Problem 47 - 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 \(a<0\) A TV station sells intervals of time for commercials of 10 seconds for \(\$ 100,20\) seconds for \(\$ 180,\) and 40 seconds for \(\$ 320 .\) It has 2 minutes for publicity during a game with a total revenue of \(\$ 1,060\) for six commercials shown. Find the number of commercials of cach length sold by the TV station if there are twice as many 10 second commercials as 40 second commercials.

Step-by-Step Solution

Answer

4 commercials of 10 seconds, 0 of 20 seconds, and 2 of 40 seconds are sold.

Step 1: Define Variables

To solve this problem, define variables for the number of commercials of each length. Let \(x\) be the number of 10-second commercials, \(y\) be the number of 20-second commercials, and \(z\) be the number of 40-second commercials.

Step 2: Write Equations Based on Conditions

We have three conditions to form equations:* Total time for commercials is \(2\) minutes, or \(120\) seconds: \(10x + 20y + 40z = 120\).* Total revenue is \(\$1060\): \(100x + 180y + 320z = 1060\).* Twice as many 10-second commercials as 40-second commercials: \(x = 2z\).

Step 3: Substitute and Simplify

Substitute the equation \(x = 2z\) into the other equations:1. \(10(2z) + 20y + 40z = 120\). Simplify to \(60z + 20y = 120\) which simplifies further to \(20y + 60z = 120\) or \(y + 3z = 6\).2. \(100(2z) + 180y + 320z = 1060\). Simplify to \(520z + 180y = 1060\) or \(26z + 9y = 53\).

Step 4: Solve the System of Equations

We now solve the system of linear equations:1. \(y + 3z = 6\)2. \(26z + 9y = 53\)Substitute \(y = 6 - 3z\) from the first into the second: \[ 26z + 9(6 - 3z) = 53 \]Simplify to find \(z\).After calculating, we find \(z = 2\). Now substitute \(z = 2\) back into \(y = 6 - 3z\) to get \(y = 0\).

Step 5: Find the Number of Each Commercial

With \(z = 2\), substitute back \(z\) into \(x = 2z\) to get \(x = 4\).Therefore, we have 4 commercials of 10 seconds, 0 commercials of 20 seconds, and 2 commercials of 40 seconds.

Key Concepts

Quadratic Equations

A Quadratic Equation is often written in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In projectile motion, the height of an object as a function of time forms a quadratic equation of the style \(h = \frac{1}{2} a t^2 + v_0 t + h_0\). This equation is essential to determine how objects move in a parabolic path due to gravity. Here, \(a\) typically represents the acceleration due to gravity, which has a value of \(-9.8\ \text{m/s}^2\) on Earth. The equation factors in:

• Initial velocity \(v_0\): the speed at which the object is projected upward.
• Initial height \(h_0\): the height from which the object is launched.
• Time \(t\): the duration for which the object is in motion.

The quadratic form of such an equation results in a parabolic trajectory, showing how the object rises, reaches a peak, and subsequently descends.

Acceleration due to Gravity

The term 'acceleration due to gravity' refers to the constant rate at which objects accelerate towards Earth. This is symbolized by \(a\), often valued at \(-9.8\ \text{m/s}^2\) due to Earth's gravitational pull. When solving problems involving vertical motion, this negative sign indicates that gravity acts downward.

Let's consider how this affects projectile motion:

• As an object ascends, gravity slows its upward velocity.
• Upon reaching the apex of its motion, the object momentarily stops before descending.
• As it descends, gravity accelerates its fall.

This consistent force alters both the speed and direction of motion, explaining the parabolic path an object follows. In terms of the quadratic function for height, \(a < 0\) signifies this downward pull.

System of Linear Equations

A System of Linear Equations involves two or more equations working together to describe relationships between variables. In solving for the TV commercial airtime, we formulated a system based on various conditions:

• Total commercial time: \(10x + 20y + 40z = 120\)
• Total revenue: \(100x + 180y + 320z = 1060\)
• Relationship between commercials: \(x = 2z\)

Using substitution, we replaced variables to simplify the system, making it easier to solve. With equations like:

• \(y + 3z = 6\)
• \(26z + 9y = 53\)

We use substitution and elimination methods to find values for \(z\), \(y\), and \(x\). These systems teach us how different variables interact under given conditions. By breaking them down, complex real-world problems become manageable.