Question

Use the following information. Blotting paper is a thick, soft paper used for absorbing fluids such as water or ink. The distance \(d\) (in centimeters) that tap water is absorbed up a strip of blotting paper at a temperature of \(28.4^{\circ} \mathrm{C}\) is given by the equation \(d=0.444 \sqrt{t}\) where \(t\) is the time (in seconds). Approximately how many minutes would it take for the water to travel a distance of 28 centimeters up the strip of blotting paper?

Short answer

It would take approximately 66.33 minutes for the water to travel a distance of 28 centimeters up the strip of blotting paper.

Step-by-step solution

Substitute the given values into the equation

Replace the variable \(d\) in the given equation with 28 to get \(28 = 0.444 \sqrt{t}\)

Solve the equation for \(t\)

First, divide both sides of the equation by 0.444 to isolate the square root on one side, giving us \(\sqrt{t} = 28 / 0.444 \approx 63.063\). Then square both sides to solve for \(t\), giving us \(t \approx (63.063)^2 \approx 3979.579\). This means that it would take approximately 3979.579 seconds for the distance to be 28cm.

Convert the time from seconds to minutes

Since there are 60 seconds in a minute, we can convert the time to minutes by dividing by 60. This gives \(t \approx 3979.579 / 60 \approx 66.326\). This means it would take approximately 66.33 minutes for the water to travel a distance of 28 centimeters up the strip of blotting paper.

Key concepts

Solving Square Roots

Solving square roots is an essential skill in algebra and finds various applications in real-world problems, including physics and engineering. The term 'square root' refers to a value that, when multiplied by itself, gives the original number. For instance, the square root of 4 is 2 because 2 times 2 equals 4. To solve square roots, there are often steps to follow.

Let's consider the equation in the exercise:
\[d = 0.444 \sqrt{t}\] To find the value of \(t\) when \(d\) is known, we must 'undo' the square root. This involves isolating the square root on one side by performing the inverse operations. In the provided exercise, you divided the distance \(d\) by the coefficient 0.444 to get the square root of \(t\) by itself. Next, to remove the squaring, you square both sides, effectively finding the value of \(t\) that, when square rooted and multiplied by 0.444, equals 28 cm. This step-by-step method of isolating the variable is a critical process in solving equations involving square roots.

Word Problems in Algebra

Word problems in algebra require translating real-world situations into mathematical equations. They often involve a bit of detective work to identify the known values and the relationships between them before forming an equation to solve. In word problems, identifying the right variables and correctly interpreting the context are as crucial as the mathematical computations themselves.

The blotting paper question is an example of such a problem. You are given a physical scenario and need to translate it into the equation \(d=0.444 \sqrt{t}\) where \(d\) stands for the distance traveled by the water along the blot paper and \(t\) represents time in seconds. To solve for the time it takes the water to travel 28 cm, one must understand the scenario, identify the information given (\(d\) and the relationship formula), and then translate this into a solvable algebraic equation.

Converting Units of Time

Often in algebra, and particularly in word problems, you will need to convert units to interpret your answer appropriately. Converting units of time is a common task. For example, we generally measure longer durations in minutes or hours instead of seconds.

In your solution, unit conversion from seconds to minutes was necessary to make the answer more comprehensible. As we know, 60 seconds make up one minute. To convert seconds into minutes, you divide the number of seconds by 60. This action is not just a mechanical step; it requires understanding that time is measured in multiple scales and units may need to be converted to present the answer in the most appropriate or requested format. In the exercise, understanding that 66.33 minutes is a more meaningful measure than approximately 3979.579 seconds for most people's perception of time was key.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operational symbols that represent mathematical relationships. In the exercise, \(d = 0.444 \sqrt{t}\) is an algebraic expression showing how distance \(d\) relates to time \(t\) through a given constant (0.444). The constant is based on the specific conditions of the blotting paper's absorption property.

Understanding how to manipulate these expressions is fundamental in algebra. It includes knowing how to isolate variables, combine like terms, and apply the correct operations to solve for an unknown value. Moreover, recognizing how constant factors, such as 0.444 in this case, affect the relationship between variables is important for understanding the problem's context and finding an accurate solution. These skills are not only useful in solving textbook exercises but are essential for analyzing and understanding patterns in various fields including science, economics, and beyond.