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Chapter 5: Problem 54

The motion of an object traveling along a straight path is given by \(s(t)=\frac{1}{2} a t^{2}+v_{0} t+s_{0},\) where \(s(t)\) is the position relative to the origin at time \(t\). For Exercises \(53-54,\) three observed data points are given. Find the values of \(a, v_{0},\) and \(s_{0}\). $$ s(1)=-7, s(2)=12, s(3)=37 $$

Short Answer

Expert verified
The values are \(a = 6\), \(v_0 = 10\), \(s_0 = -20\).

Step by step solution

01

- Plug in the Known Values

Start by plugging the known values into the equation for each given data point.For \(s(1)=-7\): \( -7 = \frac{1}{2}a(1)^2 + v_0(1) + s_0 \) For \(s(2)=12\): \( 12 = \frac{1}{2}a(2)^2 + v_0(2) + s_0 \) For \(s(3)=37\): \( 37 = \frac{1}{2}a(3)^2 + v_0(3) + s_0 \)
02

- Simplify Equations

Simplify each equation.Equation from \(s(1)\): \( -7 = \frac{1}{2}a + v_0 + s_0 \) Equation from \(s(2)\): \( 12 = 2a + 2v_0 + s_0 \) Equation from \(s(3)\): \( 37 = \frac{9}{2}a + 3v_0 + s_0 \)
03

- Create a System of Equations

Write the equations as a system:1. \( -7 = \frac{1}{2}a + v_0 + s_0 \)2. \( 12 = 2a + 2v_0 + s_0 \)3. \( 37 = \frac{9}{2}a + 3v_0 + s_0 \)
04

- Eliminate Variables

Use the system of equations to eliminate variables and solve. Start by subtracting equation 1 from equation 2 to eliminate \(s_0\): \( 19 = \frac{3}{2}a + v_0 \). Then, subtract equation 2 from equation 3: \( 25 = \frac{5}{2}a + v_0 \)
05

- Solve for One Variable

Subtract the first simplified equation from the second one to solve for \(a\): \( 25 - 19 = \frac{5}{2}a - \frac{3}{2}a \) \( 6 = a \)
06

- Substitute Back to Find Other Variables

Substitute \(a=6\) back into \( 19 = \frac{3}{2}a + v_0 \) to solve for \(v_0\): \(19 = \frac{3}{2}(6) + v_0 = 9 + v_0\) then \(v_0 = 10\). Substitute \(a=6\) and \(v_0=10\) into the first equation to solve for \(s_0\): \(-7 = \frac{1}{2}(6) + 10 + s_0 = 3 + 10 + s_0\) then \(s_0 = -20\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations
To understand motion problems in college algebra, it's essential to get familiar with kinematic equations. A kinematic equation describes the position, velocity, or acceleration of an object as a function of time. The equation in our problem is:
\[s(t) = \frac{1}{2} a t^2 + v_0 t + s_0\]
Here, \(s(t)\) represents the position of the object at time \(t\), \(a\) is the acceleration, \(v_0\) is the initial velocity, and \(s_0\) is the initial position.
This equation is derived from calculus and represents uniformly accelerated motion, where acceleration \(a\) is constant. By plugging in specific values of \(t\), we can find the object's position at those times.
solving systems of equations
In our problem, we have three observed data points: \(s(1) = -7\), \(s(2) = 12\), and \(s(3) = 37\). By substituting these points into the kinematic equation, we create a system of three equations:
\[ -7 = \frac{1}{2}a + v_0 + s_0 \ 12 = 2a + 2v_0 + s_0 \ 37 = \frac{9}{2}a + 3v_0 + s_0\]
This system of equations is essential for finding the unknown variables \(a\), \(v_0\), and \(s_0\). To solve it, we use methods such as substitution or elimination. These techniques help isolate one of the variables, making it easier to find the values of all variables involved.
algebraic simplification
Algebraic simplification helps break down complex equations into simpler forms. In our example, after substituting the data points into the kinematic equation, we simplified each equation:
\[ -7 = \frac{1}{2}a + v_0 + s_0 \ 12 = 2a + 2v_0 + s_0 \ 37 = \frac{9}{2}a + 3v_0 + s_0\]
These simpler equations are easier to work with. Next, using techniques like elimination, we subtract equations to cancel out one variable. Then, we solve for the remaining variables step-by-step.
For example, subtracting the first equation from the second gives us:
\[19 = \frac{3}{2}a + v_0 \]
This simplification process continues until all unknowns are found.
initial velocity
Initial velocity \(v_0\) represents the object's speed at time \(t = 0\). It's a crucial part of the kinematic equation. After simplifying and solving the system of equations, we find \(v_0\). In our example:
\[ 19 = \frac{3}{2}(6) + v_0 \]
Solving it gives:
\[ v_0 = 10\]
This value tells us that at the very start, the object was moving at 10 units of speed. Understanding initial velocity helps predict future positions and velocities of the object.
initial position
Initial position \(s_0\) is where the object starts, relative to an origin, at time \(t = 0\). It's another important variable in the kinematic equation. After calculating acceleration and initial velocity, we substitute them back to find \(s_0\). For example:
\[-7 = \frac{1}{2}(6) + 10 + s_0 \]
Calculating, we get:
\[-7 = 3 + 10 + s_0\]
\[s_0 = -20\]
This result indicates that the object started 20 units behind the origin. Knowing \(s_0\) helps in understanding the starting point of an object's motion, which is essential for accurate motion description.

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