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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![### Displacement Calculation Based on Velocity Function
#### Question:
What is the displacement of an object on the interval \(0 \leq t \leq 9\) if its velocity is \( v(t) = 6 - \sqrt{ 3t } \)?
#### Multiple Choice Options:
- 21.52
- 22.62
- 22.82
- 23.42
To calculate the displacement of the object, we need to integrate the given velocity function \( v(t) = 6 - \sqrt{ 3t } \) over the interval from \( t = 0 \) to \( t = 9 \).
**Detailed Solution:**
The velocity function is given by \( v(t) = 6 - \sqrt{ 3t } \). To find the displacement, we need to compute the definite integral of this function from 0 to 9.
\[
s = \int_{0}^{9} (6 - \sqrt{3t}) \, dt
\]
Let's break this down step by step:
1. **Integral of the Constant Term:**
\[
\int 6 \, dt = 6t
\]
2. **Integral of the Square Root Term:**
\[
\int -\sqrt{3t} \, dt
\]
Rewrite \(-\sqrt{3t}\) as \(-\sqrt{3} \cdot \sqrt{t}\):
\[
\int -\sqrt{3} \cdot t^{1/2} \, dt
\]
Using the power rule of integration:
\[
\int t^{1/2} \, dt = \frac{2}{3}t^{3/2}
\]
So,
\[
\int -\sqrt{3} \cdot t^{1/2} \, dt = -\sqrt{3} \cdot \frac{2}{3}t^{3/2} = -\frac{2\sqrt{3}}{3} t^{3/2}
\]
Combining the two parts, we get:
\[
\int_{0}^{9} (6 - \sqrt{3t}) \, dt = \left[ 6t - \frac{2\sqrt{3}}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5fb4dfc3-db65-4bb6-9a5b-9427f46c7a3d%2F7fb8a5ff-3853-4f93-bae9-60ddb75f1935%2F19oh75o_processed.png&w=3840&q=75)
Transcribed Image Text:### Displacement Calculation Based on Velocity Function
#### Question:
What is the displacement of an object on the interval \(0 \leq t \leq 9\) if its velocity is \( v(t) = 6 - \sqrt{ 3t } \)?
#### Multiple Choice Options:
- 21.52
- 22.62
- 22.82
- 23.42
To calculate the displacement of the object, we need to integrate the given velocity function \( v(t) = 6 - \sqrt{ 3t } \) over the interval from \( t = 0 \) to \( t = 9 \).
**Detailed Solution:**
The velocity function is given by \( v(t) = 6 - \sqrt{ 3t } \). To find the displacement, we need to compute the definite integral of this function from 0 to 9.
\[
s = \int_{0}^{9} (6 - \sqrt{3t}) \, dt
\]
Let's break this down step by step:
1. **Integral of the Constant Term:**
\[
\int 6 \, dt = 6t
\]
2. **Integral of the Square Root Term:**
\[
\int -\sqrt{3t} \, dt
\]
Rewrite \(-\sqrt{3t}\) as \(-\sqrt{3} \cdot \sqrt{t}\):
\[
\int -\sqrt{3} \cdot t^{1/2} \, dt
\]
Using the power rule of integration:
\[
\int t^{1/2} \, dt = \frac{2}{3}t^{3/2}
\]
So,
\[
\int -\sqrt{3} \cdot t^{1/2} \, dt = -\sqrt{3} \cdot \frac{2}{3}t^{3/2} = -\frac{2\sqrt{3}}{3} t^{3/2}
\]
Combining the two parts, we get:
\[
\int_{0}^{9} (6 - \sqrt{3t}) \, dt = \left[ 6t - \frac{2\sqrt{3}}{
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