4 2 A particle moving along a line has position s(t) = t - 18t'm at time t seconds (for t ≥ 0). At which times is the particle instantaneously motionless (zero velocity)? Show all work.
4 2 A particle moving along a line has position s(t) = t - 18t'm at time t seconds (for t ≥ 0). At which times is the particle instantaneously motionless (zero velocity)? Show all work.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![### Particle Motion and Instantaneous Rest
A particle moving along a line has its position described by the function \( s(t) = t^4 - 18t^2 \) meters at time \( t \) seconds (for \( t \geq 0 \)).
#### Problem
Determine the times at which the particle is instantaneously motionless (i.e., has zero velocity). Provide a detailed explanation of the solution.
#### Solution
1. **Velocity Function:**
Velocity \( v(t) \) is the first derivative of the position function \( s(t) \).
\[
v(t) = \frac{ds(t)}{dt}
\]
Compute the derivative:
\[
v(t) = \frac{d}{dt} (t^4 - 18t^2)
= 4t^3 - 36t
\]
2. **Determine where velocity is zero:**
Set the velocity function equal to zero to find the critical points.
\[
4t^3 - 36t = 0
\]
Factor out the common term \( 4t \):
\[
4t(t^2 - 9) = 0
\]
This simplifies to:
\[
4t(t - 3)(t + 3) = 0
\]
Solve for \( t \):
\[
t = 0, \quad t = 3, \quad t = -3
\]
Since \( t \geq 0 \), we ignore the negative value \( t = -3 \).
Therefore, the times at which the particle is instantaneously motionless are \( t = 0 \) seconds and \( t = 3 \) seconds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9a5f5a5-4eae-476c-bcab-4a001ac27552%2F74617cae-a686-42e1-8043-3cf4f44be45e%2F7zqqhvr_processed.png&w=3840&q=75)
Transcribed Image Text:### Particle Motion and Instantaneous Rest
A particle moving along a line has its position described by the function \( s(t) = t^4 - 18t^2 \) meters at time \( t \) seconds (for \( t \geq 0 \)).
#### Problem
Determine the times at which the particle is instantaneously motionless (i.e., has zero velocity). Provide a detailed explanation of the solution.
#### Solution
1. **Velocity Function:**
Velocity \( v(t) \) is the first derivative of the position function \( s(t) \).
\[
v(t) = \frac{ds(t)}{dt}
\]
Compute the derivative:
\[
v(t) = \frac{d}{dt} (t^4 - 18t^2)
= 4t^3 - 36t
\]
2. **Determine where velocity is zero:**
Set the velocity function equal to zero to find the critical points.
\[
4t^3 - 36t = 0
\]
Factor out the common term \( 4t \):
\[
4t(t^2 - 9) = 0
\]
This simplifies to:
\[
4t(t - 3)(t + 3) = 0
\]
Solve for \( t \):
\[
t = 0, \quad t = 3, \quad t = -3
\]
Since \( t \geq 0 \), we ignore the negative value \( t = -3 \).
Therefore, the times at which the particle is instantaneously motionless are \( t = 0 \) seconds and \( t = 3 \) seconds.
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