect to the } \hat{i}\text{-axis}. \]

Step 3: Calculate Instantaneous Velocity

The instantaneous velocity \( \vec{v} \) is the derivative of the position vector \( \vec{r} \) with respect to \( t \). Differentiate \( \vec{r}(t) = [34.0 + 2.5t^2] \hat{i} + [5.0t] \hat{\jmath} \) with respect to \( t \): \[ \vec{v}(t) = \frac{d}{dt}([34.0 + 2.5t^2] \hat{i}) + \frac{d}{dt}([5.0t] \hat{\jmath}). \] \[ = [5.0t] \hat{i} + [5.0] \hat{\jmath}. \] Now calculate \( \vec{v}(t) \) for \( t = 0, 1.0, 2.0 \text{ s}. \)At \( t = 0 \text{ s}: \) \[ \vec{v}(0) = [5.0 \times 0] \hat{i} + [5.0] \hat{\jmath} = 0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \]The magnitude is \[\|\vec{v}(0)\| = 5.0 \text{ cm/s}. \] and direction is \( 90^\circ \).At \( t = 1.0 \text{ s}: \) \[ \vec{v}(1.0) = [5.0 \times 1.0] \hat{i} + [5.0] \hat{\jmath} = 5.0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \] The magnitude is \( \|\vec{v}(1.0)\| \approx 7.1 \text{ cm/s} \) and the direction is \( 45^\circ \).At \( t = 2.0 \text{ s}: \) \[ \vec{v}(2.0) = [5.0 \times 2.0] \hat{i} + [5.0] \hat{\jmath} = 10.0 \hat{i} + 5.0 \hat{\jmath} \text{ cm/s}. \] The magnitude is \( \|\vec{v}(2.0)\| \approx 11.2 \text{ cm/s} \) and the direction is \( \tan^{-1}(0.5) \approx 26.6^\circ \).

Step 4: Sketch the Dot's Trajectory

Plot the position vector \( \vec{r}(t) \) at several points between \( t = 0 \text{ s} \) and \( t = 2.0 \text{ s} \). Notice the path starts from \( 34.0 \hat{i} \text{ cm} \) and moves in an arc towards \( 44.0 \hat{i} + 10.0 \hat{\jmath} \text{ cm} \). Along this path, show the calculated velocities: - At \( t = 0 \text{ s} \), vector pointing vertically upwards, - At \( t = 1.0 \text{ s} \), vector at \( 45^\circ \), - At \( t = 2.0 \text{ s} \), vector closer to the \( \hat{i} \) due to larger \( \hat{i} \) component.

Key Concepts

Instantaneous Velocity

Instantaneous velocity describes the velocity of an object at a specific point in time. It's like taking a snapshot of the speedometer in your car at any moment, telling you exactly how fast you're traveling right then. In mathematical terms, instantaneous velocity is the derivative of position with respect to time, \( \vec{v}(t) = \frac{d\vec{r}}{dt} \).
In our exercise, the position vector \( \vec{r}(t) \) for the dot on the screen demonstrates how the position changes over time:

• At \( t = 0 \) seconds, the velocity is \( \vec{v}(0) = 5.0 \hat{\jmath} \) cm/s, indicating the dot is moving directly upwards.
• At \( t = 1.0 \) seconds, we find \( \vec{v}(1.0) = 5.0 \hat{i} + 5.0 \hat{\jmath} \) cm/s, showing movement at a \( 45^\circ \) angle.
• At \( t = 2.0 \) seconds, the dot's velocity changes to \( 10.0 \hat{i} + 5.0 \hat{\jmath} \) cm/s, with a predominant \( \hat{i} \) direction indicating a trajectory moving more horizontally.

These calculations provide clarity on how motion is not just about getting from point A to B, but how speed and direction continuously adjust along the journey.

Vector Calculus

Vector calculus plays an essential role in understanding various physical phenomena, including the motion of objects. Vectors have both magnitude and direction, making them versatile tools for describing motion. They help in understanding how an object's location, velocity, and acceleration change in space with respect to time.
In the provided exercise, the position vector \( \vec{r}(t) \) is expressed with components \( \hat{i} \) and \( \hat{\jmath} \):

• The \( \hat{i} \) component relates to horizontal motion, formulated as \( 34.0 + 2.5t^2 \) cm.
• The \( \hat{\jmath} \) component describes vertical movement, defined as \( 5.0t \) cm.

To calculate the instantaneous velocity, we differentiate each component with respect to time \( t \), decomposing the velocity vector into \( \hat{i} \) and \( \hat{\jmath} \): \( \vec{v}(t) = [5.0t] \hat{i} + [5.0] \hat{\jmath} \).
This approach provides insight into how different forces and initial conditions influence motion in a multidimensional space, crucial for designing precise animations and simulations.

Trajectory

The concept of trajectory outlines the path that an object follows through space as a function of time. It is affected by initial positions, velocities, and any forces acting upon the object. In terms of visualization, the trajectory provides a comprehensive look at where an object has been and where it is going.
In the exercise, the dot follows a specific trajectory on the computer screen as it moves from its initial position to successive locations. By analyzing its position vector \( \vec{r}(t) = [34.0 + 2.5t^2] \hat{i} + [5.0t] \hat{\jmath} \), we can predict the dot's location at any given time between \( t = 0 \) and \( t = 2.0 \) seconds:- Initially at \( 34.0 \hat{i} \) cm.- Gradually follows an arc toward \( 44.0 \hat{i} + 10.0 \hat{\jmath} \) cm.This trajectory is plotted to understand the dot's path, pairing each point with calculated velocities that depict how its speed and direction varies. Such graphical representations are invaluable in fields like animation and engineering, where analyzing motion paths is critical for practical applications.