Problem 6 - Fundamentals of Physics

Question

An electron's position is given by \(\vec{r}=3.00 t \hat{\mathrm{i}}-4.00 t \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}}\), with \(t\) in seconds and \(\vec{r}\) in meters. (a) In unit-vector notation, what js the electron's velocity \(\vec{v}(t) ?\) At \(t=2.00 \mathrm{~s}\), what is \(\vec{v}(\mathrm{~b})\) in unitvector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis?

Step-by-Step Solution

Answer

The velocity is \(\vec{v}(t) = 3.00 \hat{i} - 4.00 \hat{j}\) m/s. At \( t = 2.00 \) s, the magnitude is 5.00 m/s and the direction angle is \( -53.1^{\circ} \) relative to the x-axis.

Step 1: Differentiate Position to Find Velocity

To find the velocity \( \vec{v}(t) \), differentiate the position vector \( \vec{r}(t) \) with respect to time \( t \). The derivative gives us:\[ \frac{d\vec{r}}{dt} = \frac{d}{dt}\left(3.00 t \hat{\mathrm{i}} - 4.00 t \hat{\mathrm{j}} + 2.00 \hat{\mathrm{k}} \right) \]Carrying out the differentiation, we obtain:\[ \vec{v}(t) = 3.00 \hat{\mathrm{i}} - 4.00 \hat{\mathrm{j}} + 0 \hat{\mathrm{k}} \]

Step 2: Calculate Velocity at Specific Time

Substitute \( t = 2.00 \) seconds into the velocity function to find the velocity at that time:\[ \vec{v}(2.00) = 3.00 \hat{\mathrm{i}} - 4.00 \hat{\mathrm{j}} \]

Step 3: Calculate Magnitude of Velocity

The magnitude of \( \vec{v} \) is determined by the formula:\[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2} \]For \( \vec{v}(2.00) = 3.00 \hat{\mathrm{i}} - 4.00 \hat{\mathrm{j}} \), the magnitude is:\[ |\vec{v}(2.00)| = \sqrt{(3.00)^2 + (-4.00)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.00 \text{ m/s} \]

Step 4: Find Direction Angle Relative to X-Axis

To find the angle \( \theta \) between \( \vec{v} \) and the positive \( x \)-axis, use the tangent function:\[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{-4.00}{3.00}\right) \]Calculating gives:\[ \theta \approx \tan^{-1}(-1.33) \approx -53.1^{\circ} \] meaning the vector is below the \( x \)-axis.

Key Concepts

Velocity Calculation

In physics, velocity is a fundamental concept that tells us the rate at which an object changes its position. When we deal with vector quantities, such as the position vector given in unit-vector notation, calculating velocity involves determining both the direction and magnitude of movement.

To find the velocity of an object, we need to determine how its position changes over time. For the electron in the problem, its position does change with time. The function \(\vec{r} = 3.00t \hat{\mathrm{i}} - 4.00t \hat{\mathrm{j}} + 2.00 \hat{\mathrm{k}}\) specifies the position at time \(t\) (measured in seconds).

For a given time, the velocity vector \(\vec{v}(t)\) is calculated by differentiating the position vector \(\vec{r}(t)\) with respect to \(t\). This gives us instantaneous velocity—speed and direction—at any time. By performing this calculation, you can accurately determine an object's velocity in real-time.

Differentiation of Vectors

Differentiation is a mathematical process that determines the rate at which a quantity changes. When applied to vectors, it allows us to find the velocity from a position vector that changes with time. The process involves computing the derivative of each component of the vector separately.

Consider the position vector \(\vec{r}(t) = 3.00t \hat{\mathrm{i}} - 4.00t \hat{\mathrm{j}} + 2.00 \hat{\mathrm{k}}\). Here, \(3.00t\), \(-4.00t\), and \(2.00\) are the vector's components in the \(i\), \(j\), and \(k\) directions respectively. By using basic rules of differentiation, such as the power rule, we compute the derivative as follows:

• \(\frac{d}{dt}(3.00t) = 3.00\) (as the derivative of \(t\) is 1)
• \(\frac{d}{dt}(-4.00t) = -4.00\)
• \(\frac{d}{dt}(2.00) = 0\) (since it's a constant)

Putting these results together, the velocity vector \(\vec{v}(t)\) becomes \(3.00 \hat{\mathrm{i}} - 4.00 \hat{\mathrm{j}}\). Differentiation reveals the speed components, showing how they are constant over time since the electron moves in a straight line.

Magnitude of Velocity

To fully understand velocity, it's not just the direction that matters, but also how fast an object is moving, which is the magnitude of velocity. After computing the velocity vector, the next step is to find its magnitude.

This is akin to finding the length of a vector and is computed using the Euclidean distance formula. For a vector \(\vec{v} = v_x \hat{\mathrm{i}} + v_y \hat{\mathrm{j}} + v_z \hat{\mathrm{k}}\), its magnitude is given by:

\[|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]

For the velocity at time \(t = 2.00 \, \text{s}\), the vector is \(\vec{v}(2.00) = 3.00 \hat{\mathrm{i}} - 4.00 \hat{\mathrm{j}}\). Thus, we calculate its magnitude as follows:

\[|\vec{v}| = \sqrt{(3.00)^2 + (-4.00)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.00 \, \text{m/s}\]

The magnitude tells us that the speed of the electron is 5 m/s, irrespective of its direction. It provides a complete picture by combining speed with direction.

Direction Angle with Axis

Understanding the direction of a velocity vector relative to a reference axis is crucial for grasping its orientation. This involves finding the direction angle, which expresses the vector's inclination.

For a two-dimensional velocity vector, like \(\vec{v}(2.00) = 3.00 \hat{\mathrm{i}} - 4.00 \hat{\mathrm{j}}\), the angle relative to the positive \(x\)-axis can be found using trigonometry.

The tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. For our vector, this translates to:

\[\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{-4.00}{3.00}\right)\approx \tan^{-1}(-1.33)\approx -53.1^\circ\]

• The negative sign indicates the vector is below the \(x\)-axis.
• This specifies the electron's movement direction as southwest in standard coordinate terms.

This angle helps visualize the movement's trajectory, completing the description of velocity for both magnitude and direction.