/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The acceleration of a particle m... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 19

The acceleration of a particle moving only on a horizontal \(x y\) plane is given by \(\vec{a}=3 t \hat{\mathrm{i}}+4 t \hat{\mathrm{j}}\), where \(\vec{a}\) is in meters per secondsquared and \(t\) is in seconds. At \(t=0\), the position vector \(\vec{r}=(20.0 \mathrm{~m}) \hat{\mathrm{i}}+(40.0 \mathrm{~m}) \hat{\mathrm{j}}\) locates the particle, which then has the velocity vector \(\vec{v}=(5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). At \(t=4.00 \mathrm{~s}\), what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the \(x\) axis?

Short Answer

Expert verified
(a) At \(t=4s\), \(\vec{r} = 88\hat{\mathrm{i}} + 69.3\hat{\mathrm{j}}\) meters. (b) The angle is approximately \(49.79^\circ\).

Step by step solution

01

Integrate the acceleration to find the velocity

The acceleration vector is given by \(\vec{a} = 3t \hat{\mathrm{i}} + 4t \hat{\mathrm{j}}\). To find the velocity vector \(\vec{v}(t)\), we integrate the acceleration with respect to time. The initial velocity at \(t=0\) is \(\vec{v}(0) = 5.00 \hat{\mathrm{i}} + 2.00 \hat{\mathrm{j}}\).\[ \vec{v}(t) = \int \vec{a} \, dt = \int (3t \hat{\mathrm{i}} + 4t \hat{\mathrm{j}}) \, dt = \left(\frac{3}{2}t^2 + C_1\right) \hat{\mathrm{i}} + \left(2t^2 + C_2\right) \hat{\mathrm{j}} \]For the constants \(C_1\) and \(C_2\), using initial conditions, \(\vec{v}(0) = 5.00 \hat{\mathrm{i}} + 2.00 \hat{\mathrm{j}}\):\[\begin{align*}\frac{3}{2}(0)^2 + C_1 &= 5.00 \2(0)^2 + C_2 &= 2.00\end{align*}\]\[\vec{v}(t) = \left(\frac{3}{2}t^2 + 5.00\right) \hat{\mathrm{i}} + (2t^2 + 2.00) \hat{\mathrm{j}}\]
02

Integrate the velocity to find the position

To find the position vector \(\vec{r}(t)\), we integrate the velocity vector. The initial position at \(t=0\) is \(\vec{r}(0) = 20.0 \hat{\mathrm{i}} + 40.0 \hat{\mathrm{j}}\).\[ \vec{r}(t) = \int \vec{v}(t) \, dt = \int \left(\left(\frac{3}{2}t^2 + 5.00\right) \hat{\mathrm{i}} + (2t^2 + 2.00) \hat{\mathrm{j}}\right) \, dt \]Calculating the integral and applying initial conditions:\[ \vec{r}(t) = \left(\frac{1}{2}t^3 + 5.00t + D_1\right) \hat{\mathrm{i}} + \left(\frac{2}{3}t^3 + 2.00t + D_2\right) \hat{\mathrm{j}} \]\(\vec{r}(0) = 20.0 \hat{\mathrm{i}} + 40.0 \hat{\mathrm{j}}\):\[\begin{align*}\frac{1}{2}(0)^3 + 5.00(0) + D_1 &= 20.0 \ \frac{2}{3}(0)^3 + 2.00(0) + D_2 &= 40.0\end{align*}\]\(\vec{r}(t) = (\frac{1}{2}t^3 + 5t + 20.0) \hat{\mathrm{i}} + (\frac{2}{3}t^3 + 2t + 40.0) \hat{\mathrm{j}}\). Evaluate at \(t = 4.00\ s\): \[\vec{r}(4) = (88.0)\hat{\mathrm{i}} + (69.3)\hat{\mathrm{j}}\]
03

Determine the angle of the position vector

The direction of travel can be found by the velocity components at \(t=4.00\ s\), \(\vec{v}(4) = ((3)(4)^2/2 + 5)\hat{\mathrm{i}} + (2(4)^2 + 2)\hat{\mathrm{j}}\): \[\vec{v}(4) = (29.0)\hat{\mathrm{i}} + (34.0)\hat{\mathrm{j}}\]The angle \(\theta\) between the velocity vector and the positive \(x\)-direction is given by: \[\theta = \tan^{-1}\left(\frac{34.0}{29.0}\right)\approx 49.79^\circ\]

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Key Concepts

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

Understanding the Velocity Vector
In kinematics, the velocity vector is a fundamental concept. It's a vector that describes both the speed and direction of an object's motion. Suppose we have an acceleration vector like in our exercise, \( \vec{a} = 3t \hat{\mathrm{i}} + 4t \hat{\mathrm{j}} \). To find the velocity vector, \( \vec{v} \), we need to perform integration over time.

This process involves determining how the object's velocity changes over time due to its acceleration. The integral of acceleration yields the velocity function: \[ \vec{v}(t) = \int (3t \hat{\mathrm{i}} + 4t \hat{\mathrm{j}}) \, dt = \left(\frac{3}{2}t^2 + C_1\right) \hat{\mathrm{i}} + \left(2t^2 + C_2\right) \hat{\mathrm{j}} \]

To find the constants \( C_1 \) and \( C_2 \), use initial conditions from the problem, like \( \vec{v}(0) = 5.00 \hat{\mathrm{i}} + 2.00 \hat{\mathrm{j}} \). This allows for precise calculation of the velocity vector at any given time.
Discovering the Position Vector
To determine where an object is at a certain time, we use the position vector \( \vec{r} \). This vector locates the particle in space. From our problem, the initial position vector is given, \( \vec{r}(0) = 20.0 \hat{\mathrm{i}} + 40.0 \hat{\mathrm{j}} \).

Finding the position vector at any time requires integrating the velocity vector: \[ \vec{r}(t) = \int \vec{v}(t) \, dt = \int \left(\left(\frac{3}{2}t^2 + 5.00\right) \hat{\mathrm{i}} + (2t^2 + 2.00) \hat{\mathrm{j}}\right) \, dt \]

After integration and applying initial conditions, the position vector becomes: \( \vec{r}(t) = (\frac{1}{2}t^3 + 5t + 20.0) \hat{\mathrm{i}} + (\frac{2}{3}t^3 + 2t + 40.0) \hat{\mathrm{j}} \). At any specific time, like \( t = 4.00 \, \mathrm{s} \), you can calculate the position.
Integration in Kinematics
Integration is a powerful tool in kinematics because it helps us transition from acceleration to velocity, and from velocity to position. In the context of motion, integration allows for the determination of the cumulative effects of acceleration on an object's movement over time.

For example, by integrating the acceleration function, we derived the velocity function. Integrating the velocity function then gave us the position function. This process is essential because it captures the continuous change in motion variables. With each integration, constants of integration arise, determined by initial conditions, such as initial velocity and initial position, specific to the problem's setup.
  • Helps find velocity from acceleration.
  • Assists in determining position from velocity.
  • Requires initial conditions to solve for constants.
If students grasp integration in kinematics, they can confidently solve many motion-related problems.

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Most popular questions from this chapter

A stairway has steps \(18.3 \mathrm{~cm}\) high and \(18.3 \mathrm{~cm}\) wide. A ball rolls horizontally off the top of the stairway with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Which step does the ball hit first?

A particle moves along a circular path over a horizontal \(x y\) coordinate system, at constant speed. At time \(t_{1}=5.00 \mathrm{~s}\), it is at point \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\) with velocity \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and acceleration in the positive \(x\) direction. At time \(t_{2}=10.0 \mathrm{~s}\), it has velocity \((-3.00\) \(\mathrm{m} / \mathrm{s}) \hat{\mathrm{i}}\) and acceleration in the positive \(y\) direction. What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path if \(t_{2}-t_{1}\) is less than one period?

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?

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?

A boat is traveling upstream in the positive direction of an \(x\) axis at \(14 \mathrm{~km} / \mathrm{h}\) with respect to the water of a river. The water is flowing at \(8.2 \mathrm{~km} / \mathrm{h}\) with respect to the ground. What are the (a) magnitude and (b) direction of the boat's velocity with respect to the ground? A child on the boat walks from front to rear at \(6.0\) \(\mathrm{km} / \mathrm{h}\) with respect to the boat. What are the (c) magnitude and (d) direction of the child's velocity with respect to the ground?
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