# Time, velocity, deceleration and the constant K

1. Jan 26, 2013

### Northbysouth

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

x = x0 + v0xt + 1/2at2

v = v0x + at

3. 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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2. Jan 26, 2013

### tms

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.

3. Jan 26, 2013

### Northbysouth

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?

4. Jan 26, 2013

### tms

Yes. This one is correct.
This one starts out correctly, but then goes astray:
$$a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$
No; simple integration is enough.

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