Problem 77 - A Graphical Approach to Precalculus with Limits

Question

The velocity \(v\) of a meteorite approaching Earth is given by $$v=\frac{k}{\sqrt{d}}$$ measured in kilometers per second, where \(d\) is its distance from the center of Earth and \(k\) is a constant. If \(k=350\) what is the velocity of a meteorite that is 6000 kilometers away from the center of Earth? Round to the nearest tenth.

Step-by-Step Solution

Answer

The velocity is approximately 4.5 km/s.

Step 1: Identify Given Information

We are given the formula \(v=\frac{k}{\sqrt{d}}\), where \(k=350\) and \(d=6000\). Our task is to find the velocity \(v\).

Step 2: Plug Values into Formula

Substitute \(k=350\) and \(d=6000\) into the formula: \(v=\frac{350}{\sqrt{6000}}\).

Step 3: Calculate the Square Root

Find \(\sqrt{6000}\). Approximate the square root using a calculator to get \(\sqrt{6000} \approx 77.46\).

Step 4: Compute the Velocity

Now calculate \(v\) by substituting the square root value into the equation: \(v = \frac{350}{77.46} \approx 4.5\).

Step 5: Round the Result

Since we need to round to the nearest tenth, we observe the value 4.5, which is already at the nearest tenth.

Key Concepts

Velocity

Velocity is a term frequently used in physics and everyday life to describe the rate at which an object changes its position. In simpler terms, it is the speed of something in a given direction. In our meteorite exercise, the velocity formula, \(v = \frac{k}{\sqrt{d}}\), provides a way to calculate the speed of a meteorite as it approaches Earth.

• Constant \( k \): Represents a fixed numerical value in the equation that would differ based on different contexts or conditions such as the gravitational effect.
• Distance \( d \): The distance from the meteorite to the center of the Earth which impacts how quickly the object might travel.
• Velocity \( v \): The resulting calculated speed at which the meteorite travels towards Earth.

Understanding velocity is crucial not only for astrophysical phenomena but also for mundane experiences like traveling by car or playing sports.

Square Root

The square root function allows us to find a number that, when multiplied by itself, gives the original number. Notated with the radical symbol \( \sqrt{} \), it's a fundamental concept in mathematics, especially prevalent in physics and engineering.

• The square root of \(6000\) (\(\sqrt{6000}\)) is crucial for computing the meteorite's velocity as shown in the solution.
• Approximating \(\sqrt{6000}\) using a calculator gives \(\approx 77.46\).

Square roots can at times seem complex, but breaking down their usage into steps can simplify their practical application. For example, breaking a number into its prime factors or estimating within a range of known perfect squares can aid in manual calculation when calculators are not available.

Rounding

Rounding numbers make calculations and results easier to understand and use. Especially in measurement, precision can become cumbersome, so simplifying a number to a desired place value such as the nearest tenth, hundredth, or whole number is often preferred.

• In our problem, the result after computing the velocity was \(4.515\) (from a calculator not visible in the steps), which was then rounded to \(4.5\).
• When rounding, observe the first digit that you cut off. Here, since \(5\) is equal to or greater than five, the preceding digit, \(1\), becomes \(2\).
• The rest of the digits are not shown as they have been omitted in the final answer which is approximated to \(4.5\).

Rounding is an essential skill, especially when dealing with measurements in physics and chemistry, or financial calculations where cent-level precision is ultimately impractical for reporting larger sums of money.

Distance from Earth

In physics problems involving celestial bodies, knowing an object's distance from a reference point (like Earth) is key to understanding its dynamics such as velocity or gravitational pull.

• Distance \( d \): The problem gives the distance of meteorite as \(6000\) kilometers from the Earth's center.
• This distance affects calculations of velocity, as shown in the formula \(v = \frac{k}{\sqrt{d}}\).
• Distance is crucial for both relative measurements and understanding the scale in astrophysics.

Having a solid understanding of distances, especially in large-scale astronomy, helps us appreciate gravitational effects, travel time for space-bound objects, and the magnitude of the universe's vastness. Recognizing these distances enhances not only scientific calculations but also interpretations of our position within the cosmos.