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An electron's position is given by \(\vec{r}=3.00 t \hat{\mathrm{i}}-4.00 t^{2} \hat{\mathrm{j}}+2.00 \hat{\mathrm{k}}\), with \(t\) in seconds and \(\vec{r}\) in meters. (a) In unit-vector notation, what is the electron's velocity \(\vec{v}(t) ?\) At \(t=3.00 \mathrm{~s}\), what is \(\vec{v}\) (b) in unitvector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the \(x\) axis?

Short Answer

Expert verified
(a) \( \vec{v}(t) = 3.00 \hat{\mathrm{i}} - 8.00t \hat{\mathrm{j}} \). (b) \( \vec{v}(3.00) = 3.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}} \). (c) Magnitude: 24.19 m/s. (d) Angle: \(-82.87^\circ\).

Step by step solution

01

Find the velocity function

The velocity of an object is the derivative of its position with respect to time. To find the velocity function \( \vec{v}(t) \), differentiate the position vector \( \vec{r}(t) = 3.00t \hat{\mathrm{i}} - 4.00t^2 \hat{\mathrm{j}} + 2.00 \hat{\mathrm{k}} \) with respect to \( t \):\[ \vec{v}(t) = \frac{d}{dt} (3.00t \hat{\mathrm{i}} - 4.00t^2 \hat{\mathrm{j}} + 2.00 \hat{\mathrm{k}}) = 3.00 \hat{\mathrm{i}} - 8.00t \hat{\mathrm{j}}. \]The \( \hat{\mathrm{k}} \) component is zero because the derivative of a constant is zero.
02

Evaluate velocity at t=3 s

Substitute \( t = 3.00 \) seconds into the velocity function found in Step 1 to find the velocity at this time point: \[ \vec{v}(3.00) = 3.00 \hat{\mathrm{i}} - 8.00(3.00) \hat{\mathrm{j}} = 3.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}}. \] Thus, the velocity in unit-vector notation at \( t = 3.00 \) seconds is \( \vec{v} = 3.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}} \).
03

Calculate the magnitude of velocity

The magnitude of the velocity vector \( \vec{v} = 3.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}} \) is found using the Pythagorean theorem: \[ |\vec{v}| = \sqrt{(3.00)^2 + (-24.00)^2} = \sqrt{9.00 + 576.00} = \sqrt{585.00} \approx 24.19 \, \text{m/s}. \]
04

Find angle with the x-axis

The angle \( \theta \) with respect to the positive x-axis is found using the inverse tangent function for the y and x components of the velocity: \[ \theta = \tan^{-1}\left(\frac{-24.00}{3.00}\right) = \tan^{-1}(-8.00). \]Calculate this angle: \[ \theta \approx -82.87^\circ. \] The negative sign indicates that the angle is measured clockwise from the positive x-axis.

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Key Concepts

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

Velocity Function
The velocity function is fundamentally important in describing how an object moves. For an electron moving in space, its position as a function of time is denoted by a vector. In this problem, the position vector given is \( \vec{r}(t) = 3.00t \hat{\mathrm{i}} - 4.00t^2 \hat{\mathrm{j}} + 2.00 \hat{\mathrm{k}} \). To find out how fast and in what direction the electron is moving at any time \( t \), we compute the velocity function.
  • The velocity \( \vec{v}(t) \) is the derivative of the position vector with respect to time \( t \).

  • We differentiate each component separately: the \( \hat{\mathrm{i}} \) component is the derivative of \( 3.00t \), the \( \hat{\mathrm{j}} \) component is the derivative of \( -4.00t^2 \), and the \( \hat{\mathrm{k}} \) component is constant, so its derivative is zero.

So, the calculated velocity function is \( \vec{v}(t) = 3.00 \hat{\mathrm{i}} - 8.00t \hat{\mathrm{j}} \). Here, - \( 3.00 \hat{\mathrm{i}} \) represents constant motion in the x-direction, and - \( -8.00t \hat{\mathrm{j}} \) indicates that the motion in the y-direction changes over time.
Magnitude of Velocity
Knowing the magnitude of a velocity vector gives you the speed of the object. Speed is the absolute value of how fast something is moving, regardless of its direction. Let's explore how we determine this magnitude from the velocity vector \( \vec{v} = 3.00 \hat{\mathrm{i}} - 24.00 \hat{\mathrm{j}} \) at \( t = 3.00 \) seconds.
  • The magnitude of the velocity vector can be calculated using the Pythagorean theorem.

  • It involves taking the square root of the sum of the squares of its components: \( \sqrt{(3.00)^2 + (-24.00)^2} \).

  • This calculation simplifies to \( \sqrt{9 + 576} = \sqrt{585} \approx 24.19 \text{ m/s} \).

This magnitude tells us how fast the electron is moving at exactly \( t = 3.00 \) seconds, without any consideration of which direction it is heading. Speed is useful in various practical situations where only how much motion exists is important, not the direction.
Angle Calculation
Determining the angle of the velocity vector helps us understand the direction of the electron's movement in relation to the axes. In this particular problem, we calculate the angle \( \theta \) relative to the positive x-axis.
  • We use the inverse tangent function to compute this angle, taking the ratio of the y-component to the x-component: \( \theta = \tan^{-1}\left(\frac{-24.00}{3.00}\right) \).

  • Mathematically, this gives \( \theta \approx \tan^{-1}(-8.00) \), which evaluates to approximately \( -82.87^\circ \).

  • The negative sign indicates that the direction is clockwise relative to the positive x-axis, suggesting the electron is moving downwards and towards the left.

Angle calculation is crucial for visualizing the trajectory of the electron and can help in comprehending complex movement patterns in multidimensional spaces.

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Most popular questions from this chapter

A stone is catapulted at time \(t=0\), with an initial velocity of magnitude \(18.0 \mathrm{~m} / \mathrm{s}\) and at an angle of \(40.0^{\circ}\) above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at \(t=1.10\) s? Repeat for the (c) horizontal and (d) vertical components at \(t=1.80 \mathrm{~s}\), and for the (e) horizontal and (f) vertical components at \(t=5.00 \mathrm{~s}\).

At \(t_{1}=2.00 \mathrm{~s}\), the acceleration of a particle in counterclockwise circular motion is \(\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). It moves at constant speed. At time \(t_{2}=5.00 \mathrm{~s}\), the particle's acceleration is \((4.00\) \(\left.\mathrm{m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). What is the radius of the path taken by the particle if \(t_{2}-t_{1}\) is less than one period?

A centripetal-acceleration addict rides in uniform circular motion with period \(T=2.0 \mathrm{~s}\) and radius \(r=3.50 \mathrm{~m}\). At \(t_{1}\) his acceleration is \(\vec{a}=\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At that instant, what are the values of (a) \(\vec{v} \cdot \vec{a}\) and (b) \(\vec{r} \times \vec{a}\) ?

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of \(3.66 \mathrm{~m} / \mathrm{s}\) and a centripetal acceleration \(\vec{a}\) of magnitude \(1.83 \mathrm{~m} / \mathrm{s}^{2}\). Position vector \(\vec{r}\) locates him relative to the rotation axis. (a) What is the magnitude of \(\vec{r}\) ? What is the direction of \(\vec{r}\) when \(\vec{a}\) is directed (b) due east and (c) due south?

Music is still available on vinyl records that are played on turntables. Such a record rotates with a period of \(1.8 \mathrm{~s}\). For a record with a radius of \(16 \mathrm{~cm}\), find the centripetal accceleration of a point on the edge of the record.

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