Problem 28 - Precalculus: Functions and Graphs

Question

Particle acceleration If a particle moves along a coordinate line with a constant acceleration \(a\) (in \(\mathrm{cm} / \mathrm{sec}^{2}\) ), then at time \(t\) (in seconds) its distance \(s(t)\) (in centimeters) from the origin is \(s(t)=\frac{1}{2} a t^{2}+v_{0} t+s_{0}\) for velocity \(v_{0}\) and distance \(s_{0}\) from the origin at \(t=0 .\) If the distances of the particle from the origin at \(t=\frac{1}{2}, t=1,\) and \(t=\frac{3}{2}\) are \(7,11,\) and \(17,\) respectively, find \(a, v_{0+}\) and \(s_{0}\)

Step-by-Step Solution

Answer

The acceleration \(a\) is 8 cm/s², \(v_0\) is 2 cm/s, and \(s_0\) is 5 cm.

Step 1: Express Given Conditions as Equations

We are given three time points with corresponding distances:1. At \( t = \frac{1}{2} \), \( s\left(\frac{1}{2}\right) = 7 \): \[ s\left(\frac{1}{2}\right) = \frac{1}{2}a\left(\frac{1}{2}\right)^2 + v_0\left(\frac{1}{2}\right) + s_0 = 7 \]2. At \( t = 1 \), \( s(1) = 11 \): \[ s(1) = \frac{1}{2}a(1)^2 + v_0(1) + s_0 = 11 \]3. At \( t = \frac{3}{2} \), \( s\left(\frac{3}{2}\right) = 17 \): \[ s\left(\frac{3}{2}\right) = \frac{1}{2}a\left(\frac{3}{2}\right)^2 + v_0\left(\frac{3}{2}\right) + s_0 = 17 \]

Step 2: Convert Equations into a System of Equations

Substitute the terms:1. \( \frac{1}{8}a + \frac{1}{2}v_0 + s_0 = 7 \)2. \( \frac{1}{2}a + v_0 + s_0 = 11 \)3. \( \frac{9}{8}a + \frac{3}{2}v_0 + s_0 = 17 \)

Step 3: Solve the System of Equations: Eliminate One Variable

Subtract the first equation from the second to eliminate \( s_0 \):\[\left(\frac{1}{2}a + v_0 + s_0\right) - \left(\frac{1}{8}a + \frac{1}{2}v_0 + s_0\right) = 11 - 7\]This simplifies to:\[\frac{3}{8}a + \frac{1}{2}v_0 = 4 \text{ (Equation 4)}\]

Step 4: Solve the System of Equations: Eliminate Another Variable

Subtract the second equation from the third:\[\left(\frac{9}{8}a + \frac{3}{2}v_0 + s_0\right) - \left(\frac{1}{2}a + v_0 + s_0\right) = 17 - 11\]This simplifies to:\[\frac{5}{8}a + \frac{1}{2}v_0 = 6 \text{ (Equation 5)}\]

Step 5: Find Acceleration \(a\)

Subtract Equation 4 from Equation 5:\[\left(\frac{5}{8}a + \frac{1}{2}v_0\right) - \left(\frac{3}{8}a + \frac{1}{2}v_0\right) = 6 - 4\]This simplifies to:\[\frac{2}{8}a = 2\] Solving for \(a\): \[a = 8\]

Step 6: Find Initial Velocity \(v_0\)

Substitute \(a = 8\) back into Equation 4:\[\frac{3}{8}(8) + \frac{1}{2}v_0 = 4\]This simplifies to:\[3 + \frac{1}{2}v_0 = 4\]Solving for \(v_0\):\[\frac{1}{2}v_0 = 1\]\[v_0 = 2\]

Step 7: Find Initial Position \(s_0\)

Substitute \(a = 8\) and \(v_0 = 2\) back into one of our original equations, say Equation 2:\[\frac{1}{2}(8) + (2) + s_0 = 11\]This simplifies to:\[4 + 2 + s_0 = 11\]Solving for \(s_0\):\[s_0 = 5\]

Step 8: Verify the Solution

Verify by substituting \(a = 8\), \(v_0 = 2\), \(s_0 = 5\) into equations and check:1. \(\frac{1}{8}(8) + \frac{1}{2}(2) + 5 = 7\) (True)2. \(\frac{1}{2}(8) + (2) + 5 = 11\) (True)3. \(\frac{9}{8}(8) + \frac{3}{2}(2) + 5 = 17\) (True)All conditions are satisfied.

Key Concepts

Acceleration

Acceleration is the rate at which the velocity of a particle changes with respect to time. In other words, it's how quickly the object speeds up or slows down. When dealing with constant acceleration, the mathematical expression becomes simpler. For a particle moving along a coordinate line with a constant acceleration \(a\), this can be described by the formula:

• \( \text{Acceleration} = \frac{\Delta \text{Velocity}}{\Delta \text{Time}} \)

When this value is positive, the particle is speeding up. Conversely, a negative acceleration indicates the particle is slowing down. In the given problem, we calculated the acceleration \(a\) by studying the change in particle distance over time. With calculated values at three different time points, we could discern the constant acceleration affecting the particle's movement.

Distance-Time Relationship

The distance-time relationship in particle motion refers to how a particle's position changes over time. With constant acceleration, this relationship is governed by the equation:

• \( s(t) = \frac{1}{2} a t^{2} + v_{0} t + s_{0} \)

Here, \(s(t)\) is the distance from the origin at any given time \(t\), \(a\) is the constant acceleration, \(v_{0}\) is the initial velocity, and \(s_{0}\) is the initial position. This quadratic equation illustrates that the distance covered by the particle increases with time and can be visualized as a parabola when plotted on a graph. Such understanding allows us to predict the future position of the particle based on known parameters.

System of Equations

A system of equations is essentially a collection of two or more equations with a shared set of unknowns. In the context of this problem, we used a system of equations to solve for the unknowns: acceleration \(a\), initial velocity \(v_{0}\), and initial position \(s_{0}\). The derived equations from the given distances at different moments in time formed:

• \( \frac{1}{8}a + \frac{1}{2}v_0 + s_0 = 7 \)
• \( \frac{1}{2}a + v_0 + s_0 = 11 \)
• \( \frac{9}{8}a + \frac{3}{2}v_0 + s_0 = 17 \)

Using algebra, we eliminated one variable at a time to find the values of \(a\), \(v_{0}\), and \(s_{0}\). Solving these equations systematically allows us to piece together the initial conditions and the nature of the particle's motion.

Initial Velocity

The initial velocity \(v_{0}\) is the speed at which the particle was moving at the start of our observation, specifically at \(t = 0\). This initial speed impacts how the particle's speed changes over time under constant acceleration. In our equation:

• \( s(t) = \frac{1}{2} a t^{2} + v_{0} t + s_{0} \)

The term \(v_{0} t\) represents the influence of this initial velocity on the entire motion. By determining \(v_{0}\) from the system of equations, we can conclude how it contributes to the distance covered by the particle in conjunction with constant acceleration. It demonstrates the direction of motion and sets the pace of change at the beginning.

Initial Position

The initial position \(s_{0}\) is where the particle was located at the time \(t = 0\). It's a crucial component of the distance equation as it establishes the starting reference point from which all subsequent positions are measured. In mathematical terms:

• \( s_{0} \) is part of \( s(t) = \frac{1}{2} a t^{2} + v_{0} t + s_{0} \)

By knowing this initial position, we can make sense of the particle's path and its trajectory over time. In solving for \(s_{0}\), we found out where exactly the particle commenced its journey before the applied acceleration and initial velocity altered its course.