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Magazine Chapter 2: Problem 61
At time \(t=0,\) a particle is at rest in the \(x\) -y plane at the coordinates \(\left(x_{0}, y_{0}\right)=(6,0)\) in. If the particle is then subjected to the acceleration components \(a_{x}=\) \(0.5-0.35 t\) in. \(/ \sec ^{2}\) and \(a_{y}=0.15 t-0.02 t^{2}\) in. \(/ \mathrm{sec}^{2}\) determine the coordinates of the particle position when \(t=6\) sec. Plot the path of the particle during this time period.
Short Answer
Expert verified
At \( t=6 \) seconds, the particle is at coordinates \((1.888, 2.79)\).
Step by step solution
01
Understand the Problem
We need to find the position of a particle with given acceleration components in the xy-plane at time \( t = 6 \) seconds, starting from rest at the point \( (6,0) \).
02
Set Up Initial Conditions
The particle is initially at rest at the coordinates \( (x_0, y_0) = (6, 0) \). This implies the initial velocities are \( v_x(0) = 0 \) and \( v_y(0) = 0 \).
03
Find Velocity Functions
To find the velocity functions, integrate the acceleration functions with respect to time:- The x-component of velocity is given by \( v_x(t) = \int (0.5 - 0.35t) \, dt \).- The y-component of velocity is given by \( v_y(t) = \int (0.15t - 0.02t^2) \, dt \).
04
Integrate to Find Velocity Components
Integrate \( a_x = 0.5 - 0.35t \):\[ v_x(t) = \int (0.5 - 0.35t) \, dt = 0.5t - 0.35\frac{t^2}{2} + C_x \]From \( v_x(0) = 0 \), solve for \( C_x \):\( 0 = 0 + 0 + C_x \Rightarrow C_x = 0 \)Thus, \( v_x(t) = 0.5t - 0.175t^2 \).Integrate \( a_y = 0.15t - 0.02t^2 \):\[ v_y(t) = \int (0.15t - 0.02t^2) \, dt = 0.075t^2 - 0.02\frac{t^3}{3} + C_y \]From \( v_y(0) = 0 \), solve for \( C_y \):\( 0 = 0 + 0 + C_y \Rightarrow C_y = 0 \)Thus, \( v_y(t) = 0.075t^2 - 0.0067t^3 \).
05
Find Position Functions
Next, integrate each velocity function to find the position functions:- The x-component of position is \( x(t) = x_0 + \int v_x(t) \, dt \).- The y-component of position is \( y(t) = y_0 + \int v_y(t) \, dt \).
06
Integrate to Find Position Components
Integrate \( v_x(t) = 0.5t - 0.175t^2 \):\[ x(t) = 6 + \int (0.5t - 0.175t^2) \, dt = 6 + 0.25t^2 - 0.175\frac{t^3}{3} \] Simplify: \( x(t) = 6 + 0.25t^2 - 0.0583t^3 \).Integrate \( v_y(t) = 0.075t^2 - 0.0067t^3 \):\[ y(t) = 0 + \int (0.075t^2 - 0.0067t^3) \, dt = 0 + 0.025t^3 - 0.0067\frac{t^4}{4} \]Simplify: \( y(t) = 0.025t^3 - 0.001675t^4 \).
07
Calculate Position at t=6 seconds
Evaluate the position functions at \( t = 6 \) seconds:- For \( x(6) \), calculate \( 6 + 0.25(6)^2 - 0.0583(6)^3 \).- For \( y(6) \), calculate \( 0.025(6)^3 - 0.001675(6)^4 \).This gives the position of the particle at \( t = 6 \) seconds.
08
Determine Final Coordinates
Calculate:- \( x(6) = 6 + 0.25(36) - 0.0583(216) = 1.888 \)- \( y(6) = 0.025(216) - 0.001675(1296) = 2.79 \)Thus, the coordinates at \( t=6 \) seconds are approximately \( (1.888, 2.79) \).
09
Plot the Path
To plot the path, create a graph with the x-axis as \( x(t) \) and the y-axis as \( y(t) \). Calculate few more points between 0 and 6 seconds using the position functions. This will show the trajectory of the particle over time.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acceleration
When a particle moves through space, understanding its acceleration is key to predicting its future position. Acceleration refers to the change in velocity over time. In this exercise, the particle's acceleration components are given as functions of time:
- \(a_x = 0.5 - 0.35t\) for the x-direction
- \(a_y = 0.15t - 0.02t^2\) for the y-direction
Velocity
Velocity is the rate at which a particle's position changes with time, incorporating both speed and direction. Initially, the particle is at rest, which means its velocity at time 0, \(v_x(0)\) and \(v_y(0)\), is 0. By integrating our acceleration functions:
- From \(a_x = 0.5 - 0.35t\), we get \(v_x(t) = 0.5t - 0.175t^2\)
- From \(a_y = 0.15t - 0.02t^2\), we get \(v_y(t) = 0.075t^2 - 0.0067t^3\)
Integral Calculus
Integral calculus is a powerful tool used to calculate accumulations such as area under a curve, total distance, or in this case, the velocity and position of a moving particle.
- First, by integrating the given acceleration functions, we determine the velocity functions.
- Then, integrating these velocity functions gives the position functions.
Trajectory Analysis
Trajectory analysis involves mapping the path taken by the particle as it travels through space over time. By applying the position functions:
Moreover, plotting these positions at incremental moments helps visualize the entire trajectory. This visualization is vital for understanding particle dynamics and predicting future locations and spatial behavior. Such trajectory plots are especially helpful in engineering applications where precision is critical, like projectile motions or orbital paths in physics.
- \(x(t) = 6 + 0.25t^2 - 0.0583t^3\)
- \(y(t) = 0.025t^3 - 0.001675t^4\)
Moreover, plotting these positions at incremental moments helps visualize the entire trajectory. This visualization is vital for understanding particle dynamics and predicting future locations and spatial behavior. Such trajectory plots are especially helpful in engineering applications where precision is critical, like projectile motions or orbital paths in physics.
