12. For an object whose velocity in ft/sec is given by v(t) = cos(t), what is its distance, in feet, travelled on the interval t = 1 tot = 5? 0.75 0.29 1.8 2.20
12. For an object whose velocity in ft/sec is given by v(t) = cos(t), what is its distance, in feet, travelled on the interval t = 1 tot = 5? 0.75 0.29 1.8 2.20
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 12: Velocity and Distance Calculation
#### Problem Statement:
For an object whose velocity in ft/sec is given by \( v(t) = \cos(t) \), what is its distance, in feet, travelled on the interval \( t = 1 \) to \( t = 5 \)?
#### Answer Choices:
- \( \circ \) 0.75
- \( \circ \) 0.29
- \( \circ \) 1.8
- \( \circ \) 2.20
#### Explanation:
To find the distance travelled by the object, we need to find the definite integral of the velocity function \( v(t) \) over the interval from \( t = 1 \) to \( t = 5 \).
This involves calculating:
\[
\text{Distance} = \int_{1}^{5} \cos(t) \, dt
\]
To solve this integral:
1. Find the antiderivative of \( \cos(t) \).
2. Evaluate this antiderivative from \( t = 1 \) to \( t = 5 \).
The antiderivative of \( \cos(t) \) is \( \sin(t) \). Thus:
\[
\int_{1}^{5} \cos(t) \, dt = \sin(t) \bigg|_{1}^{5}
\]
\[
= \sin(5) - \sin(1)
\]
#### Numerical Calculation:
Evaluating \( \sin(5) \) and \( \sin(1) \) and subtracting the results gives the distance.
Use a calculator to find the precise values of \( \sin(5) \) and \( \sin(1) \):
\[
\sin(5) \approx -0.9589
\]
\[
\sin(1) \approx 0.8415
\]
Therefore:
\[
\sin(5) - \sin(1) \approx -0.9589 - 0.8415 = -1.8004
\]
Since the context is physical distance which is always positive, we take the absolute value:
\[
\text{Distance} \approx 1.8004
\]
Thus, the correct answer is:
\[
\circ \ 1.8
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4bf8d9f8-9f3c-4f11-9f6a-4feeeac00eb3%2Fc85fcb10-a325-4271-81d7-8e5d882991c8%2Fm3mu2bi_processed.png&w=3840&q=75)
Transcribed Image Text:### Question 12: Velocity and Distance Calculation
#### Problem Statement:
For an object whose velocity in ft/sec is given by \( v(t) = \cos(t) \), what is its distance, in feet, travelled on the interval \( t = 1 \) to \( t = 5 \)?
#### Answer Choices:
- \( \circ \) 0.75
- \( \circ \) 0.29
- \( \circ \) 1.8
- \( \circ \) 2.20
#### Explanation:
To find the distance travelled by the object, we need to find the definite integral of the velocity function \( v(t) \) over the interval from \( t = 1 \) to \( t = 5 \).
This involves calculating:
\[
\text{Distance} = \int_{1}^{5} \cos(t) \, dt
\]
To solve this integral:
1. Find the antiderivative of \( \cos(t) \).
2. Evaluate this antiderivative from \( t = 1 \) to \( t = 5 \).
The antiderivative of \( \cos(t) \) is \( \sin(t) \). Thus:
\[
\int_{1}^{5} \cos(t) \, dt = \sin(t) \bigg|_{1}^{5}
\]
\[
= \sin(5) - \sin(1)
\]
#### Numerical Calculation:
Evaluating \( \sin(5) \) and \( \sin(1) \) and subtracting the results gives the distance.
Use a calculator to find the precise values of \( \sin(5) \) and \( \sin(1) \):
\[
\sin(5) \approx -0.9589
\]
\[
\sin(1) \approx 0.8415
\]
Therefore:
\[
\sin(5) - \sin(1) \approx -0.9589 - 0.8415 = -1.8004
\]
Since the context is physical distance which is always positive, we take the absolute value:
\[
\text{Distance} \approx 1.8004
\]
Thus, the correct answer is:
\[
\circ \ 1.8
\]
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