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 Statement:**
Find the total distance traveled by a particle that has velocity \( V(t) = 2t - 4 \) over the interval \([0, 5]\).
**Solution Overview:**
To find the total distance traveled by the particle, evaluate the integral of the speed \(|V(t)|\) over the given interval.
1. Identify when the velocity \( V(t) = 2t - 4 \) is zero to determine intervals where the particle changes direction. Set the equation to zero and solve for \( t \):
\[
2t - 4 = 0 \implies t = 2
\]
This indicates that the particle changes direction at \( t = 2 \).
2. Calculate the distance for each interval separately.
**Interval [0, 2]:**
\[
V(t) = 2t - 4 \implies |V(t)| = -(2t - 4) = 4 - 2t
\]
Find the integral:
\[
\int_0^2 (4 - 2t) \, dt = \left[ 4t - t^2 \right]_0^2 = (8 - 4) - (0) = 4
\]
**Interval [2, 5]:**
\[
V(t) = 2t - 4 \implies |V(t)| = 2t - 4
\]
Find the integral:
\[
\int_2^5 (2t - 4) \, dt = \left[ t^2 - 4t \right]_2^5 = (25 - 20) - (4 - 8) = 5 + 4 = 9
\]
3. Add the distances from each interval:
\[
\text{Total Distance} = 4 + 9 = 13
\]
The total distance traveled by the particle is 13 units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcdc8b149-2467-45b8-88fa-7bc1ca3ec305%2Fe51d25bc-910d-4fc8-a9c3-7a4a98d1f258%2Fnq8qcsa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the total distance traveled by a particle that has velocity \( V(t) = 2t - 4 \) over the interval \([0, 5]\).
**Solution Overview:**
To find the total distance traveled by the particle, evaluate the integral of the speed \(|V(t)|\) over the given interval.
1. Identify when the velocity \( V(t) = 2t - 4 \) is zero to determine intervals where the particle changes direction. Set the equation to zero and solve for \( t \):
\[
2t - 4 = 0 \implies t = 2
\]
This indicates that the particle changes direction at \( t = 2 \).
2. Calculate the distance for each interval separately.
**Interval [0, 2]:**
\[
V(t) = 2t - 4 \implies |V(t)| = -(2t - 4) = 4 - 2t
\]
Find the integral:
\[
\int_0^2 (4 - 2t) \, dt = \left[ 4t - t^2 \right]_0^2 = (8 - 4) - (0) = 4
\]
**Interval [2, 5]:**
\[
V(t) = 2t - 4 \implies |V(t)| = 2t - 4
\]
Find the integral:
\[
\int_2^5 (2t - 4) \, dt = \left[ t^2 - 4t \right]_2^5 = (25 - 20) - (4 - 8) = 5 + 4 = 9
\]
3. Add the distances from each interval:
\[
\text{Total Distance} = 4 + 9 = 13
\]
The total distance traveled by the particle is 13 units.
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