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Chapter 2: Problem 91

A fast-food restaurant uses a conveyor belt to send the burgers through a grilling machine. If the grilling machine is \(1.1 \mathrm{~m}\) long and the burgers require \(2.5 \mathrm{~min}\) to cook, how fast must the conveyor belt travel? If the burgers are spaced \(15 \mathrm{~cm}\) apart, what is the rate of burger production (in burgers/min)?

Short Answer

Expert verified
The conveyor belt must travel at 0.44 m/min and the burger production rate is approximately 2.93 burgers/min.

Step by step solution

01

Understand the Problem

The problem asks for two things: the speed of the conveyor belt required for the burgers to take exactly 2.5 minutes to travel 1.1 meters, and the rate of burger production if burgers are 15 cm apart. We need to find the conveyor speed in meters per minute first.
02

Calculate Conveyor Speed

To find the speed of the conveyor belt, we use the formula: Speed = Distance / Time. Here, the distance the conveyor must cover is 1.1 meters, and the time is 2.5 minutes. Therefore: \[ Speed = \frac{1.1 \text{ meters}}{2.5 \text{ minutes}} = 0.44 \text{ meters/minute} \] The conveyor belt needs to travel at 0.44 meters per minute.
03

Convert Burger Spacing to Meters

Before we calculate the burger production rate, convert the spacing between burgers from centimeters to meters. 15 cm is equal to 0.15 meters.
04

Calculate Burger Production Rate

The production rate is calculated by dividing the speed of the conveyor belt by the distance between the burgers. The conveyor belt moves 0.44 meters per minute and the burgers are spaced 0.15 meters apart. Therefore: \[ \text{Rate of Burger Production} = \frac{0.44 \text{ meters/minute}}{0.15 \text{ meters/burger}} \approx 2.93 \text{ burgers/minute} \] Thus, the restaurant produces approximately 2.93 burgers per minute.

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Key Concepts

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

Burger Production Rate
The burger production rate is an essential metric for any fast-food restaurant as it determines how efficiently the kitchen operates. It tells us how many burgers can be produced within a certain time frame, which in this case is one minute. Calculating the rate involves understanding the speed at which the conveyor belt moves and the spacing between each burger. Once we know the speed of the conveyor belt, we can find how many burgers pass through the grilling machine per minute by dividing this speed by the space between the burgers.
  • Conveyor Speed: 0.44 meters/minute
  • Burger Spacing: 0.15 meters
  • Production Rate: Approximately 2.93 burgers/minute
This calculation helps optimize production and ensure the kitchen can meet customer demand efficiently.
Grilling Machine
The grilling machine in a fast-food restaurant is used to consistently cook the burgers to the desired level. In this scenario, its length is an important factor since it determines how long the burgers will cook, given the speed of the conveyor belt. The grilling machine is 1.1 meters long, and the burgers need 2.5 minutes to pass through it completely. This setup requires precise speed control of the conveyor belt to ensure each burger cooks for the exact duration. Proper timing ensures uniform cooking, maintaining constant quality standards for every burger served.
Physics of Motion
Understanding the physics of motion is crucial when determining how a conveyor belt needs to operate to deliver consistent results. The conveyor belt鈥檚 movement is a clear example of linear motion, and calculating its speed involves using basic physics equations. The formula used is:\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]Here, we see that knowing the distance and the time it takes to cover that distance is key to finding how fast the conveyor belt needs to move. This speed directly affects how quickly the burgers move through the grilling machine, illustrating a direct real-world application of physics in everyday situations.
Unit Conversion in Physics
Unit conversion is part of physics that ensures all calculations are consistent and accurate. In this exercise, the spacing between the burgers was initially given in centimeters, which needed to be converted to meters to match the unit of the conveyor speed. Conversion is achieved using the factor of 1 meter = 100 centimeters, so:
  • 15 centimeters = 0.15 meters
This conversion is essential because, without consistent units, calculations can become incorrect. Physics often involves multiple unit types, and proficiency in converting them ensures correctness and precision in solving real-life problems.

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Most popular questions from this chapter

(I) If you are driving \(110 \mathrm{~km} / \mathrm{h}\) along a straight road and you look to the side for \(2.0 \mathrm{~s}\), how far do you travel during this inattentive period?

(II) A car slows down uniformly from a speed of \(18.0 \mathrm{~m} / \mathrm{s}\) to rest in 5.00 s. How far did it travel in that time?

(I) A particle at \(t_{1}=-2.0 \mathrm{~s}\) is at \(x_{1}=4.3 \mathrm{~cm}\) and at \(t_{2}=4.5 \mathrm{~s}\) is at \(x_{2}=8.5 \mathrm{~cm} .\) What is its average velocity? Can you calculate its average speed from these data?

(II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about \(0.28 \mu \mathrm{m} .\) A CD player's readout laser scans along the spiral's sequence of bits at a constant speed of about \(1.2 \mathrm{~m} / \mathrm{s}\) as the CD spins. (a) Determine the number \(N\) of digital bits that a CD player reads every second. (b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits and so one would (at first glance) think the required bit rate for a CD player is \(N_{0}=2\left(44,100 \frac{\text { samplings }}{\text { second }}\right)\left(16 \frac{\text { bits }}{\text { sampling }}\right)=1.4 \times 10^{6} \frac{\text { bits }}{\text { second }}\) where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that \(N_{0}\) is less than the number \(N\) of bits actually read per second by a CD player. The excess number of bits \(\left(=N-N_{0}\right)\) is needed for encoding and error-correction. What percentage of the bits on a \(\mathrm{CD}\) are dedicated to encoding and error-correction?

(II) The Table below gives the speed of a particular drag racer as a function of time. (a) Calculate the average acceleration \(\left(\mathrm{m} / \mathrm{s}^{2}\right)\) during each time interval. \((b)\) Using numerical integration (see Section \(2-9\) ) estimate the total distance traveled \((\mathrm{m})\) as a function of time. [Hint: for \(\bar{v}\) in each interval sum the velocities at the beginning and end of the interval and divide by \(2 ;\) for example, in the second interval use \(\bar{v}=(6.0+13.2) / 2=9.6]\) (c) Graph each of these. $$ \begin{array}{lccccccccccc} \hline t(\mathrm{~s}) & 0 & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 & 4.50 & 5.00 \\ v(\mathrm{~km} / \mathrm{h}) & 0.0 & 6.0 & 13.2 & 22.3 & 32.2 & 43.0 & 53.5 & 62.6 & 70.6 & 78.4 & 85.1 \\ \hline \end{array} $$
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