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user100 [1]
3 years ago
6

[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.

1 ft/s2 of deceleration for each ft per s (foot per second) of the car’s velocity. Set up an Initial Value Problem (IVP) for the velocity of the car. Solve the IVP and use the solution to find the car’s limiting velocity. If the car starts at position x(0) = x0 = 0, ft, what position will it be at after 1 2 hour? Be sure to set up your coordinate system and include units in all of your answers.

Physics
1 answer:
kumpel [21]3 years ago
4 0

Answer:

a) Initial Value Problem

dv/dt = 4 - 0.1v

v(0) = 0

b) solution to the IVP

v(t) = 40(1 - e^(-t/10))

c) Limiting velocity

Vo = 40 ft/s

Position of the car after 12 hours

X = 14,390 ft

Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

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So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

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