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Chapter 4: Problem 11

The position \(\vec{r}\) of a particle moving in an \(x y\) plane is given by \(\vec{r}=\left(2.00 t^{3}-5.00 t\right) \hat{\mathrm{i}}+\left(6.00-7.00 t^{4}\right) \hat{\mathrm{j}},\) with \(\vec{r}\) in meters and \(t\) in seconds. In unit-vector notation, calculate (a) \(\vec{r},(\mathrm{~b}) \vec{v},\) and (c) \(\vec{a}\) for \(t=2.00 \mathrm{~s}\). (d) What is the angle between the positive direction of the \(x\) axis and a line tangent to the particle's path at \(t=2.00 \mathrm{~s} ?\)

Short Answer

Expert verified
At t=2 s, \(\vec{r}=6 \hat{\mathrm{i}} - 106 \hat{\mathrm{j}}\), \(\vec{v}=19 \hat{\mathrm{i}} - 224 \hat{\mathrm{j}}\), \(\vec{a}=24 \hat{\mathrm{i}} - 336 \hat{\mathrm{j}}\), angle = -84.2掳.

Step by step solution

01

Calculate the Position Vector at t=2.00 seconds

To find \(\vec{r}\) at \(t = 2.00\) s, substitute \(t = 2.00\) into the expression for \(\vec{r}\):\[ \vec{r} = \left(2.00 \times (2.00)^3 - 5.00 \times 2.00 \right) \hat{\mathrm{i}} + \left(6.00 - 7.00 \times (2.00)^4 \right) \hat{\mathrm{j}} \]Calculate each component:- \(x\)-component: \(2.00 \times 8 - 10 = 16 - 10 = 6\)- \(y\)-component: \(6.00 - 7.00 \times 16 = 6.00 - 112 = -106\)So, \(\vec{r} = 6 \hat{\mathrm{i}} - 106 \hat{\mathrm{j}}\).
02

Calculate the Velocity Vector

To find the velocity \(\vec{v}\), take the derivative of the position vector \(\vec{r}\) with respect to time \(t\):\[ \vec{v} = \frac{d\vec{r}}{dt} = \left(6t^2 - 5\right) \hat{\mathrm{i}} + \left(-28t^3\right) \hat{\mathrm{j}} \]Substitute \(t = 2.00\) into \(\vec{v}\):- \(x\)-component: \(6 \times 4 - 5 = 24 - 5 = 19\)- \(y\)-component: \(-28 \times 8 = -224\)So, \(\vec{v} = 19 \hat{\mathrm{i}} - 224 \hat{\mathrm{j}}\).
03

Calculate the Acceleration Vector

To find the acceleration \(\vec{a}\), take the derivative of the velocity vector \(\vec{v}\) with respect to time \(t\):\[ \vec{a} = \frac{d\vec{v}}{dt} = \left(12t\right) \hat{\mathrm{i}} + \left(-84t^2\right) \hat{\mathrm{j}} \]Substitute \(t = 2.00\) into \(\vec{a}\):- \(x\)-component: \(12 \times 2 = 24\)- \(y\)-component: \(-84 \times 4 = -336\)So, \(\vec{a} = 24 \hat{\mathrm{i}} - 336 \hat{\mathrm{j}}\).
04

Calculate the Angle of the Tangent with the x-axis

The angle \(\theta\) between the tangent to the path (which is in the direction of the velocity vector \(\vec{v}\)) and the positive x-axis can be found using the inverse tangent (arctan) of the y-component and x-component of \(\vec{v}\):\[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{-224}{19}\right) \]Calculate the angle:\[ \theta \approx \tan^{-1}\left(-11.79\right) \approx -84.2^\circ \]This indicates that the vector points below the positive x-axis.

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Key Concepts

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

Position Vector
The position vector \( \vec{r} \) describes the precise location of a particle in a coordinate system, often represented in terms of unit vectors. In this case, \( \vec{r} \) in terms of time \( t \) is given by:
\[ \vec{r} = \left(2.00 t^3 - 5.00 t\right) \hat{\mathrm{i}} + \left(6.00 - 7.00 t^4\right) \hat{\mathrm{j}} \]
To find the position vector at a particular time, such as \( t = 2.00 \) seconds, we substitute \( t \) into the equation:
  • x-component: \( 2.00 \times (2.00)^3 - 5.00 \times 2.00 = 6 \)
  • y-component: \( 6.00 - 7.00 \times (2.00)^4 = -106 \)
This results in the position vector \( \vec{r} = 6 \hat{\mathrm{i}} - 106 \hat{\mathrm{j}} \), indicating the particle's location in the plane.
Velocity Vector
The velocity vector \( \vec{v} \) provides information about the speed and direction of a particle鈥檚 movement. It is derived by taking the derivative of the position vector \( \vec{r} \) with respect to time \( t \).
\[ \vec{v} = \frac{d\vec{r}}{dt} = \left(6t^2 - 5\right) \hat{\mathrm{i}} + \left(-28t^3\right) \hat{\mathrm{j}} \]
At \( t = 2.00 \) seconds, calculating components yields:
  • x-component: \( 6 \times 4 - 5 = 19 \)
  • y-component: \( -28 \times 8 = -224 \)
Thus, the velocity vector is \( \vec{v} = 19 \hat{\mathrm{i}} - 224 \hat{\mathrm{j}} \), showing the rate of change of the position vector.
Acceleration Vector
The acceleration vector \( \vec{a} \) describes how the velocity of a particle changes over time. This can be found by differentiating the velocity vector \( \vec{v} \) with respect to time \( t \).
\[ \vec{a} = \frac{d\vec{v}}{dt} = \left(12t\right) \hat{\mathrm{i}} + \left(-84t^2\right) \hat{\mathrm{j}} \]
For \( t = 2.00 \) seconds, we find the components as:
  • x-component: \( 12 \times 2 = 24 \)
  • y-component: \( -84 \times 4 = -336 \)
Therefore, the acceleration vector is \( \vec{a} = 24 \hat{\mathrm{i}} - 336 \hat{\mathrm{j}} \), which represents the change in velocity at that moment.
Tangent Angle
The tangent angle \( \theta \) refers to the angle between the velocity vector \( \vec{v} \) and the positive x-axis. This can be calculated using the arctangent (\( \tan^{-1} \)) of the ratio of the y-component to the x-component of the velocity vector.
\[ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{-224}{19}\right) \]
Plugging in the numbers, we get:
  • \( \theta \approx \tan^{-1}(-11.79) \approx -84.2^\circ \)
This negative angle signifies that the velocity vector points below the x-axis, providing insights into the direction of motion along with its speed.

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Most popular questions from this chapter

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