Problem 56 - Beginning and Intermediate Algebra

Question

A model rocket is projected vertically upward from the ground. Its distance \(s\) in feet above the ground after t seconds is given by the quadratic function $$ s(t)=-16 t^{2}+256 t $$ Work Exercise in order, to see how quadratic equations and inequalities are related. At what times will the rocket be more than \(624 \mathrm{ft}\) above the ground? (Hint: Let \(s(t)>624\) and solve the quadratic inequality.)

Step-by-Step Solution

Answer

The rocket will be more than 624 feet above the ground between 3 and 13 seconds.

Step 1: Set Up the Inequality

Given is the function for the height of the rocket: \[ s(t) = -16t^2 + 256t \]We need to find the times when the rocket is more than 624 feet above the ground. Set up the inequality: \[ -16t^2 + 256t > 624 \]

Step 2: Move All Terms to One Side

Rearrange the inequality to set it to zero:\[ -16t^2 + 256t - 624 > 0 \]

Step 3: Simplify the Inequality

Factor the inequality. Divide the entire equation by -16. Remember to reverse the inequality sign when dividing by a negative number:\[ t^2 - 16t + 39 < 0 \]

Step 4: Solve the Quadratic Equation

To find the critical points, solve the corresponding quadratic equation:\[ t^2 - 16t + 39 = 0 \]Use the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 1\), \(b = -16\), and \(c = 39\):\[ t = \frac{16 \pm \sqrt{256 - 156}}{2} = \frac{16 \pm 10}{2} \]The solutions are:\[ t = \frac{26}{2} = 13 \]\[ t = \frac{6}{2} = 3 \]

Step 5: Determine the Interval

The critical points partition the number line into intervals. Test each interval to see where the inequality holds:Test a point in the interval \(t < 3\), \(3 < t < 13\), and \(t > 13\):\[ t = 0: 0^2 - 16(0) + 39 = 39 (fails) \]\[ t = 5: 5^2 - 16(5) + 39 = -16 (true) \]\[ t = 14: 14^2 - 16(14) + 39 = 235 (fails) \]

Step 6: State the Solution

The solution to the inequality is therefore:\[ 3 < t < 13 \]That means the rocket is more than 624 feet above the ground between 3 and 13 seconds.

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial where the highest degree of the variable is squared. The general form of a quadratic function is \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. Quadratic functions graph as parabolas.They can open upwards or downwards depending on the sign of \( a \).If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.In our problem, \( s(t) = -16t^2 + 256t \) is a quadratic function for the height of a rocket.Here, \( a = -16 \) and the parabola opens downwards.
Quadratic functions are essential in modeling various real-world situations, especially any context involving projectile motion, like rocket launches.

Solving Inequalities

When solving quadratic inequalities, we are looking for the values of the variable that make the inequality true.In this exercise, the inequality is \( -16t^2 + 256t > 624 \). We must rearrange and simplify the inequality to solve it.
First, move all terms to one side to set the inequality to zero: \( -16t^2 + 256t - 624 > 0 \).Next, simplify the inequality. Dividing by \( -16 \) and reversing the inequality sign: \( t^2 - 16t + 39 < 0 \).
Then, solve the corresponding quadratic equation to find the critical points.The solutions help us determine the intervals to test for where the inequality holds true.
This process of solving involves both algebraic manipulation and number line testing to determine the correct intervals.

Factoring Quadratics

Factoring quadratics is a method used to solve quadratic equations.In the standard form of a quadratic equation \( at^2 + bt + c = 0 \), we can use the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \( t^2 - 16t + 39 = 0 \), we have \( a = 1 \), \( b = -16 \), and \( c = 39 \).Applying the quadratic formula gives us the solutions: \( t = 13 \) and \( t = 3 \).
These solutions (or roots) help us identify the intervals on the number line.and check which intervals satisfy the original inequality.This method of solving by factoring is crucial in finding specific values needed to interpret real-world problems.

Rocket Motion

In the context of rocket motion, we often use quadratic functions to model the height of the rocket over time.The height function is typically a result of the initial velocity and the acceleration due to gravity.For this exercise, the function is\( s(t) = -16t^2 + 256t \).Each term has physical significance: -16 represents the effect of gravity, and 256 represents the initial velocity at launch.
We needed to find when the rocket is more than 624 feet above the ground.We set up and solved the inequality:\( -16t^2 + 256t > 624 \).The solution, \( 3 < t < 13 \), indicates the rocket is above 624 feet between 3 and 13 seconds after launch.
Understanding the motion through quadratic functions helps us predict the behavior of the rocket more accurately, providing crucial insights for time-specific events.