Problem 56 - College Algebra with Corequisite Support

Question

A falling object travels a distance given by the formula \(d=5 t+16 t^{2} \mathrm{ft},\) where \(t\) is measured in seconds. How long will it take for the object to travel \(74 \mathrm{ft}\) ?

Step-by-Step Solution

Answer

It takes 2 seconds for the object to travel 74 feet.

Step 1: Understanding the Problem

The problem asks us to determine the time it takes for a falling object to travel a certain distance. We are given a formula: \( d = 5t + 16t^2 \), where \(d\) is the distance in feet and \(t\) is the time in seconds. We need to find the value of \(t\) when \(d = 74\) feet.

Step 2: Setting Up the Equation

We need to replace \(d\) with 74 in the equation. Thus, the equation becomes: \[ 74 = 5t + 16t^2 \]. Our goal is to solve this equation for \(t\).

Step 3: Reordering the Equation

To solve the quadratic equation, rearrange it in standard form: \[ 16t^2 + 5t - 74 = 0 \]. This equation is now in the form \( ax^2 + bx + c = 0 \) where \(a = 16\), \(b = 5\), and \(c = -74\).

Step 4: Using the Quadratic Formula

The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \(a = 16\), \(b = 5\), and \(c = -74\) into the formula:\[ t = \frac{-5 \pm \sqrt{5^2 - 4(16)(-74)}}{2(16)} \].

Step 5: Calculating the Discriminant

First, calculate the discriminant \( \Delta = b^2 - 4ac \): \[ \Delta = 5^2 - 4 \times 16 \times (-74) = 25 + 4736 = 4761 \].

Step 6: Solving for Time \( t \)

With the discriminant \( \Delta = 4761 \), find the square root: \( \sqrt{4761} = 69 \). Now substitute back into the formula: \[ t = \frac{-5 \pm 69}{32} \]. This gives two potential solutions: 1) \( t = \frac{64}{32} = 2 \) seconds 2) \( t = \frac{-74}{32} = -2.3125 \) seconds.

Step 7: Selecting the Realistic Solution

Since time cannot be negative, we choose the positive solution. Therefore, the time it takes for the object to travel 74 feet is 2 seconds.

Key Concepts

Physics applications

Quadratic equations often serve as a bridge between mathematics and physics, helping to describe natural phenomena such as the motion of falling objects. In physics, these equations are vital for solving problems related to motion, force, and energy.

For instance, in our exercise, the given equation \(d = 5t + 16t^2\) comes from the context of gravitational motion. Here, it represents the distance \(d\) traveled by a falling object over a period \(t\) in seconds. Each component of the equation correlates with physical aspects like initial velocity and acceleration due to gravity:

• The term \(5t\) indicates an initial velocity component, where the object starts moving with some initial speed.
• The \(16t^2\) is related to the acceleration due to gravity, showing how quickly the object speeds up as it falls.

Understanding such equations assists students in applying mathematical solutions to real-world physics problems, making abstract concepts more tangible.

Time calculation

In physics problems, particularly those involving equations of motion, calculating time is crucial. Time calculation in quadratic equations involves solving for \(t\), which often helps determine how long a particular event or motion occurs.

Given the equation \(74 = 5t + 16t^2\), our objective was to solve for \(t\) when the distance \(d\) is 74 feet. Rearranging the equation into its standard quadratic form, \(16t^2 + 5t - 74 = 0\), presents it clearly for solution.

When solving for \(t\), remember:

• Every term reflects a different aspect of the movement; \(16t^2\) affects acceleration, while \(5t\) represents initial velocity.
• The actual calculation requires careful manipulation of numbers, attention to detail, and logical reasoning to ensure only physically meaningful solutions are considered.

By determining \(t\), these calculations reveal the time frame for an event—in this problem, how quickly a distance of 74 feet is crossed.

Quadratic formula

The quadratic formula is a mathematical tool designed to find solutions to quadratic equations of the form \(ax^2 + bx + c = 0\). It's particularly useful when factorization or simple algebraic methods don't yield easily obtainable solutions.

The formula itself, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is versatile and universally applicable to any quadratic, provided you know the coefficients \(a\), \(b\), and \(c\).

For the problem at hand with the equation \(16t^2 + 5t - 74 = 0\):

• Replace \(a = 16\), \(b = 5\), and \(c = -74\) into the formula to find \(t\).
• The discriminant, \(b^2 - 4ac\), dictates the nature of the roots. In this exercise, a positive discriminant \(4761\) ensures two real solutions.
• From the roots, select the physically meaningful one (as time cannot be negative in practical scenarios).

Mastering the quadratic formula equips students with the ability to tackle a broad array of quadratic problems across various scientific disciplines.