/*! 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 53 Blo Bacterial Motion Approximate... [FREE SOLUTION] | 91Ó°ÊÓ

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Blo Bacterial Motion Approximately 0.1\% of the bacteria in an adult human's intestines are Esclerichia coll. These bacteria have been observed to move with speeds up to \(15 \mathrm{mm} / \mathrm{s}\) and maximum accelerations of \(166 \mu \mathrm{m} / \mathrm{s}^{2}\). Suppose an \(E\). coli bacterium in your intestines starts at rest and accelerates at \(156 \mu \mathrm{m} / \mathrm{s}^{2}\). How much (a) time and (b) distance are required for the bacterium to reach a speed of \(12 \mu \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
Time: \(0.0769\) seconds; Distance: \(0.4615\) \(\mu m\).

Step by step solution

01

Understand the Problem

The problem asks us to determine how much time and distance it takes for an E. coli bacterium that starts from rest, and accelerates at a constant rate, to reach a specific speed. We can use the physics equations of motion for constant acceleration to solve this.
02

Identify Given Values

We know the following: initial velocity, \(u = 0\, \mu m/s\); acceleration, \(a = 156 \, \mu m/s^2\); final velocity, \(v = 12 \, \mu m/s\). We need to find time \(t\) and distance \(s\).
03

Use Velocity Equation to Find Time

The equation to use is: \(v = u + at\). Substitute the known values: \(12 = 0 + 156t\). Solve for \(t\):\[t = \frac{12}{156} = \frac{1}{13} \approx 0.0769 \text{ seconds}\]
04

Use Distance Equation to Find Distance

The equation to use is: \(s = ut + \frac{1}{2}at^2\). Since the initial velocity \(u = 0\), we have:\[s = \frac{1}{2} \times 156 \times \left(\frac{1}{13}\right)^2\]\[s = \frac{1}{2} \times 156 \times \frac{1}{169} \approx \frac{78}{169} \approx 0.4615 \, \mu m\]
05

Conclusion

We've calculated that the time required for the bacterium to reach the speed of \(12 \, \mu m/s\) is approximately \(0.0769\) seconds, and the distance traveled during this acceleration is approximately \(0.4615 \, \mu m\).

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Key Concepts

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

Constant Acceleration
When we talk about constant acceleration, we're referring to a situation where a body's velocity changes by the same amount every second. It's like pushing a toy car on a uniform surface – if the car speeds up by the same amount each second, it's under constant acceleration.
In our case with the Escherichia coli, this bacterium accelerates from rest at a constant rate of \(156 \, \mu m/s^2\). This means that every second, the bacterium's velocity increases by \(156 \, \mu m/s\).
Constant acceleration is fundamental in many natural and artificial systems, from speeding cars to falling raindrops. By understanding this concept, we can predict how objects move under various forces.
Equations of Motion
Equations of motion are crucial tools in physics that help describe the movement of objects under constant acceleration. They allow us to find various parameters like displacement, velocity, and time.
For our Escherichia coli problem, we used two key equations:
  • Velocity Equation: \(v = u + at\)
  • Distance Equation: \(s = ut + \frac{1}{2}at^2\)
These equations help us solve for different unknowns. In the exercise, we used the velocity equation to find the time taken to achieve a certain speed. Meanwhile, the distance equation gave us how far the bacterium travels during that time.
Understanding these equations is vital in predicting and analyzing motion in a constant acceleration framework.
Escherichia coli
Escherichia coli, often abbreviated as E. coli, is a type of bacteria that's commonly found in the intestines of humans and animals. Although most strains are harmless, some can cause illness.
For our physics problem, we focus on E. coli's unique ability for motion. Surprisingly, these tiny organisms can propel themselves at speeds up to \(15 \, mm/s\) and display impressive acceleration capabilities.
The study of bacterial motion, like that of E. coli, sheds light on biological processes and can inspire technological innovations, such as microscopic machines that mimic their swimming movements.
Velocity and Time Relationship
In physics, the relationship between velocity and time is crucial for understanding motion. This is particularly true when dealing with acceleration, as velocity changes over time.
With our friendly E. coli, we observe how its velocity increases from rest to \(12 \, \mu m/s\) over a certain time under constant acceleration. Here, the velocity at any given moment is determined by the initial velocity, the rate of acceleration, and time.
The velocity equation \(v = u + at\) embodies this relationship clearly. By knowing the initial velocity (in this case, \(0\)), acceleration, and final velocity, we can precisely determine how long it will take for the bacterium to reach its desired speed.
Understanding this relationship helps us predict movement in everyday life, from planning trips to assessing how fast muscles must contract for specific motions.

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Most popular questions from this chapter

You drive in a straight line at \(20.0 \mathrm{m} / \mathrm{s}\) for \(10.0 \mathrm{minutes}\), then at \(30.0 \mathrm{m} / \mathrm{s}\) for another \(10.0 \mathrm{minutes}\). (a) Is your average speed 25.0 \(\mathrm{m} / \mathrm{s},\) more than \(25.0 \mathrm{m} / \mathrm{s},\) or less than \(25.0 \mathrm{m} / \mathrm{s}\) ? Explain. (b) Verify your answer to part (a) by calculating the average speed.

A model rocket rises with constant acceleration to a height of \(3.2 \mathrm{m},\) at which point its speed is \(26.0 \mathrm{m} / \mathrm{s} .\) (a) How much time does it take for the rocket to reach this height? (b) What was the magnitude of the rocket's acceleration? (c) Find the height and speed of the rocket \(0.10 \mathrm{s}\) after launch.

An Extraterrestrial Volcano The first active volcano observed outside the Earth was discovered in 1979 on lo, one of the moons of Jupiter. The volcano was observed to be ejecting material to a height of about \(2.00 \times 10^{5} \mathrm{m}\). Given that the acceleration of gravity on lo is \(1.80 \mathrm{m} / \mathrm{s}^{2}\), find the initial velocity of the ciected material.

At the starting gun, a runner accelerates at \(1.9 \mathrm{m} / \mathrm{s}^{2}\) for \(5.2 \mathrm{s}\). The runner's acceleration is zero for the rest of the race. What is the speed of the runner (a) at \(t=2.0 \mathrm{s},\) and (b) at the end of the race?

In heavy rush-hour traffic you drive in a straight line at \(12 \mathrm{m} / \mathrm{s}\) for 1.5 minutes, then you have to stop for 3.5 minutes, and finally you drive at \(15 \mathrm{m} / \mathrm{s}\) for another \(2.5 \mathrm{minutes}\). (a) Plot a positionversus-time graph for this motion. Your plot should extend from \(t=0\) to \(t=7.5 \mathrm{minutes}\) (b) Use your plot from part (a) to calculate the average velocity between \(t=0\) and \(t=7.5\) minutes.

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