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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 \(\mathrm{t}\) in seconds. In unit-vector notation, calculate (a) \(\vec{r},(\mathrm{~b}) \vec{v}\), and \((\mathrm{c}) \vec{a}\) 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
\(\vec{r} = 6.00 \hat{\mathrm{i}} - 106.00 \hat{\mathrm{j}}\), \(\vec{v} = 19.00 \hat{\mathrm{i}} - 224.00 \hat{\mathrm{j}}\), \(\vec{a} = 24.00 \hat{\mathrm{i}} - 336.00 \hat{\mathrm{j}}\), \(\theta \approx -85^{\circ}\).

Step by step solution

01

Calculate Position Vector at t = 2.00 s

To find \(\vec{r}\) at \(t = 2.00\, \text{s}\), substitute \(t = 2.00\) into the position equation.\[\vec{r} = (2.00\cdot(2.00)^3 - 5.00\cdot2.00) \hat{\mathrm{i}} + (6.00 - 7.00\cdot(2.00)^4) \hat{\mathrm{j}}\]\[\vec{r} = (16.00 - 10.00) \hat{\mathrm{i}} + (6.00 - 112.00) \hat{\mathrm{j}}\]\[\vec{r} = 6.00 \hat{\mathrm{i}} - 106.00 \hat{\mathrm{j}}\].
02

Calculate Velocity Vector at t = 2.00 s

Differentiate the position vector \(\vec{r}\) with respect to time \(t\) to find the velocity vector \(\vec{v}\):\[\vec{v} = \frac{d}{dt} \left((2.00 t^3 - 5.00 t)\hat{\mathrm{i}} + (6.00 - 7.00 t^4)\hat{\mathrm{j}}\right)\]\[\vec{v} = (6.00 t^2 - 5.00) \hat{\mathrm{i}} + (-28.00 t^3) \hat{\mathrm{j}}\]At \(t = 2.00\) s,\[\vec{v} = (6.00 \cdot(2.00)^2 - 5.00) \hat{\mathrm{i}} + (-28.00 \cdot(2.00)^3) \hat{\mathrm{j}}\]\[\vec{v} = (24.00 - 5.00) \hat{\mathrm{i}} - 224.00 \hat{\mathrm{j}}\]\[\vec{v} = 19.00 \hat{\mathrm{i}} - 224.00 \hat{\mathrm{j}}\].
03

Calculate Acceleration Vector at t = 2.00 s

Differentiate the velocity vector \(\vec{v}\) with respect to time \(t\) to find the acceleration vector \(\vec{a}\):\[\vec{a} = \frac{d}{dt} ((6.00 t^2 - 5.00)\hat{\mathrm{i}} + (-28.00 t^3)\hat{\mathrm{j}})\]\[\vec{a} = (12.00 t) \hat{\mathrm{i}} + (-84.00 t^2) \hat{\mathrm{j}}\]At \(t = 2.00\) s,\[\vec{a} = (12.00 \cdot 2.00) \hat{\mathrm{i}} + (-84.00 \cdot (2.00)^2) \hat{\mathrm{j}}\]\[\vec{a} = 24.00 \hat{\mathrm{i}} - 336.00 \hat{\mathrm{j}}\].
04

Calculate Angle of Tangent Line to Path

The tangent's direction is given by the velocity vector. Find the angle \(\theta\) between the positive x-axis and the velocity vector \(\vec{v}\) using:\[\tan \theta = \frac{v_y}{v_x} = \frac{-224.00}{19.00}\]Calculate \(\theta\):\[\theta = \tan^{-1}\left(\frac{-224.00}{19.00}\right)\]\[\theta \approx -85^{\circ}\].

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Key Concepts

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

Position Vector
Imagine tracking the movement of a particle in a two-dimensional space using coordinates. Similar to how coordinates locate points on a map, position vectors help locate points in space. Given by the equation \( \vec{r} = (2.00 t^3 - 5.00 t) \hat{\mathrm{i}} + (6.00 - 7.00 t^4) \hat{\mathrm{j}} \), the position vector \( \vec{r} \) tells us the particle's location at any time \( t \).
  • The \( \hat{\mathrm{i}} \) and \( \hat{\mathrm{j}} \) are unit vectors along the x-axis and y-axis respectively.
  • At \( t = 2 \text{s} \), substitute \( t \) in the equation to find the position. For instance, \( \vec{r} = 6.00 \hat{\mathrm{i}} - 106.00 \hat{\mathrm{j}} \), which means the particle is at 6 meters along the x-axis and \(-106\) meters along the y-axis.
The position vector is crucial as it serves as the starting point for other calculations, like velocity and acceleration, which are derived from it.
Velocity Vector
Velocity vectors describe how fast and in which direction a particle's position changes. Think of it like looking at how fast a car is moving and which road it's taking. To find the velocity vector \( \vec{v} \), differentiate the position vector \( \vec{r} \) with respect to time. This involves finding the rate of change of each component in the position vector:
  • For our problem, the velocity is \( \vec{v} = (6.00 t^2 - 5.00) \hat{\mathrm{i}} + (-28.00 t^3) \hat{\mathrm{j}} \).
  • At \( t = 2 \text{s} \), calculate to get \( \vec{v} = 19.00 \hat{\mathrm{i}} - 224.00 \hat{\mathrm{j}} \), meaning the particle is moving 19 meters per second along the x-axis and \(-224\) meters per second along the y-axis.
The velocity vector does more than tell us speed; it also explains direction, helping to understand the path the particle follows over time.
Acceleration Vector
Acceleration vectors give insight into how a particle's velocity changes over time, similar to how a car speeds up or slows down. To obtain the acceleration vector \( \vec{a} \), further differentiate the velocity vector \( \vec{v} \) with respect to time:
  • For this problem, the acceleration vector is \( \vec{a} = (12.00 t) \hat{\mathrm{i}} + (-84.00 t^2) \hat{\mathrm{j}} \).
  • At \( t = 2 \text{s} \), calculate to find \( \vec{a} = 24.00 \hat{\mathrm{i}} - 336.00 \hat{\mathrm{j}} \), indicating the particle's velocity is increasing at 24 meters per second squared along the x-axis, and decreasing at \(-336\) meters per second squared along the y-axis.
Acceleration vectors are key in understanding changes in motion, predicting future positions and speeds, which is essential for complex dynamic systems.
Tangent Line Angle
The tangent line angle helps determine the direction a particle is heading compared to the x-axis. This is akin to finding out which way a road bends. Use the velocity vector \( \vec{v} \) to find this angle, as it defines the direction of motion:
  • The angle \( \theta \) is found using \( \tan \theta = \frac{v_y}{v_x} \), where \( v_y \) and \( v_x \) are the y and x components of the velocity vector.
  • For the given problem, at \( t = 2 \text{s} \), calculate \( \theta \) using the formula \( \theta = \tan^{-1}\left(\frac{-224.00}{19.00}\right) \).
  • This results in \( \theta \approx -85^{\circ} \), showing the particle is moving almost vertically downwards.
Understanding tangent line angles is vital for visualizing a particle's trajectory and making predictions about its future path. It connects vector components to real-world direction.

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Most popular questions from this chapter

In \(3.50 \mathrm{~h}\), a balloon drifts \(21.5 \mathrm{~km}\) north, \(9.70 \mathrm{~km}\) east, and \(2.88 \mathrm{~km}\) upward from its release point on the ground. Find (a) the magnitude of its average velocity and (b) the angle its average velocity makes with the horizontal.

(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator due to the rotation of Earth? (b) What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration of magnitude \(9.8 \mathrm{~m} / \mathrm{s}^{2} ?\)

The velocity \(\vec{v}\) of a particle moving in the \(x y\) plane is given by \(\vec{v}=\left(6.0 t-4.0 t^{2}\right) \hat{\mathrm{i}}+8.0 \hat{\mathrm{j}}\), with \(\vec{v}\) in meters per second and \(t(>0)\) in seconds. (a) What is the acceleration when \(t=3.0 \mathrm{~s}\) ? (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal \(10 \mathrm{~m} / \mathrm{s} ?\)

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A cameraman on a pickup truck is traveling westward at \(20 \mathrm{~km} / \mathrm{h}\) while he records a cheetah that is moving westward \(30 \mathrm{~km} / \mathrm{h}\) faster than the truck. Suddenly, the cheetah stops, turns, and then runs at \(45 \mathrm{~km} / \mathrm{h}\) eastward, as measured by a suddenly nervous crew member who stands alongside the cheetah's path. The change in the animal's velocity takes \(2.0 \mathrm{~s}\). What are the (a) magnitude and (b) direction of the animal's acceleration according to the cameraman and the (c) magnitude and (d) direction according to the nervous crew member?
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