/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 A person sees a lightning bolt p... [FREE SOLUTION] | 91Ó°ÊÓ

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A person sees a lightning bolt pass close to an airplane that is flying in the distance. The person hears thunder \(5.0 \mathrm{~s}\) after seeing the bolt and sees the airplane overhead \(10 \mathrm{~s}\) after hearing the thunder. The speed of sound in air is \(1100 \mathrm{ft} / \mathrm{s}\). (a) Find the distance of the airplane from the person at the instant of the bolt. (Neglect the time it takes the light to travel from the bolt to the eye.) (b) Assuming the plane travels with a constant speed toward the person, find the velocity of the airplanc. (c) Look up the speed of light in air and defend the approximation used in part (a).

Short Answer

Expert verified
a) The airplane was 5500 feet from the person at the moment of the lightning. b) The airplane velocity is 367 feet per second. c) Given the speed of light in air, it is reasonable to neglect the time it takes for light to travel from the bolt to the eye.

Step by step solution

01

Calculate the Distance

The person hears the thunder \(5.0 \mathrm{~s}\) after they see the lightning. This delay is caused by the time it takes for the sound of the thunder to travel the distance from the lightening to the person. We can calculate this distance (which is the distance from the person to the plane at the moment of the lightning) using the formula for the speed of sound. The speed of sound in air is \(1100 \mathrm{ft} / \mathrm{s}\), so multiply the speed of sound by the time delay to get the distance. \[Distance = Speed \times Time = 1100 \ \mathrm{ft/s} \times 5.0 \ \mathrm{s} = 5500 \ \mathrm{ft}\]
02

Calculate the Velocity

The plane covers the distance calculated in Step 1 in \(5.0 \mathrm{~s} + 10.0 \mathrm{~s} = 15.0 \mathrm{~s}\). The velocity is then the distance covered divided by this its time to cover that distance. \[Velocity = \frac{Distance}{Time} = \frac{5500 \ \mathrm{ft}}{15.0 \ \mathrm{s}} = 367 \ \mathrm{ft/s} \]
03

Analyze the Approximation

The speed of light in air is roughly \(299792458 \ \mathrm{m/s}\) or \(983571088.9 \ \mathrm{ft/s}\), which is many orders of magnitude greater than the speed of sound (\(1100 \ \mathrm{ft/s}\)). Given this vast difference, it is reasonable to neglect the time it takes for the light from the lightning bolt to reach the person as compared to the time taken for the sound of the thunder. The time for light to travel the same distance as sound is essentially instantaneous on this scale, hence the approximation used in part (a) is justified.

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Key Concepts

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

Speed of Sound
In this exercise, the speed of sound is a crucial part of determining how far the airplane is from an observer. Typically, the speed of sound in dry air at 20 degrees Celsius is about 343 meters per second or roughly 1125 feet per second. However, for this specific exercise, we've rounded it to 1100 feet per second. Variations might happen due to humidity or temperature changes.
When you hear thunder after seeing a lightning flash, it's because the sound takes longer to travel. Sound waves move much slower than light waves, so there's often a noticeable delay. By using the speed of sound, you can calculate the distance of an object based on how long it takes for sound to travel from it to you. Use the formula:
  • Distance = Speed of Sound imes Time
So, if you know the speed and how long the sound took to reach you, you can deduce the distance. In this case, multiplying 1100 ft/s by 5.0 seconds gives a distance of 5500 feet.
Speed of Light
The speed of light is incredibly fast compared to sound. In air, light typically travels at around 983,571,088.9 feet per second. This speed is so great that for short distances, like seeing a light from a lightning strike, it's essentially instantaneous to the human eye.
Because of this, when calculating the distance to something using sound, you can usually ignore the time it takes for light to travel over the same distance. Unlike sound, light waves don't slow significantly when passing through air, making them far superior for instantaneous signal detection. For our exercise, this means we can focus on the sound delay without worrying about the light delay. That explains why we haven't included the speed of light in those calculations. Its contribution to the delay is practically meaningless in human experience in situations like these.
Distance and Velocity Calculations
Understanding how to calculate distance and velocity is fundamental in physics. Distance is how far an object has moved, calculated with formulas involving speed and time. Velocity, however, not only considers speed but also the direction.In our exercise:
  • First, we calculated the distance from the observer to the plane at the time of lighting using the speed of sound and time delay.
  • Next, the plane's velocity was determined by dividing the distance by the time taken (15 seconds, here).
The formula for velocity is:
  • Velocity = \( \frac{Distance}{Time} \)
By applying this method, we found the plane's velocity to be approximately 367 ft/s. These calculations demonstrate how closely related understanding of distance, speed, and time is when solving physics problems.

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Most popular questions from this chapter

Two boats start together and race across a \(60-\mathrm{km}\) -wide lake and back. Boat A goes across at \(60 \mathrm{~km} / \mathrm{h}\) and returns at \(60 \mathrm{~km} / \mathrm{h}\). Boat \(\mathrm{B}\) goes across at \(30 \mathrm{~km} / \mathrm{h}\), and its crew, realizing how far behind it is getting, returns at \(90 \mathrm{~km} / \mathrm{h}\). Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?

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