Problem 59 - Algebra 1

Question

P. BioLoGY Many birds drop clams or other shellfish in order to break the shell and get the food inside. The time \(t\) (in seconds) it takes for an object such as a clam to fall a certain distance \(d\) (in feet) is given by the equation $$ t=\frac{\sqrt{d}}{4} $$ A gull drops a clam from a height of 50 feet. A second gull drops a clam from a height of 32 feet. Find the difference in the times that it takes for the clams to reach the ground. Round your answer to the nearest hundredth.

Step-by-Step Solution

Answer

The difference in the times that it takes for the clams to reach the ground is approximately \(\Delta t = 0.44\) seconds.

Step 1: Calculate the fall time for the first clam

Insert the first height \(d = 50\) feet into the formula \(t=\frac{\sqrt{d}}{4}\) which results in \(t_1=\frac{\sqrt{50}}{4}\).

Step 2: Calculate the fall time for the second clam

Insert the second height \(d = 32\) feet into the formula \(t=\frac{\sqrt{d}}{4}\) which results in \(t_2=\frac{\sqrt{32}}{4}\).

Step 3: Calculate the difference

Subtract the second time from the first time to get the difference: \(\Delta t = t_1 - t_2 = \frac{\sqrt{50}}{4} - \frac{\sqrt{32}}{4}\)

Step 4: Simplify the difference

Now simplify the expression and round to the nearest hundredth to obtain the final answer.

Key Concepts

Equations

In algebra, equations are like puzzles that represent statements of equality between two expressions. They are essential in expressing relationships between variables. Equations such as the one in the exercise, allow us to calculate unknown values easily. For example, the equation is given as: \[ t = \frac{\sqrt{d}}{4} \]Here, \( t \) represents time, and \( d \) represents distance. By substituting different values of \( d \), we can determine the time \( t \) it takes for an object to fall a certain distance. Equations enable us to solve real-world problems by giving us a way to quantify and manipulate relationships.

Square Roots

Square roots are a fundamental concept in algebra, represented by the symbol \( \sqrt{} \). When you see \( \sqrt{d} \), it means you are looking for a number which, when multiplied by itself, gives you \( d \). For instance, \( \sqrt{4} = 2 \) because \( 2 \times 2 = 4 \).

• Square roots simplify expressions involving distance in physics as seen in the example.
• They help in solving equations where a variable is squared by "undoing" the square.

In our problem, we compute square roots of 50 and 32 to find times. This demonstrates how manipulating square roots is necessary to find concrete numeric solutions.

Problem Solving

Problem-solving in mathematics often involves breaking down complex questions into simpler parts. Our goal is to understand what the problem is asking and how to use given formulas to find the solution. With the clam-dropping exercise:

• Identify what you need to find: the different times it takes for each clam to reach the ground.
• Use the provided formula appropriately by carefully inserting the known values.
• Calculate each part step-by-step, ensuring rounding is accurate and as specified, to arrive at the correct final answer.

Whether it's clams or other situations, problem-solving is a systematic approach to finding answers, keeping computations clear and organized.

Physics Applications

In physics, mathematical concepts often describe natural phenomena. The fall of objects, such as clams dropped by gulls, can be explained using physics equations dealing with gravity and motion. The exercise applies: \[ t = \frac{\sqrt{d}}{4} \]This formula relates time \( t \) with distance \( d \), indicating how far an object will fall under gravity over time.Physics applications:

• Make abstract concepts tangible, like measuring how fast a clam falls.
• Allow predictions and comparisons, such as calculating the difference in fall times.

By integrating physics with algebra, students can explore how variables like height impact an object's fall, enriching their understanding of how mathematics describes the physical world.