A particle is moving on a straight line with the acceleration and initial velocity a(t) = 2t - 2, v(0) = -3. Find the distance travelled by this particle during 0 ≤ t ≤ 4.
A particle is moving on a straight line with the acceleration and initial velocity a(t) = 2t - 2, v(0) = -3. Find the distance travelled by this particle during 0 ≤ t ≤ 4.
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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![### Physics Problem: Particle Motion
**Problem Statement:**
A particle is moving on a straight line with the acceleration and initial velocity given by:
\[ a(t) = 2t - 2 \]
\[ v(0) = -3 \]
Find the distance traveled by this particle during \( 0 \le t \le 4 \).
---
**Solution:**
The acceleration function \( a(t) = 2t - 2 \) is given. To find the velocity function \( v(t) \), we integrate the acceleration with respect to time \( t \):
\[ v(t) = \int (2t - 2) \, dt \]
\[ v(t) = t^2 - 2t + C \]
We use the initial condition \( v(0) = -3 \) to find \( C \):
\[ v(0) = 0^2 - 2(0) + C = -3 \Rightarrow C = -3 \]
So, the velocity function is:
\[ v(t) = t^2 - 2t - 3 \]
Next, to find the position function \( x(t) \), we integrate the velocity function with respect to time \( t \):
\[ x(t) = \int (t^2 - 2t - 3) \, dt \]
\[ x(t) = \frac{t^2}{3} - t^2 - 3t + D \]
We'll assume the initial position \( x(0) = 0 \) to find \( D \):
\[ x(0) = 0 \Rightarrow D = 0 \]
So, the position function is:
\[ x(t) = \frac{1}{3}t^3 - t^2 - 3t \]
To find the distance traveled during \( 0 \le t \le 4 \):
We should calculate \( x(4) - x(0) \):
\[ x(4) = \frac{1}{3}(4)^3 - (4)^2 - 3(4) \]
\[ x(4) = \frac{1}{3}(64) - 16 - 12 \]
\[ x(4) = 21.33 - 16 - 12 \]
\[ x(4) = -3.67 \]
Thus, the total distance traveled by the particle is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc07cb2f-6e43-45a8-b045-75d992a79e43%2F8e0fa607-bdf9-49cb-bc61-f2a3a7f11135%2Fgx73rlt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Particle Motion
**Problem Statement:**
A particle is moving on a straight line with the acceleration and initial velocity given by:
\[ a(t) = 2t - 2 \]
\[ v(0) = -3 \]
Find the distance traveled by this particle during \( 0 \le t \le 4 \).
---
**Solution:**
The acceleration function \( a(t) = 2t - 2 \) is given. To find the velocity function \( v(t) \), we integrate the acceleration with respect to time \( t \):
\[ v(t) = \int (2t - 2) \, dt \]
\[ v(t) = t^2 - 2t + C \]
We use the initial condition \( v(0) = -3 \) to find \( C \):
\[ v(0) = 0^2 - 2(0) + C = -3 \Rightarrow C = -3 \]
So, the velocity function is:
\[ v(t) = t^2 - 2t - 3 \]
Next, to find the position function \( x(t) \), we integrate the velocity function with respect to time \( t \):
\[ x(t) = \int (t^2 - 2t - 3) \, dt \]
\[ x(t) = \frac{t^2}{3} - t^2 - 3t + D \]
We'll assume the initial position \( x(0) = 0 \) to find \( D \):
\[ x(0) = 0 \Rightarrow D = 0 \]
So, the position function is:
\[ x(t) = \frac{1}{3}t^3 - t^2 - 3t \]
To find the distance traveled during \( 0 \le t \le 4 \):
We should calculate \( x(4) - x(0) \):
\[ x(4) = \frac{1}{3}(4)^3 - (4)^2 - 3(4) \]
\[ x(4) = \frac{1}{3}(64) - 16 - 12 \]
\[ x(4) = 21.33 - 16 - 12 \]
\[ x(4) = -3.67 \]
Thus, the total distance traveled by the particle is
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