particle is moving with acceleration a(t) = 18t + 8. its position at time t = 0 is s(0) = 6 and its velocity at time t = 0 is v(0) = 5. What is its position at time t = 12?
particle is moving with acceleration a(t) = 18t + 8. its position at time t = 0 is s(0) = 6 and its velocity at time t = 0 is v(0) = 5. What is its position at time t = 12?
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 54CR
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![### Problem Description:
A particle is moving with acceleration given by the function \( a(t) = 18t + 8 \). The position of the particle at \( t = 0 \) is \( s(0) = 6 \) and its velocity at \( t = 0 \) is \( v(0) = 5 \).
### Objective:
To determine the position of the particle at time \( t = 12 \).
### Given:
- Acceleration function: \( a(t) = 18t + 8 \)
- Initial position: \( s(0) = 6 \)
- Initial velocity: \( v(0) = 5 \)
### Required:
Find the position of the particle at \( t = 12 \).
### Solution:
1. **Integration to find the velocity**:
Since acceleration \( a(t) = \frac{dv}{dt} \):
\[
v(t) = \int a(t) \, dt = \int (18t + 8) \, dt = 18 \int t \, dt + \int 8 \, dt
\]
\[
v(t) = 18 \left(\frac{t^2}{2}\right) + 8t + C_1
\]
\[
v(t) = 9t^2 + 8t + C_1
\]
Using the initial velocity \( v(0) = 5 \):
\[
5 = 9(0)^2 + 8(0) + C_1 \implies C_1 = 5
\]
Therefore, the velocity function is:
\[
v(t) = 9t^2 + 8t + 5
\]
2. **Integration to find the position**:
Since velocity \( v(t) = \frac{ds}{dt} \):
\[
s(t) = \int v(t) \, dt = \int (9t^2 + 8t + 5) \, dt
\]
\[
s(t) = 9 \int t^2 \, dt + 8 \int t \, dt + 5 \int dt
\]
\[
s(t) = 9 \left(\frac{t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F276d7283-01d9-45ed-a762-7e45846115eb%2Fc82a3da5-ee05-4a04-bbba-92f79cd9a251%2Fc2j0dfy_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Description:
A particle is moving with acceleration given by the function \( a(t) = 18t + 8 \). The position of the particle at \( t = 0 \) is \( s(0) = 6 \) and its velocity at \( t = 0 \) is \( v(0) = 5 \).
### Objective:
To determine the position of the particle at time \( t = 12 \).
### Given:
- Acceleration function: \( a(t) = 18t + 8 \)
- Initial position: \( s(0) = 6 \)
- Initial velocity: \( v(0) = 5 \)
### Required:
Find the position of the particle at \( t = 12 \).
### Solution:
1. **Integration to find the velocity**:
Since acceleration \( a(t) = \frac{dv}{dt} \):
\[
v(t) = \int a(t) \, dt = \int (18t + 8) \, dt = 18 \int t \, dt + \int 8 \, dt
\]
\[
v(t) = 18 \left(\frac{t^2}{2}\right) + 8t + C_1
\]
\[
v(t) = 9t^2 + 8t + C_1
\]
Using the initial velocity \( v(0) = 5 \):
\[
5 = 9(0)^2 + 8(0) + C_1 \implies C_1 = 5
\]
Therefore, the velocity function is:
\[
v(t) = 9t^2 + 8t + 5
\]
2. **Integration to find the position**:
Since velocity \( v(t) = \frac{ds}{dt} \):
\[
s(t) = \int v(t) \, dt = \int (9t^2 + 8t + 5) \, dt
\]
\[
s(t) = 9 \int t^2 \, dt + 8 \int t \, dt + 5 \int dt
\]
\[
s(t) = 9 \left(\frac{t
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