Time, velocity, deceleration and the constant K

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Homework Help Overview

The problem involves a scenario where a jet aircraft's touchdown speed must be reduced while landing on a lake. The deceleration is described as being proportional to the square of the velocity, leading to a need to find a design parameter K and the time elapsed during the deceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of equations of motion and the implications of variable acceleration on the problem-solving approach. There is a suggestion to solve symbolically before substituting numerical values, and a question arises about the necessity of differential equations for the solution.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the use of first principles and the nature of acceleration in this context, indicating a productive direction without reaching a consensus on the solution.

Contextual Notes

Participants note that the equations typically used apply to constant acceleration, which is not the case here due to the velocity-dependent nature of the acceleration. This raises questions about the correct approach to take in solving the problem.

Northbysouth
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Homework Statement



A certain lake is proposed as a landing area for large jet aircraft. The touchdown speed of 110 mi/hr upon contact with the water is to be reduced to 24 mi/hr in a distance of 1370 ft. If the deceleration is proportional to the square of the velocity of the aircraft through the water, a = -Kv2, find the value of the design parameter K, which would be a measure of the size and shape of the landing gear vanes that plow through the water. Also find the time t elapsed during the specified interval.


Homework Equations



x = x0 + v0xt + 1/2at2

v = v0x + at

The Attempt at a Solution



v = v0x + at

24 mi/hr = 110 mi/hr - (kv2)t
24 mi/hr = 110 mi/hr -576kt
576kt = 86 mi/hr

t = 86/576k

Taking this I plugged it into

x = x0 + v0xt + 1/2at2

1370ft = 0 + (110mi/hr)t + 1/2(-kv2)t2

Plugged in v = 24 mi/hr

1370ft = (110 mi/hr)5 - 288 kt2

1370 ft = (110 mi/hr)(86/576k) - 288(86/576)

k = 0.01162

t = 12/845

I realize that I entered my value for k into the system incorrectly, but I used that value to calculate t, which was incorrect. With that said, I'm not sure what to do. Suggestions are welcome and appreciated. THanks
 

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Northbysouth said:

Homework Statement



A certain lake is proposed as a landing area for large jet aircraft. The touchdown speed of 110 mi/hr upon contact with the water is to be reduced to 24 mi/hr in a distance of 1370 ft. If the deceleration is proportional to the square of the velocity of the aircraft through the water, a = -Kv2, find the value of the design parameter K, which would be a measure of the size and shape of the landing gear vanes that plow through the water. Also find the time t elapsed during the specified interval.


Homework Equations



x = x0 + v0xt + 1/2at2

v = v0x + at

The Attempt at a Solution



v = v0x + at

24 mi/hr = 110 mi/hr - (kv2)t
24 mi/hr = 110 mi/hr -576kt
576kt = 86 mi/hr
It is best to solve all problems symbolically before plugging in any numbers. Each time one plugs in a number, one loses information that can help solve the problem or check one's answer for sanity.

However, your difficulty here is that the equations you are using only apply when the acceleration is constant, whereas in this problem the acceleration depends on velocity. You have to start from first principles, that is, the definitions of velocity and acceleration.
 
When you say first principles do you mean:

v = dx/dt

a = dv/dt = dv/dx*dx/dt = v dv/dx

Will I need differential equations to solve this problem?
 
Northbysouth said:
When you say first principles do you mean:

v = dx/dt
Yes. This one is correct.
a = dv/dt = dv/dx*dx/dt = v dv/dx
This one starts out correctly, but then goes astray:
a = \frac{dv}{dt} = \frac{d^2x}{dt^2}
Will I need differential equations to solve this problem?
No; simple integration is enough.

You are given
a = \frac{dv}{dt} = -Kv^2.
Start from there.
 

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