Problem 58 - College Physics

Question

BIO Bacterial motion. A bacterium using its flagellum as propulsion can move through liquids at a rate of 0.003 \(\mathrm{m} / \mathrm{s} .\) For a \(50 \mu \mathrm{m}-\) long bacterium, that is the equivalent of 60 cell lengths per second. Bacteria of that size have a mass of approximately \(1 \times 10^{-12} \mathrm{g}\). The viscous drag on a swimming bacterium is so great that if it stops beating its flagellum it will stop within a distance of 0.01 \(\mathrm{nm} .\) What is the acceleration that stops the bacterium? A. \(1.2 \times 10^{4} \mathrm{m} / \mathrm{s}^{2}\) B. \(5 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\) C. \(6 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\) D. \(9 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}\)

Step-by-Step Solution

Answer

Option A: \(1.2 \times 10^{4} \, \mathrm{m/s^2}\).

Step 1: Understand the Problem

We need to find the acceleration that stops the bacterium when it ceases to move its flagellum. This involves using its initial speed and the distance over which it stops.

Step 2: Identify Known Values

Initially, the bacterium moves at a speed of \(0.003 \, \mathrm{m/s}\). The distance it moves before stopping is \(0.01 \, \mathrm{nm}\), which is equivalent to \(0.01 \times 10^{-9} \, \mathrm{m}\).

Step 3: Apply the Kinematic Equation

We can use the equation \(v^2 = u^2 + 2as\), where \(v\) is the final velocity (\(0 \, \mathrm{m/s}\)), \(u\) is the initial velocity (\(0.003 \, \mathrm{m/s}\)), \(a\) is the acceleration, and \(s\) is the stopping distance (\(0.01 \times 10^{-9} \, \mathrm{m}\)).

Step 4: Rearrange to Solve for Acceleration

Rearranging the equation for acceleration gives us \(a = \frac{v^2 - u^2}{2s}\). Substituting the known values: \(v = 0\), \(u = 0.003\), and \(s = 0.01 \times 10^{-9} \, \mathrm{m}\).

Step 5: Calculate Acceleration

Plugging the values into the equation, \(a = \frac{0 - (0.003)^2}{2 \times 0.01 \times 10^{-9}}\), we find \(a = \frac{-0.000009}{2 \times 0.01 \times 10^{-9}} = 1.2 \times 10^{4} \, \mathrm{m/s^2}\).

Step 6: Select the Correct Answer

The calculated acceleration is \(1.2 \times 10^{4} \, \mathrm{m/s^2}\), which corresponds to option A in the problem statement.

Key Concepts

Flagellum Propulsion

Bacteria, although minuscule, possess the remarkable ability to navigate their liquid environments efficiently. This movement is primarily due to their flagellum - a whip-like appendage that propels them forward. Imagine the flagellum as a tiny propeller; it twists and turns, generating thrust that moves the bacterial cell through its watery surroundings.
In essence, flagellum propulsion operates on the principle similar to a boat propeller, just on a much smaller scale. The motion is not just about linear propulsion but involves complex rotations that create a helical wave, pushing the bacterium ahead. This form of travel is vital for bacteria to reach nutrients, survive in adverse conditions, or find new environments.
This relentless rotation and propulsion ensure bacteria can achieve a surprising mobility despite their small size and the thick, syrupy medium they swim through. Understanding flagellum propulsion opens up insights into how even the smallest organisms harness incredible natural engineering for survival.

Viscous Drag

Viscous drag is a crucial force to consider when understanding bacterial movement. Because bacteria live in liquid environments where they constantly encounter resistance, this is where viscous drag comes into play.
Viscous drag refers to the resistance faced by an object moving through a fluid. It's like trying to swim through honey; the thicker the liquid, the more effort is needed to move through it. For bacteria, the fluid surrounding them is their lifeline, but also their greatest challenge.
This resistance is so pronounced that if the bacterium's flagellum stops moving, it halts almost instantaneously. The viscous properties of their environment do not allow them to glide to a stop over a long distance as a car would on a highway. Rather, they come to an abrupt stop due to the high drag, making their propulsion all the more critical to maintain momentum.

Kinematic Equation

Calculating the acceleration needed to stop a bacterium when it stops moving its flagellum involves some intriguing physics. This brings us to the use of the kinematic equation, specifically focusing on constant acceleration, even at the microscopic level.
The equation used is often written as: \[v^2 = u^2 + 2as\]where:

• \(v\) is the final velocity (which is zero when the bacterium stops),
• \(u\) is the initial velocity,
• \(a\) is the acceleration,
• \(s\) is the displacement during stopping.

By rearranging this formula, we can solve for acceleration \(a\) as shown in the exercise. This equation is a basic tool in physics that helps us understand motion not just on our scale, but microscopically for tiny organisms, affirming that physics principles apply uniformly across scales.

Microorganism Dynamics

Understanding how microorganisms like bacteria move and stop sparks appreciation for the complexity hidden in simplicity. The dynamics of microorganisms involve several intertwined factors—physical, biological, and environmental—that influence their movement.
Microorganism dynamics not only refer to their locomotion but also include how they interact with their environment. Factors like temperature, fluid viscosity, and chemical gradients can alter how effectively a bacterium propels itself with its flagellum.
The dynamics investigated in these exercises show how critical it is for bacteria to adapt quickly. Since they often dwell in richly viscous environments, their ability to respond promptly to changing conditions through rapid acceleration or halting quickly showcases nature's evolutionary problem-solving. These dynamics are a testament to the intricate design and efficiency found in microscopic life.