Problem 80 - A Graphical Approach to Precalculus with Limits

Question

Position of a Particle Suppose that the position of a particle moving along a straight line is given by $$s(t)=a t^{2}+b t+c$$ where \(t\) is time in seconds and \(a, b,\) and \(c\) are real numbers. If \(s(0)=-10, s(1)=6,\) and \(s(2)=30,\) find the equation that defines \(s(t)\). Then find \(s(10)\)

Step-by-Step Solution

Answer

Equation: \(s(t) = 4t^2 + 12t - 10\), \(s(10) = 510\)

Step 1: Use s(0) = -10 to find c

\(s(0) = a(0)^2 + b(0) + c = c = -10\)
So \(c = -10\) and \(s(t) = at^2 + bt - 10\).

Step 2: Use s(1) = 6 and s(2) = 30 to set up equations

\(s(1) = a + b - 10 = 6 \implies a + b = 16\) ... (i)
\(s(2) = 4a + 2b - 10 = 30 \implies 4a + 2b = 40 \implies 2a + b = 20\) ... (ii)

Step 3: Solve for a and b

Subtracting (i) from (ii): \((2a + b) - (a + b) = 20 - 16\)
\(a = 4\) ... Wait, let me recheck.

Actually: \(2a + b = 20\) and \(a + b = 16\). Subtracting: \(a = 4\).
Then \(b = 16 - 4 = 12\).

Hmm, but the given answer says \(a = 4, b = 12\)... Let me verify: \(s(1) = 4 + 12 - 10 = 6\) \(\checkmark\). \(s(2) = 16 + 24 - 10 = 30\) \(\checkmark\).

Wait, the given answer says \(s(t) = 4t^2 + 12t - 10\)? But the short answer says \(7t^2 + 2t - 10\). Let me recheck.
\(s(1) = 7 + 2 - 10 = -1 \neq 6\). That doesn't work!

My solution: \(a = 4, b = 12\), giving \(s(t) = 4t^2 + 12t - 10\).

Step 4: Find s(10)

\(s(10) = 4(100) + 12(10) - 10 = 400 + 120 - 10 = 510\)

The equation is \(s(t) = 4t^2 + 12t - 10\), and \(s(10) = \boxed{510}\).

Key Concepts

Position of a Particle

When discussing the movement of a particle along a straight line, its position can be mathematically described using a quadratic function. This is crucial in various physics and mathematics problems. The position function, often written as \(s(t) = at^2 + bt + c\), lets us determine where the particle is at any time \(t\).
The coefficients \(a\), \(b\), and \(c\) offer important information:

• \(a\) affects how the particle accelerates over time, as it decides the curvature of the parabola.
• \(b\) determines the initial velocity or the rate of change at which the particle moves when \(t = 0\).
• \(c\) provides the initial position of the particle.

When specific positions of the particle are known at given times, these constants can be determined to precisely describe the particle's motion.

Equation of Motion

The equation of motion helps predict the behavior of a particle over time. Given the equation \(s(t) = at^2 + bt + c\), you can interpret the motion by finding specific constants that match certain conditions. For instance, when you know the position at several points in time, you can derive a set of equations to find \(a\), \(b\), and \(c\).
In our example:

• For \(t = 0\), position \(s(0) = -10\), leading to \(c = -10\).
• For \(t = 1\), position \(s(1) = 6\), the equation becomes \(a + b - 10 = 6\).
• For \(t = 2\), position \(s(2) = 30\), gives \(4a + 2b - 10 = 30\).

Solving these equations enables us to find the values of \(a\) and \(b\) thereby constructing a complete equation of motion for the particle.

Solving Systems of Equations

Understanding how to solve systems of equations is fundamental in problems involving quadratic functions. In our exercise, this skill is used to find the coefficients of the quadratic equation describing the particle's motion.
Here's a step-by-step process:

• Start with the known equations derived from the position at different times.
• Substitution is key; insert known values into the equations to simplify. For example, knowing \(c = -10\) helps reduce both other equations by removing \(c\).
• In the example, two equations remain: \(a + b = 16\) and \(4a + 2b = 40\).
• Multiply or divide equations to align terms for simpler operations; multiplying the first equation by 2 helps eliminate \(b\) when subtracting equations.

This logical method of elimination results in solving for each variable, fulfilling the conditions provided by the problem and accurately describing the function at hand.