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A small piece of Styrofoam packing material is dropped from a height of \(2.00 \mathrm{m}\) above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by \(a=g-b v .\) After falling \(0.500 \mathrm{m},\) the Styrofoam effectively reaches terminal speed, and then takes \(5.00 \mathrm{s}\) more to reach the ground. (a) What is the value of the constant \(b ?\) (b) What is the acceleration at \(t=0 ?\) (c) What is the acceleration when the speed is \(0.150 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
The value of the constant b is 32.7 s^-1, the acceleration at t=0 is 9.81 m/s^2, and the acceleration when the speed is 0.150 m/s is 4.86 m/s^2.

Step by step solution

01

Determining the time taken to reach terminal velocity

Firstly, identify the total time it takes for the Styrofoam to hit the ground. This is given as the time it takes to fall after reaching terminal velocity (5 seconds) plus the time it takes to reach terminal velocity. As the total distance fallen is 2.00 m, and the distance fallen before reaching terminal velocity is 0.500 m, the distance fallen at terminal speed is \(2.00 \, m - 0.500 \, m = 1.50 \, m\). Terminal speed, \(V_t\), can therefore be calculated using the formula \(Distance = Speed \times Time\), rearranging for Speed gives: \(V_t= \frac{Distance}{Time} = \frac{1.50 \, m}{5.00 \, s} = 0.300 \, m/s.\)
02

Calculate the value of the constant b

Once terminal velocity has been reached, the acceleration becomes zero as there are no more changes in velocity. Therefore, we can write the given equation \(a = g - bv\) at terminal speed as \(0 = g - bV_t\). By rearranging, we can solve for \(b = g/V_t = \frac{9.81 \, m/s^2}{0.300 \, m/s} =32.7 \, s^-1\)
03

Calculate acceleration at t=0

The acceleration at the time instant when the object is just dropped (t=0) will be the acceleration due to gravity, because initially velocity is zero and hence the term \(bv\) in the equation \(a=g-bv\) will be zero. Therefore \(a = g = 9.81 \, m/s^2 \) at t=0.
04

Calculate acceleration when speed is 0.150 m/s

We can substitute the velocity and our calculated \(b\) value into the equation \(a = g - bv\). This gives: \(a = 9.81 \, m/s^2 - 32.7 \, s^-1 \times 0.150 \, m/s = 4.86 \, m/s^2\).

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly an object's velocity changes over time. When something speeds up or slows down, it's accelerating. In this scenario, we look at how the Styrofoam piece's speed changes when dropped from a height.
Here's what you need to understand about acceleration in this context:- **Initial Acceleration**: At the moment of release, the Styrofoam is subjected only to gravity, so its acceleration is equal to the gravitational acceleration, 9.81 m/s².- **Effect of Terminal Velocity**: When an object reaches terminal velocity, it means it's no longer accelerating. The air resistance balances the force of gravity, resulting in zero net acceleration.- **Changing Acceleration**: As the Styrofoam falls and its speed changes, so does its acceleration, until it finally reaches terminal velocity and stops accelerating. This is expressed by the equation \( a = g - bv \), where \( b \) is a constant related to resistance, \( v \) is velocity, and \( g \) is acceleration due to gravity.
Gravity
Gravity is the force that pulls objects towards each other, and it's what causes the Styrofoam to fall to the ground. In this exercise, gravity provides the initial force that sets the Styrofoam in motion.
A few things to keep in mind about gravity:- **Gravity Constant**: On Earth, the acceleration due to gravity is a constant \( 9.81 \, m/s^2 \). This constant acts on the Styrofoam, causing it to accelerate as it falls.- **Impact on Motion**: Gravity acts downward, giving objects a constant acceleration unless acted upon by another force (like air resistance in our scenario).- **Role in Reaching Terminal Velocity**: Initially, the only force acting on the Styrofoam is gravity, making it accelerate. However, as it speeds up, air resistance starts to increase until it balances out with gravity, leading to terminal velocity.
Velocity
Velocity refers to the speed of an object in a given direction. In this exercise, it plays a crucial role as the Styrofoam falls through the air.
Understanding velocity in this context involves: - **Starting from Rest**: Initially, the Styrofoam has zero velocity when it's dropped. It gains velocity as gravity pulls it down. - **Reaching Terminal Velocity**: Terminal velocity is a steady speed that occurs when the force of air resistance is equal to the gravitational force, meaning no further acceleration. Thus, the velocity becomes constant. - **Calculating Velocity**: By knowing the distance the Styrofoam traveled at terminal velocity (1.50 m over 5 seconds), we calculated its terminal velocity to be 0.300 m/s, which is the speed at which it was falling once it stopped accelerating.

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Most popular questions from this chapter

In the Bohr model of the hydrogen atom, the speed of the electron is approximately \(2.20 \times 10^{6} \mathrm{m} / \mathrm{s} .\) Find \((\mathrm{a})\) the force acting on the electron as it revolves in a circular orbit of radius \(0.530 \times 10^{-10} \mathrm{m}\) and (b) the centripetal acceleration of the electron.

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