Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 2: Problem 2
The position of a particle is given by \(s=0.27 t^{3}-\) \(0.65 t^{2}-2.35 t+4.4,\) where \(s\) is in feet and the time \(t\) is in seconds. Plot the displacement, velocity, and acceleration as functions of time for the first 5 seconds of motion. Determine the positive time when the particle changes its direction.
Short Answer
Expert verified
The particle changes direction at \(t \approx 2.7\) seconds.
Step by step solution
01
Understand the Function for Position
The given function for the position of the particle is \(s(t) = 0.27t^3 - 0.65t^2 - 2.35t + 4.4\). We will use this to find the velocity and acceleration by differentiating it with respect to time \(t\).
02
Find the Velocity Function
Velocity \(v(t)\) is the derivative of the position function \(s(t)\) with respect to time. Therefore: \[ v(t) = \frac{d}{dt}(0.27t^3 - 0.65t^2 - 2.35t + 4.4) = 0.81t^2 - 1.3t - 2.35 \]
03
Find the Acceleration Function
Acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\) with respect to time. Therefore: \[ a(t) = \frac{d}{dt}(0.81t^2 - 1.3t - 2.35) = 1.62t - 1.3 \]
04
Plot Displacement, Velocity, and Acceleration
To plot the functions, evaluate \(s(t)\), \(v(t)\), and \(a(t)\) for \(t\) ranging from 0 to 5 seconds, and graph these using a graphing tool or software. Each plot should have \(t\) on the x-axis and the function value on the y-axis.
05
Determine When the Particle Changes Direction
A particle changes direction when its velocity changes sign. Solve \(v(t) = 0.81t^2 - 1.3t - 2.35 = 0\) to find the times when the velocity is zero:Using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 0.81\), \(b = -1.3\), and \(c = -2.35\).\[t = \frac{-(-1.3) \pm \sqrt{(-1.3)^2 - 4 \cdot 0.81 \cdot (-2.35)}}{2 \cdot 0.81}\]Calculate the discriminant and solve for \(t\).
06
Calculate Times for Change of Direction
Calculate the exact times using the solution from the previous step to find the positive time as: \[t = \frac{1.3 \pm \sqrt{1.69 + 7.596}}{1.62}\]This simplifies to \[t = \frac{1.3 \pm 3.093087}{1.62}\], thus giving two solutions for \(t\). Choose the positive value that occurs within 5 seconds.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Position Function
In particle motion analysis, the position function gives us the location of a particle at any given time. The position function here is given by: \[ s(t) = 0.27t^3 - 0.65t^2 - 2.35t + 4.4 \]where \( s \) is the position in feet, and \( t \) is the time in seconds. This equation is a cubic polynomial in terms of time and will provide the path of the particle as it moves. To analyze the motion, it serves as the basis from which velocity and acceleration are derived. Understanding the position function helps in visualizing the particle's trajectory over the specified interval of time.
Velocity Function
The velocity function describes how fast and in which direction the particle's position changes concerning time. To find the velocity from the position function, we differentiate the position function with respect to time:\[ v(t) = \frac{d}{dt}(0.27t^3 - 0.65t^2 - 2.35t + 4.4) \]This results in:\[ v(t) = 0.81t^2 - 1.3t - 2.35 \]Velocity can tell us if the particle is moving forward or backward. Positive values mean forward movement, while negative values describe backward movement. This function is a quadratic, meaning it can change from positive to negative and vice versa over time.
Acceleration Function
Acceleration is the rate at which the velocity of a particle changes with time. It's the derivative of the velocity function:\[ a(t) = \frac{d}{dt}(0.81t^2 - 1.3t - 2.35) \]Calculating this derivative yields:\[ a(t) = 1.62t - 1.3 \]The acceleration function tells us how much the velocity changes per unit of time. Whether a particle is speeding up or slowing down depends on the sign of the acceleration. Positive acceleration implies an increase in velocity, while negative acceleration suggests a decrease. Tracking these changes helps understand the particle's behavior dynamically over time.
Change of Direction
A particle changes direction when its velocity shifts from positive to negative or vice versa. To find when this occurs, we solve:\[ v(t) = 0.81t^2 - 1.3t - 2.35 = 0 \]Using the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]with \(a = 0.81\), \(b = -1.3\), \(c = -2.35\). This gives:\[ t = \frac{1.3 \pm \sqrt{1.69 + 7.596}}{1.62} \]Calculating yields two values for \( t \); choose the positive one within the 5-second interval where:
- Velocity crosses zero.
- The particle changes direction.
Graphical Representation
Graphical representation involves plotting the position, velocity, and acceleration functions over the interval of interest 鈥� here, 0 to 5 seconds. This provides a visual understanding of how the particle moves.
- Position graph: Shows how the particle's location changes over time.
- Velocity graph: Indicates the speed and direction of travel.
- Acceleration graph: Highlights changes in speed over time.
