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Chapter 4: Problem 11

A particle that is moving in an \(x y\) plane has a position vector given by \(\vec{r}=\left(3.00 t^{3}-6.00 t\right) \hat{i}+\left(7.00-8.00 t^{4}\right) \hat{j}\), where \(\vec{r}\) is measured in meters and \(t\) is measured in seconds. For \(t=3.00 \mathrm{~s}\), in unitvector notation, find (a) \(\vec{r}\), (b) \(\vec{v}\), and (c) \(\vec{a}\). (d) Find the angle between the positive direction of the \(x\) axis and a line that is tangent to the path of the particle at \(t=3.00 \mathrm{~s}\).

Short Answer

Expert verified
\(\vec{r} = 63.00 \hat{i} - 641.00 \hat{j}\), \(\vec{v} = 75.00 \hat{i} - 864.00 \hat{j}\), \(\vec{a} = 54.00 \hat{i} - 864.00 \hat{j}\), angle = \(95.03^\circ\).

Step by step solution

01

Find the Position Vector \(\vec{r}\)

To find \(\vec{r}\) at \(t = 3.00\, \text{s}\), substitute \(t = 3.00\) into the position vector equation \(\vec{r}(t) = (3.00t^3 - 6.00t) \hat{i} + (7.00 - 8.00t^4) \hat{j}\). Calculate:\[ \vec{r}(3.00) = (3.00 \times (3.00)^3 - 6.00 \times 3.00) \hat{i} + (7.00 - 8.00 \times (3.00)^4) \hat{j} \]\[ = (81.00 - 18.00) \hat{i} + (7.00 - 648.00) \hat{j} \]\[ = 63.00 \hat{i}- 641.00 \hat{j} \]. Thus, the position \(\vec{r} = 63.00 \hat{i} - 641.00 \hat{j}\).
02

Calculate the Velocity Vector \(\vec{v}\)

The velocity vector \(\vec{v}(t)\) is the derivative of the position vector \(\vec{r}(t)\) with respect to time:\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \left(\frac{d}{dt}(3.00t^3 - 6.00t)\right) \hat{i} + \left(\frac{d}{dt}(7.00 - 8.00t^4)\right) \hat{j} \]Calculate:\[ = (9.00t^2 - 6.00) \hat{i} - (32.00t^3) \hat{j} \]Substitute \(t = 3.00\):\[ \vec{v}(3.00) = (9.00 \times 3.00^2 - 6.00) \hat{i} - (32.00 \times 3.00^3) \hat{j} \]\[ = (81.00 - 6.00) \hat{i} - 864.00 \hat{j} \]\[ = 75.00 \hat{i} - 864.00 \hat{j} \]. Thus, \(\vec{v} = 75.00 \hat{i} - 864.00 \hat{j}\).
03

Find the Acceleration Vector \(\vec{a}\)

The acceleration vector \(\vec{a}(t)\) is the derivative of the velocity vector \(\vec{v}(t)\):\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \left(\frac{d}{dt}(9.00t^2 - 6.00)\right) \hat{i} + \left(\frac{d}{dt}(-32.00t^3)\right) \hat{j} \]Calculate:\[ = (18.00t) \hat{i} - (96.00t^2) \hat{j} \]Substitute \(t = 3.00\):\[ \vec{a}(3.00) = (18.00 \times 3.00) \hat{i} - (96.00 \times 3.00^2) \hat{j} \]\[ = 54.00 \hat{i} - 864.00 \hat{j} \]. Thus, \(\vec{a} = 54.00 \hat{i} - 864.00 \hat{j}\).
04

Calculate the Angle \(\theta\)

The angle \(\theta\) between the positive x-axis and the velocity vector can be found using:\[ \tan \theta = \frac{v_y}{v_x} \]where \(v_y = -864.00\) and \(v_x = 75.00\).Thus:\[ \tan \theta = \frac{-864.00}{75.00} = -11.52 \]Calculate \(\theta\):\(\theta = \tan^{-1}(-11.52)\).Using a calculator, find \(\theta \approx -84.97^\circ\) or \(\theta = 180^\circ - 84.97^\circ = 95.03^\circ\) \( \) when considering the direction.

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Key Concepts

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

Understanding the Position Vector
A position vector specifies the location of a particle in the plane relative to an origin. It tells us where a particle is at any given time.
In this example, the position vector \( \vec{r} \) is given by the equation:
  • \( \vec{r}(t) = (3.00t^3 - 6.00t) \hat{i} + (7.00 - 8.00t^4) \hat{j} \)
To find where the particle is at \( t = 3.00\, \text{s} \), we substitute 3 into the expression:
\[ \vec{r}(3.00) = (81.00 - 18.00) \hat{i} + (7.00 - 648.00) \hat{j} = 63.00 \hat{i} - 641.00 \hat{j} \]This shows us the particle's coordinates \( 63.00 \) meters in the \( x \)-direction and \( -641.00 \) meters in the \( y \)-direction.
Exploring the Velocity Vector
The velocity vector represents the rate at which the position changes over time. It is essentially the derivative of the position vector with respect to time.
Mathematically, it gives us:
  • \( \vec{v}(t) = \frac{d\vec{r}}{dt} \)
For the given problem, we find:
  • \( \vec{v}(t) = (9.00t^2 - 6.00) \hat{i} - (32.00t^3) \hat{j} \)
Substituting \( t = 3.00 \), we derive:\[ \vec{v}(3.00) = 75.00 \hat{i} - 864.00 \hat{j} \]This vector tells us that at \( t = 3.00 \text{s} \), the particle's velocity components are \( 75.00 \) meters/second in the \( x \)-direction and \( -864.00 \) meters/second in the \( y \)-direction.
Diving into the Acceleration Vector
Acceleration describes how quickly the velocity of a particle changes with time. It's the derivative of the velocity vector.
For our moving particle, the acceleration vector is:
  • \( \vec{a}(t) = \frac{d\vec{v}}{dt} \)
Upon differentiating, we get:
  • \( \vec{a}(t) = (18.00t) \hat{i} - (96.00t^2) \hat{j} \)
Evaluating at \( t = 3.00 \), the acceleration vector becomes:\[ \vec{a}(3.00) = 54.00 \hat{i} - 864.00 \hat{j} \]Thus, at \( t = 3.00 \text{s} \), the particle undergoes an acceleration of \( 54.00 \) meters/second2 in the \( x \)-direction and \( -864.00 \) meters/second2 in the \( y \)-direction.
Angle Calculation in Motion
To determine the angle \( \theta \) between the positive \( x \)-axis and the tangent line to the path, it's essential to consider the velocity vector. Since the tangent to the path aligns with the velocity, the angle is found using:
  • \( \tan \theta = \frac{v_y}{v_x} \)
Where:
  • \( v_y = -864.00 \)
  • \( v_x = 75.00 \)
Hence:\[ \tan \theta = \frac{-864.00}{75.00} = -11.52 \]To find \( \theta \), calculate:\( \theta = \tan^{-1}(-11.52) \), which approximately equals \( -84.97^\circ \).
Considering direction, the correct angle relative to the positive \( x \)-axis is \( 95.03^\circ \). This shows the angle at which the path of the particle is tangent at \( t = 3.00 \text{s} \).

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Most popular questions from this chapter

You throw a ball toward a wall at speed \(25.0 \mathrm{~m} / \mathrm{s}\) and at angle \(\theta_{0}=\) \(40.0^{\circ}\) above the horizontal (Fig. 426). The wall is distance \(d=22.0 \mathrm{~m}\) from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

A projectile's launch speed is \(6.00\) times that of its speed at its maximum height. Find the launch angle \(\theta_{0}\).

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Ship \(A\) is located \(4.0 \mathrm{~km}\) north and \(2.5 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(22 \mathrm{~km} / \mathrm{h}\) toward the south, and \(\operatorname{ship} B\) has a velocity of \(40 \mathrm{~km} / \mathrm{h}\) in a direction \(37^{\circ}\) north of east. (a) What is the velocity of \(A\) relative to \(B\) in unit-vector notation with \(\hat{i}\) toward the east? (b) Write an expression (in terms of \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) ) for the position of \(A\) relative to \(B\) as a function of \(t\), where \(t=0\) when the ships are in the positions described above. (c) At what time is the separation between the ships least? (d) What is that least separation?

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking \(2.50 \mathrm{~s}\). Then security agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking \(10.0 \mathrm{~s}\). What is the ratio of the man's running speed to the sidewalk's speed?
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