III. Solve using a system of equations.neeitib 9ewlt gnoms be129vni 26w 000,S2 to sontherni nA (8 6.) The height at time t of an object that is moving in a vertical line with constant acceleration, a, is given by the position equation S = ½ at2 + vot + So. The height, s, is measured in feet, the acceleration, a, is measured in feet/sec?, t is measured in seconds, v, is the initial velocity (t = 0), and s, is the initial height (t = 0). Find the values of a, vo, and s, if s 52 att 1, s = 52 at t 2, and s = 20 att 3, and interpret the result.
III. Solve using a system of equations.neeitib 9ewlt gnoms be129vni 26w 000,S2 to sontherni nA (8 6.) The height at time t of an object that is moving in a vertical line with constant acceleration, a, is given by the position equation S = ½ at2 + vot + So. The height, s, is measured in feet, the acceleration, a, is measured in feet/sec?, t is measured in seconds, v, is the initial velocity (t = 0), and s, is the initial height (t = 0). Find the values of a, vo, and s, if s 52 att 1, s = 52 at t 2, and s = 20 att 3, and interpret the result.
Calculus: Early Transcendentals
8th Edition
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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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![### Solving a System of Equations for Vertical Motion
**Problem Statement:**
The height at time \( t \) of an object moving in a vertical line with constant acceleration \( a \) is given by the position equation:
\[ S = \frac{1}{2} a t^2 + v_0 t + s_0 \]
Where:
- \( S \) is the height in feet.
- \( a \) is the acceleration in feet per second squared (\( \text{ft/s}^2 \)).
- \( t \) is the time in seconds.
- \( v_0 \) is the initial velocity (when \( t = 0 \)).
- \( s_0 \) is the initial height (when \( t = 0 \)).
Given values:
- \( S = 52 \) at \( t = 1 \)
- \( S = 52 \) at \( t = 2 \)
- \( S = 20 \) at \( t = 3 \)
**Objective:**
Find the values of \( a \), \( v_0 \), and \( s_0 \), and interpret the result.
**Explanation:**
The problem provides conditions at three different times which will help us set up a system of equations.
1. At \( t = 1 \), the equation becomes:
\[
52 = \frac{1}{2}a(1)^2 + v_0(1) + s_0
\]
2. At \( t = 2 \), the equation becomes:
\[
52 = \frac{1}{2}a(2)^2 + v_0(2) + s_0
\]
3. At \( t = 3 \), the equation becomes:
\[
20 = \frac{1}{2}a(3)^2 + v_0(3) + s_0
\]
Using these equations, you can solve for the unknowns \( a \), \( v_0 \), and \( s_0 \) by substituting and eliminating variables as needed.
**Note:**
This problem explores the dynamics of vertical motion under constant acceleration—a fundamental concept in physics involving kinematic equations. Solving such systems develops problem-solving skills and understanding of mathematical modeling in physics contexts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92b9fa95-2e7d-446b-a380-41f972519da6%2F3c311f50-d1d4-482c-9ba6-92b998c874b7%2Fwdm7a9f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving a System of Equations for Vertical Motion
**Problem Statement:**
The height at time \( t \) of an object moving in a vertical line with constant acceleration \( a \) is given by the position equation:
\[ S = \frac{1}{2} a t^2 + v_0 t + s_0 \]
Where:
- \( S \) is the height in feet.
- \( a \) is the acceleration in feet per second squared (\( \text{ft/s}^2 \)).
- \( t \) is the time in seconds.
- \( v_0 \) is the initial velocity (when \( t = 0 \)).
- \( s_0 \) is the initial height (when \( t = 0 \)).
Given values:
- \( S = 52 \) at \( t = 1 \)
- \( S = 52 \) at \( t = 2 \)
- \( S = 20 \) at \( t = 3 \)
**Objective:**
Find the values of \( a \), \( v_0 \), and \( s_0 \), and interpret the result.
**Explanation:**
The problem provides conditions at three different times which will help us set up a system of equations.
1. At \( t = 1 \), the equation becomes:
\[
52 = \frac{1}{2}a(1)^2 + v_0(1) + s_0
\]
2. At \( t = 2 \), the equation becomes:
\[
52 = \frac{1}{2}a(2)^2 + v_0(2) + s_0
\]
3. At \( t = 3 \), the equation becomes:
\[
20 = \frac{1}{2}a(3)^2 + v_0(3) + s_0
\]
Using these equations, you can solve for the unknowns \( a \), \( v_0 \), and \( s_0 \) by substituting and eliminating variables as needed.
**Note:**
This problem explores the dynamics of vertical motion under constant acceleration—a fundamental concept in physics involving kinematic equations. Solving such systems develops problem-solving skills and understanding of mathematical modeling in physics contexts.
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