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Joe_1234

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In summary,The landing speed of an airplane is 360 kph. When it touches down, it puts on its brakes and reverses its engines. The retardation in its speed is 0.2 times the square root of its speed. Determine the time elapsed in seconds from the point of touchdown until the plane comes to a complete stop.

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Joe_1234

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skeeter

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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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Joe_1234

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Sir thank you.skeeter said:

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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Joe_1234

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skeeter

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Joe_1234 said:

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.

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Joe_1234

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Thanks a lot sirskeeter said: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.

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Joe_1234

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Sir, please help me how to get v(t)skeeter said: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.

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skeeter

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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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MarkFL

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Joe_1234 said:Sir, please help me how to get v(t)

As skeeter posted, finding the velocity involves solving the following IVP:

\(\displaystyle \d{v}{t} = -k\sqrt{v}\) 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...

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Kyosh

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skeeter said:

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

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

can you finish from here?

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!

- #11

skeeter

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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 ...

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

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