Problem 107 - Algebra A Combined Function

Question

At this writing, the world's tallest building is the Taipei 101 in Taipei, Taiwan, at a height of 1671 feet. (Source: Council on Tall Buildings and Urban Habitat) Suppose a worker is suspended 71 feet below the top of the pinnacle atop the building, at a height of 1600 feet above the ground. If the worker accidentally drops a bolt, the height of the bolt after tseconds is given by the expression \(1600-16 t^{2}\). a. Find the height of the bolt after 3 seconds. b. Find the height of the bolt after 7 seconds. c. To the nearest whole second, estimate when the bolt hits the ground. d. Factor \(1600-16 t^{2}\).

Step-by-Step Solution

Answer

a) 1456 feet; b) 816 feet; c) 10 seconds; d) \((40 + 4t)(40 - 4t)\).

Step 1: Substitute t for 3 seconds

To find the height of the bolt after 3 seconds, substitute \( t=3 \) into the expression \( 1600 - 16t^2 \).

Step 2: Calculate the height for 3 seconds

Evaluate the expression: \( 1600 - 16 imes 3^2 = 1600 - 16 imes 9 = 1600 - 144 = 1456 \). The height of the bolt after 3 seconds is 1456 feet.

Step 3: Substitute t for 7 seconds

To find the height of the bolt after 7 seconds, substitute \( t=7 \) into the expression \( 1600 - 16t^2 \).

Step 4: Calculate the height for 7 seconds

Evaluate the expression: \( 1600 - 16 imes 7^2 = 1600 - 16 imes 49 = 1600 - 784 = 816 \). The height of the bolt after 7 seconds is 816 feet.

Step 5: Set height to zero for ground impact

When the bolt hits the ground, its height is 0. Set the equation \( 1600 - 16t^2 = 0 \) and solve for \( t \).

Step 6: Solve quadratic equation for t

Rearrange to \( 16t^2 = 1600 \), then divide by 16 to get \( t^2 = 100 \). Solve for \( t \) to find \( t = 10 \) seconds. To the nearest whole second, the bolt hits the ground at 10 seconds.

Step 7: Factor the expression

Factor \( 1600 - 16t^2 \) as a difference of squares: \( (40)^2 - (4t)^2 \) can be factored as \((40 + 4t)(40 - 4t)\).

Key Concepts

Factorization

When we talk about factorization in quadratic expressions like \(1600 - 16t^2\), we are referring to breaking down the expression into a product of simpler expressions. This process is particularly helpful in solving equations and understanding the roots or solutions of the equations. In this exercise, we want to factor \(1600 - 16t^2\).

The expression \(1600 - 16t^2\) is a "difference of squares". It can be rewritten using the identity \(a^2 - b^2 = (a+b)(a-b)\). Here, we recognize \(1600\) as \((40)^2\) and \(16t^2\) as \((4t)^2\). Therefore, \(1600 - 16t^2\) can be factored as:

• \((40 + 4t)(40 - 4t)\)

Factorization like this reveals that the solutions to the equation \(1600 - 16t^2 = 0\) are the values that make either \(40 + 4t = 0\) or \(40 - 4t = 0\). Recognizing patterns in factoring can greatly enhance your problem-solving skills in algebra.

Height Calculation

Calculating the height of an object in motion, like a bolt being dropped, involves substituting the time of fall into a given equation. In this example, the equation \(1600 - 16t^2\) describes the height \(h\) of the bolt after \(t\) seconds. This equation is a quadratic equation, where \(1600\) is the initial height at \(t = 0\) and \(-16t^2\) represents the effect of gravity over time.

To find the height of the bolt after a specific time, say 3 seconds, simply substitute \(t = 3\) into the equation:

• \(1600 - 16 imes 3^2 = 1600 - 144 = 1456\) feet.

For 7 seconds, substitute \(t = 7\):

• \(1600 - 16 imes 7^2 = 1600 - 784 = 816\) feet.

Such calculations illustrate how height changes over time due to gravity's constant acceleration downwards. Practicing height calculations helps build a strong foundation in both physics and algebra.

Time Estimation

Estimating when a dropped object, such as a bolt, will hit the ground is an important task in physics. Using the expression \(1600 - 16t^2\), which gives the height at time \(t\), we set the height to zero to find out when the object hits the ground.

To determine this, solve \(1600 - 16t^2 = 0\). Rearrange to find \(16t^2 = 1600\), then simplify to \(t^2 = 100\). Taking the square root, we get \(t = 10\) seconds. Hence, the bolt will take 10 seconds to reach the ground.

This exercise demonstrates how time estimation requires an understanding of quadratic equations. Recognizing when to apply these equations in real-world scenarios helps improve problem-solving skills.

Physics in Algebra

Physics and algebra often intertwine, especially in problems involving motion under gravity. The equation \(h = 1600 - 16t^2\) models the height of an object over time, incorporating concepts from both fields: initial position, gravitational acceleration, and time.

Key aspects of this intersection include:

• Understanding how quadratic equations describe parabolic motion (like a free-falling object).
• Recognizing that \(-16t^2\) represents half the gravitational acceleration, simplifying calculations of vertical motion.
• Factoring expressions to reveal time intervals at important positions, such as when reaching the ground.

By learning how physics is used in algebra, students can see the practical applications of these mathematical concepts, making them more tangible and easier to understand.