5. The position function of a particle is given by s (t) = t³ - 6t² + 9t, 0≤ t ≤ 4. Position units are meters, time units are seconds. (a) Find the velocity v (t) and acceleration a (t) of the particle. (b) Determine the times when the velocity is zero. (c) Sketch the graph of velocity v (t). When is the particle moving right? When is the particle moving left?
5. The position function of a particle is given by s (t) = t³ - 6t² + 9t, 0≤ t ≤ 4. Position units are meters, time units are seconds. (a) Find the velocity v (t) and acceleration a (t) of the particle. (b) Determine the times when the velocity is zero. (c) Sketch the graph of velocity v (t). When is the particle moving right? When is the particle moving left?
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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![### Problem 5
The position function of a particle is given by \( s(t) = t^3 - 6t^2 + 9t \), for \( 0 \le t \le 4 \).
- Position units are meters, time units are seconds.
#### (a) Find the velocity \( v(t) \) and acceleration \( a(t) \) of the particle.
#### (b) Determine the times when the velocity is zero.
#### (c) Sketch the graph of velocity \( v(t) \). When is the particle moving right? When is the particle moving left?
##### Solution:
1. **Finding the velocity and acceleration:**
- Velocity \( v(t) \) is the first derivative of the position function \( s(t) \):
\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9
\]
- Acceleration \( a(t) \) is the first derivative of the velocity function \( v(t) \):
\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12
\]
2. **Determining the times when the velocity is zero:**
- Set the velocity function equal to zero and solve for \( t \):
\[
3t^2 - 12t + 9 = 0
\]
- Simplify and solve the quadratic equation:
\[
t^2 - 4t + 3 = 0
\]
\[
(t-1)(t-3) = 0
\]
Thus, \( t = 1 \) and \( t = 3 \).
3. **Sketching the graph of velocity \( v(t) \):**
- The velocity \( v(t) = 3t^2 - 12t + 9 \) is a quadratic function that opens upwards (since the coefficient of \( t^2 \) is positive).
- The roots of the equation \( v(t) = 0 \) are at \( t = 1 \) and \( t =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6484a2ba-7b6c-4144-bd9b-1f2d784a131b%2Fbb390721-da7c-4e13-bfe6-dc415654fdc5%2Fsjlz4by_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 5
The position function of a particle is given by \( s(t) = t^3 - 6t^2 + 9t \), for \( 0 \le t \le 4 \).
- Position units are meters, time units are seconds.
#### (a) Find the velocity \( v(t) \) and acceleration \( a(t) \) of the particle.
#### (b) Determine the times when the velocity is zero.
#### (c) Sketch the graph of velocity \( v(t) \). When is the particle moving right? When is the particle moving left?
##### Solution:
1. **Finding the velocity and acceleration:**
- Velocity \( v(t) \) is the first derivative of the position function \( s(t) \):
\[
v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t) = 3t^2 - 12t + 9
\]
- Acceleration \( a(t) \) is the first derivative of the velocity function \( v(t) \):
\[
a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12
\]
2. **Determining the times when the velocity is zero:**
- Set the velocity function equal to zero and solve for \( t \):
\[
3t^2 - 12t + 9 = 0
\]
- Simplify and solve the quadratic equation:
\[
t^2 - 4t + 3 = 0
\]
\[
(t-1)(t-3) = 0
\]
Thus, \( t = 1 \) and \( t = 3 \).
3. **Sketching the graph of velocity \( v(t) \):**
- The velocity \( v(t) = 3t^2 - 12t + 9 \) is a quadratic function that opens upwards (since the coefficient of \( t^2 \) is positive).
- The roots of the equation \( v(t) = 0 \) are at \( t = 1 \) and \( t =
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