Find the position of a body moving on a coordinate line at time t for an acceleration a= dt² initial velocity v(0) = 20, and initial position s(0) = 6. ... The position of the body at time t is given by s =
Find the position of a body moving on a coordinate line at time t for an acceleration a= dt² initial velocity v(0) = 20, and initial position s(0) = 6. ... The position of the body at time t is given by s =
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
![**Problem Statement:**
Find the position of a body moving on a coordinate line at time \( t \) for an acceleration \( a = \dfrac{d^2 s}{dt^2} = e^t \), initial velocity \( v(0) = 20 \), and initial position \( s(0) = 6 \).
---
**Solution:**
The position of the body at time \( t \) is given by \( s = \text{____} \).
---
To solve this problem:
1. **Determine the velocity function:** We start with the given acceleration \( a = e^t \). Since \( a = \dfrac{d^2 s}{dt^2} \), we integrate to find the velocity \( v(t) \):
\[
\dfrac{dv}{dt} = e^t \implies v(t) = \int e^t \, dt = e^t + C_1
\]
2. **Apply the initial velocity condition:** Given \( v(0) = 20 \),
\[
v(0) = e^0 + C_1 = 1 + C_1 = 20 \implies C_1 = 19
\]
Therefore, the velocity function is:
\[
v(t) = e^t + 19
\]
3. **Determine the position function:** Next, we integrate to find the position \( s(t) \):
\[
v(t) = \dfrac{ds}{dt} = e^t + 19 \implies s(t) = \int (e^t + 19) \, dt = e^t + 19t + C_2
\]
4. **Apply the initial position condition:** Given \( s(0) = 6 \),
\[
s(0) = e^0 + 19 \cdot 0 + C_2 = 1 + C_2 = 6 \implies C_2 = 5
\]
Therefore, the position function is:
\[
s(t) = e^t + 19t + 5
\]
Thus, the position of the body at time \( t \) is \( \boxed{e^t + 19t + 5} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d31b151-adde-452e-9f9e-014f79fed582%2Fb5f2c07c-851c-4e6f-b871-fbd9dad5fd27%2Ff3fyyqd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the position of a body moving on a coordinate line at time \( t \) for an acceleration \( a = \dfrac{d^2 s}{dt^2} = e^t \), initial velocity \( v(0) = 20 \), and initial position \( s(0) = 6 \).
---
**Solution:**
The position of the body at time \( t \) is given by \( s = \text{____} \).
---
To solve this problem:
1. **Determine the velocity function:** We start with the given acceleration \( a = e^t \). Since \( a = \dfrac{d^2 s}{dt^2} \), we integrate to find the velocity \( v(t) \):
\[
\dfrac{dv}{dt} = e^t \implies v(t) = \int e^t \, dt = e^t + C_1
\]
2. **Apply the initial velocity condition:** Given \( v(0) = 20 \),
\[
v(0) = e^0 + C_1 = 1 + C_1 = 20 \implies C_1 = 19
\]
Therefore, the velocity function is:
\[
v(t) = e^t + 19
\]
3. **Determine the position function:** Next, we integrate to find the position \( s(t) \):
\[
v(t) = \dfrac{ds}{dt} = e^t + 19 \implies s(t) = \int (e^t + 19) \, dt = e^t + 19t + C_2
\]
4. **Apply the initial position condition:** Given \( s(0) = 6 \),
\[
s(0) = e^0 + 19 \cdot 0 + C_2 = 1 + C_2 = 6 \implies C_2 = 5
\]
Therefore, the position function is:
\[
s(t) = e^t + 19t + 5
\]
Thus, the position of the body at time \( t \) is \( \boxed{e^t + 19t + 5} \).
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