Problem 37 - Precalculus

Question

Involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will cventually reach a maximum height and then fall to the ground. The equation that determines the height \(h\) of a projectile \(t\) seconds after it is shot upward is given by $$ h=\frac{1}{2} a t^{2}+v_{0} t+h_{0} $$ where \(a\) is the acceleration due to gravity, \(h_{0}\) is the initial height of the object at time \(t=0,\) and \(v_{0}\) is the initial velocity of the object at time \(t=0 .\) Note that a projectile follows the path of a parabola opening down, so \(a<0\) The number of minutes that an average person of age \(x\) spends driving a car can be modeled by a quadratic function \(y=a x^{2}+b x+c,\) where \(a<0\) and \(18 \leq x \leq 65\) The following table gives the average number of minutes per day that a person spends driving a car. Determine the quadratic function that models this quantity. $$\begin{array}{|l|c|} \hline \text { Aae } & \begin{array}{c} \text { Averacae Daner } \\ \text { Minutes Dervina } \end{array} \\ \hline 20 & 30 \\\ \hline 40 & 60 \\ \hline 60 & 40 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

The quadratic function is \( y = -\frac{1}{16}x^2 + 5.25x - 50 \).

Step 1: Understanding the Quadratic Model

To find the quadratic function that models the average driving minutes based on age, we recognize that the function takes the form \( y = ax^2 + bx + c \), where \( y \) represents the number of minutes driven, and \( x \) is the age. We are provided with points \( (20, 30), (40, 60), (60, 40) \).

Step 2: Setting Up the System of Equations

For each age and minute pair, substitute the values into the quadratic equation to form three equations: 1. For \( x = 20, y = 30 \): \[ 30 = a(20)^2 + b(20) + c \]2. For \( x = 40, y = 60 \): \[ 60 = a(40)^2 + b(40) + c \]3. For \( x = 60, y = 40 \): \[ 40 = a(60)^2 + b(60) + c \]

Step 3: Expressing Equations and Simplification

Simplify each equation:1. \( 30 = 400a + 20b + c \)2. \( 60 = 1600a + 40b + c \)3. \( 40 = 3600a + 60b + c \)

Step 4: Solving the System of Equations Using Elimination or Substitution

Subtract equation 1 from equation 2: \[ 30 = 1200a + 20b \] Subtract equation 2 from equation 3: \[ -20 = 2000a + 20b \] Now we have two simpler equations:1. \( 30 = 1200a + 20b \)2. \( -20 = 2000a + 20b \)

Step 5: Solve for Variables

Subtract the second simplified equation from the first: \[ 50 = -800a \] Solve for \( a \): \[ a = -\frac{1}{16} \]Substitute \( a \) back into equation 1 and solve for \( b \): \( 30 = 1200(-\frac{1}{16}) + 20b \) \[ 30 = -75 + 20b \] \[ 105 = 20b \] \[ b = 5.25 \]Substitute \( a \) and \( b \) values into the original equation to solve for \( c \): \[ 30 = 400(-\frac{1}{16}) + 20(5.25) + c \] \[ 30 = -25 + 105 + c \] \[ c = -50 \]

Step 6: Formulating the Final Quadratic Function

With the values calculated, the quadratic function is given by: \[ y = -\frac{1}{16}x^2 + 5.25x - 50 \]

Key Concepts

Projectile Motion

Projectile motion is a fascinating topic that involves objects moving through the air after being launched. Imagine kicking a ball or shooting an arrow into the sky. The path that these projectiles follow is influenced by several factors including initial speed, angle of launch, and gravity. The resulting path is a curved arc known as a parabolic trajectory. This is because all projectiles follow a quadratic path, similar to a mathematical function where they rise to a peak before falling back down.

Key elements of projectile motion include:

• Initial velocity (\(v_0\)): This is the speed at which the projectile is thrown or shot upwards. It greatly impacts how high and how far the projectile will travel.
• Initial height (\(h_0\)): If the projectile starts above ground level, this will affect the overall path.
• Parabolic trajectory: The motion is governed by a quadratic equation: \(h = \frac{1}{2} a t^{2}+v_{0} t+h_{0}\), where \(t\) is time, and \(a\) is the acceleration due to gravity.

The projectile's highest point is known as the apex or the maximum height, which is where the projectile momentarily stops going up and starts descending. Understanding projectile motion requires understanding both the physics behind the motion and the math used to calculate it.

Acceleration Due to Gravity

Gravity is a fundamental force that pulls objects towards the Earth. When dealing with projectile motion, gravity plays a crucial role, as it is the reason why objects fall back down after being propelled up. On Earth, the acceleration due to gravity is denoted by \(a\), and it has a standard value of approximately \(-9.8\, \text{m/s}^2\).

When a projectile is launched upward, gravity's downward pull gradually slows it down until it reaches the highest point. Afterward, gravity accelerates the projectile back towards the Earth, following the parabolic path of projectile motion. This is why in the equation for projectile motion, \(a\) is always negative, indicating the downward force: \(h = \frac{1}{2} a t^{2}+v_{0} t+h_{0}\).

Key points to remember about gravity in projectile motion:

• It acts downwards, influencing the peak height and descent of the projectile.
• The value of gravity can slightly change with location (e.g., it is different on other planets!), but for most calculations on Earth, we use \(-9.8\, \text{m/s}^2\).
• Gravity affects all objects equally, regardless of their mass, which is a key principle of physics.

Understanding gravity's impact is crucial for predicting how a projectile will move and where it will land.

System of Equations

A system of equations consists of multiple equations that share common variables, requiring simultaneous solutions. In the context of quadratic functions and projectile motion, we often use systems of equations to find unknown variables, such as initial speed or time taken to reach the ground. Solving a system of equations means finding values for the variables that satisfy all the given equations at once.

There are several methods to solve these systems:

• Substitution: Isolate one variable in one equation, then substitute it into another. This helps in finding a solution step by step.
• Elimination: Add or subtract equations to eliminate a variable, simplifying the system into smaller, manageable parts.
• Graphical: Plot the equations on a graph to see where they intersect; the intersection represents the solution.

In our exercise, we find the quadratic model for the given points using these systems. By substituting the points into the quadratic equation \(y = ax^2 + bx + c\), we form several equations. Solving these gives us the values of \(a\), \(b\), and \(c\), leading us to the final function. This method is essential in fields like physics and engineering, where such problems frequently arise.