/*! 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 54 On a Saturday morning, pickup tr... [FREE SOLUTION] | 91Ó°ÊÓ

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On a Saturday morning, pickup trucks and sport utility vehicles carrying garbage to the town dump form a nearly steady procession on a country road, all traveling at \(19.7 \mathrm{m} / \mathrm{s} .\) From one direction, two trucks arrive at the dump every 3 min. A bicyclist is also traveling toward the dump, at \(4.47 \mathrm{m} / \mathrm{s}\). (a) With what frequency do the trucks pass him? (b) What If? A hill does not slow down the trucks, but makes the out-of-shape cyclist's speed drop to \(1.56 \mathrm{m} / \mathrm{s} .\) How often do noisy, smelly, inefficient, garbage-dripping, roadhogging trucks whiz past him now?

Short Answer

Expert verified
For the first part, a truck passes the cyclist about once every 23 minutes. For the second part, when the cyclist slows down to 1.56 m/s, a truck passes him about once every 32 minutes.

Step by step solution

01

Understand the concept of relative speed

The relative velocity of two moving objects in the same direction is the difference between their velocities. When the two objects are moving in the opposite direction, their relative velocity is the sum of their velocities. You would use the former for the case where both are traveling towards the dump and the latter for the case where the cyclist is heading away from the dump.
02

Calculate the relative speed for the first part

Two trucks arrive every 3 minutes, so they arrive with a frequency of \(2 \text{ trucks} / 180 \text{ s} = 0.0111 \text{ trucks/s} \). The relative speed of the cyclist and the truck is \(19.7 \text{ m/s} - 4.47 \text{ m/s} = 15.23 \text{ m/s}\). To find out how often the trucks pass the cyclist, divide the relative speed of the cyclist and the truck by the frequency at which the trucks pass the cyclist. This gives \( \frac{15.23 \text{ m/s}}{0.0111 \text{ trucks/s}} = 1371 \text{ s/truck} \) or about one truck every 23 minutes.
03

Calculate the relative speed for the second part

Now that the cyclist is heading away from the dump and his speed has dropped to \(1.56 \text{ m/s}\), use the sum of the velocities instead of their difference to calculate the relative speed. This gives a relative speed of \(19.7 \text{ m/s} + 1.56 \text{ m/s} = 21.26 \text{ m/s}\). Again, divide the relative speed of the cyclist and the truck by the frequency at which the trucks pass the cyclist. This results in \( \frac{21.26 \text{ m/s}}{0.0111 \text{ trucks/s}} = 1914 \text{ s/truck} \) or about one truck every 32 minutes.

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Key Concepts

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

Relative Speed Calculation
Understanding the relative speed calculation is crucial for solving problems where two objects are moving with respect to each other. In physics, when we talk about the relative speed, we're typically referring to the speed of one object as observed from another moving object.

For example, if you are traveling in a car and another car passes you, the speed at which this car seems to be moving past you is its relative speed to you. If both cars are moving in the same direction, the relative speed is calculated by subtracting the speed of your car from the speed of the passing car. However, if the cars are moving in opposite directions, you'd add the speeds to get the relative speed.

Let's look at an everyday scenario for better comprehension. Suppose two joggers are running in the same direction along a track. If one jogger runs at 6 meters per second and the other at 4 meters per second, the slower jogger would see the faster one moving away at a relative speed of 2 meters per second (6 m/s - 4 m/s). Conversely, if they run towards each other from opposite ends of the track, their relative speed would be 10 meters per second (6 m/s + 4 m/s).
Motion in One Dimension
Motion in one dimension, or linear motion, refers to movement in a straight line. This concept is foundational in physics as it helps us understand and predict the behavior of objects traveling along a single axis, like a car moving down a road or an athlete sprinting on a straight track.

In this context, we usually deal with variables such as displacement, velocity, acceleration, and time. Displacement tells us the change in position of an object, velocity refers to the speed of an object in a given direction, and acceleration is the change in velocity over time. Time, naturally, is the duration over which motion is observed.

When solving problems involving motion in one dimension, it's crucial to pay attention to reference points and direction of motion. Two objects moving with the same speed in opposite directions would have different velocities due to their contrary directional components. Always remember that velocity is a vector quantity, which means it has both magnitude and direction, whereas speed is just a scalar, having magnitude only.
Frequency and Period
Frequency and period are two fundamental concepts in physics tied to the cyclical nature of many phenomena. For instance, the rotation of the Earth, the orbit of a planet around the sun, or the back and forth swing of a pendulum. Frequency refers to how often something happens within a specified amount of time, while period is the duration it takes for one cycle to occur.

Consider a heart beating, the frequency of a heartbeat might be 60 beats per minute, meaning it beats once every second. Consequently, the period of the heartbeat is 1 second. These concepts are vital in studying not only mechanical systems but also electromagnetic waves, sound waves, and even quantum mechanics.

In terms of formula, the frequency (f) is the reciprocal of the period (T): \( f = \frac{1}{T} \). Therefore, if you're dealing with a periodic event and you are given the period, you can easily find the frequency, and vice versa. It's crucial to be comfortable with converting between these two, as they are commonly used in both theoretical and practical physics problems.

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Most popular questions from this chapter

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