91Ó°ÊÓ

In a psychological study, it was determined that the proper length \(L\) of the letters of a word printed on pavement is given by $$ L=\frac{0.000169 d^{2.27}}{h} $$ where \(d\) is the distance of a car from the lettering and \(h\) is the height of the eye above the surface of the road. All units are in meters. This formula says that if a person is \(h\) meters above the surface of the road and is to be able to recognize a message \(d\) meters away, that message will be the most recognizable if the length of the letters is \(L .\) Find \(L\) to the nearest tenth of a meter, given \(d\) and \(h\) a) \(h=1 \mathrm{m}, d=60 \mathrm{m}\) b) \(h=0.9906 \mathrm{m}, d=75 \mathrm{m}\) c) \(h=2.4 \mathrm{m}, d=80 \mathrm{m}\) d) \(h=1.1 \mathrm{m}, d=100 \mathrm{m}\)

Step by step solution

01

Understand the formula

The formula to find the proper length of the letters is given by: \[ L = \frac{0.000169 d^{2.27}}{h} \]. This means we'll plug in the values of \(d\) and \(h\) given in each part of the question to find \(L\).

02

Part (a): Plug in values for part (a)

For part (a), \(h = 1\) meter and \(d = 60\) meters. Substitute these values into the formula: \[ L = \frac{0.000169 (60)^{2.27}}{1} \].

03

Calculate the power and product for part (a)

\(60^{2.27} \approx 27658.7\). Now multiply by 0.000169: \[ L = 0.000169 \times 27658.7 \approx 4.6733 \].

04

Round to the nearest tenth for part (a)

The length of the letters \(L\) is approximately 4.7 meters.

05

Part (b): Plug in values for part (b)

For part (b), \(h = 0.9906\) meters and \(d = 75\) meters. Substitute these values into the formula: \[ L = \frac{0.000169 (75)^{2.27}}{0.9906} \].

06

Calculate the power and product for part (b)

\(75^{2.27} \approx 52401.3\). Now multiply by 0.000169: \[ L = 0.000169 \times 52401.3 \approx 8.8578 \]. Divide by 0.9906: \[ L \approx 8.939 \].

07

Round to the nearest tenth for part (b)

The length of the letters \(L\) is approximately 8.9 meters.

08

Part (c): Plug in values for part (c)

For part (c), \(h = 2.4\) meters and \(d = 80\) meters. Substitute these values into the formula: \[ L = \frac{0.000169 (80)^{2.27}}{2.4} \].

09

Calculate the power and product for part (c)

\(80^{2.27} \approx 78732.3\). Now multiply by 0.000169: \[ L = 0.000169 \times 78732.3 \approx 13.308 \]. Divide by 2.4: \[ L \approx 5.545 \].

10

Round to the nearest tenth for part (c)

The length of the letters \(L\) is approximately 5.5 meters.

11

Part (d): Plug in values for part (d)

For part (d), \(h = 1.1\) meters and \(d = 100\) meters. Substitute these values into the formula: \[ L = \frac{0.000169 (100)^{2.27}}{1.1} \].

12

Calculate the power and product for part (d)

\(100^{2.27} \approx 146779.9\). Now multiply by 0.000169: \[ L = 0.000169 \times 146779.9 \approx 24.799 \]. Divide by 1.1: \[ L \approx 22.545 \].

13

Round to the nearest tenth for part (d)

The length of the letters \(L\) is approximately 22.5 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

lettering recognition distance

To grasp the concept of 'lettering recognition distance,' we need to understand how the visibility of letters on a pavement is influenced by the distance of the viewer. In our exercise, the formula used to determine the optimal letter length is: \[ L = \frac{0.000169 d^{2.27}}{h} \] where \(d\) is the distance (in meters) from the lettering, and \(h\) is the height of the person's eyes above the road surface. This formula ensures the letters are large enough for someone at a certain distance to recognize them clearly.

• **Distance \(d\)**: This specifies how far the person is from the letters on the pavement. For example, in part (a), \(d = 60\) meters.
• **Height \(h\)**: This represents the height of the eyes from the ground. For instance, in part (a), \(h = 1\) meter.
• **Letter Length \(L\)**: This is the optimal length of the letters for them to be recognizable at the given distance and height.

With these parameters, you can predict the appropriate letter size for different distances and eye heights, enhancing the overall readability.

calculation of powers

The calculation of powers is a critical step in using the formula to find the letter length. When using exponents, the base number is raised to the power specified. In simple terms, an exponent indicates how many times you multiply the base by itself. Take for example the power calculation in step 6: Calculate \(75^{2.27}\).

• **Base (75):** The number you are raising to a power.
• **Exponent (2.27):** Indicates how many times the base is multiplied by itself.
• **Result:** Using a calculator, \(75^{2.27} \approx 52401.3\). This is an exact function of exponential calculation.

Without understanding how to perform these power calculations, applying the formula accurately would be very difficult. By mastering this step, you ensure you can handle the essential calculations needed for problem-solving in algebra.

unit conversions

Unit conversions are essential in ensuring that the formula's inputs are in the correct units. In this problem, it is crucial all measurements are in meters to maintain the integrity of the formula: \[ L = \frac{0.000169 d^{2.27}}{h} \]

• Ensure all distances (\(d\)) are measured in meters.
• Ensure that the height (\(h\)) is also measured in meters.

Thankfully, in the given exercise, all units were already in meters. Remember, if they were not, you would need to convert them to meters to use the formula correctly. For instance, if distances were given in kilometers or inches, you'd need to convert them to meters first before plugging them into the formula. Consistency in units ensures accuracy and helps avoid errors in calculations.

problem-solving with formulas

Problem-solving with formulas involves using the given mathematical models to find unknown values. In our problem, the formula to derive the letter length (\(L\)) based on visibility relates to the viewer's distance and eye height: \[ L = \frac{0.000169 d^{2.27}}{h} \] Steps to solve these problems typically involve:

• **Identify given variables:** Known values for \(d\) and \(h\).
• **Plug in values:** Substitute \(d\) and \(h\) into the formula.
• **Calculate exponents:** Compute the power of \(d\), for instance, \(75^{2.27}\).
• **Final calculation:** Multiply by 0.000169 and divide by \(h\) to find \(L\).
• **Round the result:** Round the result to the nearest tenth for practical use.

By following these steps, you can systematically arrive at the right answer. Moreover, being methodical and thorough ensures you do not miss any critical aspects of the calculation. The more you practice, the more proficient you'll become at using formulas to solve problems.