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(III) You dive straight down into a pool of water. You hit the water with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\), and your mass is \(75 \mathrm{~kg}\). Assuming a drag force of the form \(F_{\mathrm{D}}=-\left(1.00 \times 10^{4} \mathrm{~kg} / \mathrm{s}\right) v,\) how long does it take you to reach \(2 \%\) of your original speed? (Ignore any effects of buoyancy.)

Short Answer

Expert verified
The time to reach 2% of the initial speed is approximately 0.017 seconds.

Step by step solution

01

Understanding the Forces

When you dive into the water, the main force acting against your motion is the drag force, which is given by \( F_D = -k v \), where \( k = 1.00 \times 10^4 \text{ kg/s} \) is the drag coefficient and \( v \) is the velocity. This force slows you down over time.
02

Determine the Final Velocity

We need to determine the velocity when you slow down to \( 2\% \) of your initial velocity, which was \( 5.0 \text{ m/s} \). Therefore, the final velocity \( v_f \) is given by \( 0.02 \times 5.0 \text{ m/s} = 0.1 \text{ m/s} \).
03

Set Up the Differential Equation

Apply Newton's second law where the net force \( F = ma \) becomes \( m \frac{dv}{dt} = -kv \) because the only force acting is the drag force. This sets up the differential equation for the velocity \( \frac{dv}{dt} = -\frac{k}{m}v \).
04

Solve the Differential Equation

This linear differential equation can be solved using separation of variables. Rewriting, we get \( \frac{dv}{v} = -\frac{k}{m} dt \). Integrate both sides to obtain \( \ln|v| = -\frac{k}{m}t + C \). Exponentiating both sides, \( v(t) = v_0 e^{-\frac{k}{m} t} \), where \( v_0 = 5.0 \text{ m/s} \) is the initial velocity.
05

Substitute Known Values

Use the known values, \( v_0 = 5.0 \text{ m/s} \), \( v_f = 0.1 \text{ m/s} \), \( k = 1.00 \times 10^4 \text{ kg/s} \), and \( m = 75 \text{ kg} \). Substitute them into the equation \( v_f = v_0 e^{-\frac{k}{m} t} \) to find \( 0.1 = 5.0 e^{-\frac{1.00 \times 10^4}{75} t} \).
06

Solve for Time

Divide both sides of the equation by \( 5.0 \) to get \( 0.02 = e^{-\frac{1.00 \times 10^4}{75} t} \). Take the natural logarithm: \( \ln(0.02) = -\frac{1.00 \times 10^4}{75} t \). Solve for \( t \) to get \( t = -\frac{75}{1.00 \times 10^4} \ln(0.02) \). Calculate to get \( t \approx 0.017 \, \text{s} \).

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Key Concepts

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

Differential Equation
When tackling problems involving forces and motion, differential equations often come into play. In this scenario, as you dive into a pool, you face a force known as drag force. This force can vary according to how fast you are moving. Mathematically, we express this as a differential equation, which helps us understand how your velocity changes with time. The equation for this situation is \( \frac{dv}{dt} = -\frac{k}{m}v \), where \( v \) is velocity, \( k \) is the drag coefficient, and \( m \) is your mass. By studying differential equations like this one, you can discover how different parameters affect your motion.
  • The left side of the equation, \( \frac{dv}{dt} \), is the change in velocity over time, also known as acceleration.
  • The right side indicates how this change is proportionate to the velocity itself, multiplied by a factor \( -\frac{k}{m} \).
Solving this type of differential equation generally involves methods like separation of variables or integrating factors, leading to an exponential solution. Understanding these solutions allows you to predict future velocities.
Newton's Second Law
Newton's Second Law is fundamental to understanding motion. It states that the force on an object is equal to its mass multiplied by its acceleration \( F = ma \). In this problem, the force is a drag force that resists your motion as you dive into the water. This drag force is expressed as \( F_D = -kv \), making it dependent on the velocity. Here, the negative sign indicates that the force acts in the opposite direction to your movement.
  • When applying Newton's Second Law \( F = ma \), you substitute for the force, leading to \( ma = -kv \).
  • This gives rise to the differential equation \( m \frac{dv}{dt} = -kv \).
By setting up the equation in this way, you can explore how an object's speed changes over time due to applied forces, and ultimately solve for the time required for certain velocity changes.
Velocity
Velocity plays a critical role in analyzing motion problems like this. Initial velocity is what you start with when entering the pool, which is given as \( 5.0 \text{ m/s} \). The goal is to calculate the time it takes for you to reach a much slower speed of \( 2\% \) of your initial speed. In mathematical terms, you are looking to find when your velocity drops to \( 0.1 \text{ m/s} \).
  • The exponential decay formula, derived from solving the differential equation, provides \( v(t) = v_0 e^{-\frac{k}{m} t} \).
  • Here, \( v_0 \) is the initial speed, and \( v(t) \) is your velocity at any time \( t \).
This exponential function shows how velocity decreases over time under the influence of forces like drag. Calculating the point when your speed reaches a given fraction of starting velocity involves logarithms, giving insights into the dynamics involved.

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Most popular questions from this chapter

(II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every \(2.5 \mathrm{~s}\). If we assume their arms are each \(0.80 \mathrm{~m}\) long and their individual masses are \(60.0 \mathrm{~kg}\), how hard are they pulling on one another?

(II) Is it possible to whirl a bucket of water fast enough in a vertical circle so that the water won't fall out? If so, what is the minimum speed? Define all quantities needed.

(III) A 3.0 -kg block sits on top of a \(5.0-\mathrm{kg}\) block which is on a horizontal surface. The 5.0 -kg block is pulled to the right with a force \(\vec{\mathbf{F}}\) as shown in Fig. \(39 .\) The coefficient of static friction between all surfaces is 0.60 and the kinetic coeffi- cient is \(0.40 .\) (a) What is the minimum value of \(F\) needed to move the two blocks? (b) If the force is 10\(\%\) greater than your answer for \((a),\) what is the acceleration of each block?

(II) \((a)\) Show that the minimum stopping distance for an automobile traveling at speed \(v\) is equal to \(v^{2} / 2 \mu_{\mathrm{s}} g,\) where \(\mu_{\mathrm{s}}\) is the coefficient of static friction between the tires and the road, and \(g\) is the acceleration of gravity. (b) What is this distance for a \(1200-\mathrm{kg}\) car traveling \(95 \mathrm{~km} / \mathrm{h}\) if \(\mu_{\mathrm{s}}=0.65 ?\) (c) What would it be if the car were on the Moon (the acceleration of gravity on the Moon is about \(g / 6\) ) but all else stayed the same?

(I) Suppose you are standing on a train accelerating at \(0.20 \mathrm{~g}\). What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide?

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