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- Mar 1, 2012

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solve the separable differential equation for velocity as a function of time.

you are given $v_0 = 360 \, km/hr$ (you’ll need to convert to m/s). use it to determine the constant of integration.

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Sir thank you.

solve the separable differential equation for velocity as a function of time.

you are given $v_0 = 360 \, km/hr$ (you’ll need to convert to m/s). use it to determine the constant of integration.

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- Mar 1, 2012

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once you've determined $v(t)$ and solved for the time required to stop ...Thank you Sir. Please help me solve again for the length of runway it will from the point of touchdown until it comes to a complete stop.

$\displaystyle \Delta x = \int_0^T v(t) \, dt$ , where $T$ is the time required to come to a full stop.

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Thanks a lot sironce you've determined $v(t)$ and solved for the time required to stop ...

$\displaystyle \Delta x = \int_0^T v(t) \, dt$ , where $T$ is the time required to come to a full stop.

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Sir, please help me how to get v(t)once you've determined $v(t)$ and solved for the time required to stop ...

$\displaystyle \Delta x = \int_0^T v(t) \, dt$ , where $T$ is the time required to come to a full stop.

- Mar 1, 2012

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$\dfrac{dv}{\sqrt{v}} = -0.2 \, dt$

$2\sqrt{v} = -0.2t + C$

can you finish from here?

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As skeeter posted, finding the velocity involves solving the following IVP:Sir, please help me how to get v(t)

\(\displaystyle \d{v}{t} = -k\sqrt{v}\) where \(v(0)=v_0\)

The ODE associated with this IVP is separable, and I would next write:

\(\displaystyle \int_{v_0}^{v(t)} \frac{1}{\sqrt{a}}\,da=-k\int_0^t\,db\)

Try seeing if you can proceed from there and get the same result you get from skeeter's post made above just now...

Hello! If it's not too much to ask, can you please dumb this down for me (aka someone who's good with Physics concepts but absolute trash with Mathematics)? Thank you. I hope you have a nice day!

$\dfrac{dv}{\sqrt{v}} = -0.2 \, dt$

$2\sqrt{v} = -0.2t + C$

can you finish from here?

- Mar 1, 2012

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note $v_0 = 360 \text{ km/hr } = 100 \text{ m/s}$ and $v_f = 0$

$a = \dfrac{dv}{dt} = -0.2 \sqrt{v}$

separating variables yields ...

$v^{-1/2} \, dv = -0.2 \, dt$

integrating both sides ...

$2v^{1/2} = -0.2t + C$, where $C$ is a constant of integration

$v_0 = 100 \text{ m/s } \implies C = 20 \implies v = (10 - 0.1t)^2$

$v_f = 0 \implies t = 100 \text{ s}$

to get the distance the plane travels before it comes to a stop ...

$\displaystyle D = \int_0^{100} (10 - 0.1 t)^2 \, dt \approx 3333 \text{ m}$