lilg132
  • lilg132
help with question
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lilg132
  • lilg132
\[v = {g \over k} (1 - e ^{-kt})\] v = velocity g = gravitational constant which is \[9.8 ms ^{-2}\] k = (measured in \[s^{-1}\] t = seconds what is e? in the equation
lilg132
  • lilg132
lanaaa to the rescue :p
lalaly
  • lalaly
which course are u taking lilg/?

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lilg132
  • lilg132
you know already ee can you help me?
lalaly
  • lalaly
differentiation?
lilg132
  • lilg132
i think it involves differentiation
lalaly
  • lalaly
this is a differential equation and e^-kt is the homogeneous solution
lilg132
  • lilg132
the value for k (s^-1) is 0.15 the value of t you can say is 1 second
lilg132
  • lilg132
what would it be -kte^-kt?
lilg132
  • lilg132
0? :s
lalaly
  • lalaly
I still dont get what ur question is, why did u differentiate that? write down the full question lilg
lilg132
  • lilg132
its really long one minute
lalaly
  • lalaly
Take ur time im not leavin:)
lilg132
  • lilg132
It is already known that objects falling under the influence of gravity obey the equations of constant acceleration, with a gravitational constant g (usually taken to be 9.8 ms ^-2) that applies to all objects independent of their size or mass Under more realistic conditions, objects falling through air will quickly reach a state of constant velocity, this velocity, which depends on the size, shape, and mass of the falling object, is called terminal velocity. you are on a team developing parachutes for skydivers the velocity v (in ms^-1) of a typical skydiver after t seconds is modelled by your team using the equation \[v = {g \over k} (1 - e ^{-kt})\] the value of k (measured in s^-1) is used to model the different stages of the skydivers decent. The table shows some values found through the experimentation: TYPE OF MOTION k(s^-1) Human in free fall 0.15 Parachute deployed 2.00 1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist 2) what is the terminal velocity for the skydiver? theres more questions but yes im stuck on 1 :p
lilg132
  • lilg132
anyone?
anonymous
  • anonymous
do u k anything abt Euler's formula ?
lilg132
  • lilg132
no
anonymous
  • anonymous
e^x=e^z(cosy + i siny) this is the formula
lalaly
  • lalaly
Jop euler cant be used here:S
lilg132
  • lilg132
http://en.wikipedia.org/wiki/Euler%27s_formula
anonymous
  • anonymous
WHY?
lilg132
  • lilg132
e^2.00(1 second) = cos(1) + 2sin(1)
anonymous
  • anonymous
did it work with u?
lilg132
  • lilg132
that would give me -1.223244275 (btw i made a mistake it should -2sin(1))
lilg132
  • lilg132
so \[{9.8 \over 2.00} (-1.223244275)\]
lilg132
  • lilg132
but shoudn't it be \[{9.8^{-2} \over 2.00^{-1}}(-1.223244275)\]
lilg132
  • lilg132
which gives you -0.02547364171
lilg132
  • lilg132
-0.02547364171^-1
lilg132
  • lilg132
is that right?
lilg132
  • lilg132
which is -39.25626384
anonymous
  • anonymous
i think it makes sense But i'm nt sure
lilg132
  • lilg132
but i also forgot to take in account that 2.00 should 2.00^s-1 :S that confuses me the most
anonymous
  • anonymous
Is your question really just what e is?
lilg132
  • lilg132
thats one of my questions but i can see gets taken out using the eulers formula, if thats what im supposed to use in this. but i need help working out question 1 really
lilg132
  • lilg132
if someone can show me how they would write the equation out with the values that would help me understand alot better
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=y+%3D+10%281+-+e^%28-2t%29%29+from+t%3D0+to+5
lilg132
  • lilg132
where did you get 10 from? and what about when it says k is measured in s^-1?
anonymous
  • anonymous
Okay okay. Well, the question of what e is is not too hard. e is a mathematical constant. Its value is approximately 2.71828
lilg132
  • lilg132
ok thanks for that smoothmath
lilg132
  • lilg132
also g = 9.8 ms^-2
anonymous
  • anonymous
Yup. If you want to read more about it or understand why it's important, wikipedia is a good source, as always. http://en.wikipedia.org/wiki/E_(mathematical_constant)
experimentX
  • experimentX
just putting arbitrary values ... e is Euler's constant ... your velocity should be increasing with constant acceleration ... but since air resistance is directly proportional to velocity .... after certain velocity .. your velocity remains constant ... just another decay equation!!
lilg132
  • lilg132
{lilg scratches his head}
anonymous
  • anonymous
1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist
anonymous
  • anonymous
I mean... I'm not fantastic at spreadsheets, but this is a simple problem as long as you are inputting the equation correctly.
lilg132
  • lilg132
what im mainly confused about is when i use these values in the equation FOR EXAMPLE g = 9.8 ms^-2 to put it in the equation as 9.8^-2 or just 9.8
anonymous
  • anonymous
just 9.8 =)
lilg132
  • lilg132
and the same for k when it says it is measured in s^-1 so i should just leave it as 0.15 or 2.00 instead of 0.15^-1
anonymous
  • anonymous
When they say "ms^-2" that's the unit only. \[ms^{-2} = \frac{m}{s^2}\] Which is the saem as "meters per second squared."
lilg132
  • lilg132
also the end result v is in ms^-1 so the end answer e.g. is 20 would i write it as 20^-1 or just 20
anonymous
  • anonymous
Yup. Don't worry too much about the units here =)
lilg132
  • lilg132
ok :)
lilg132
  • lilg132
soooo if you can just check one answer for me if its correct im going to use k = 0.15 and t = 1 second so i would get \[v = {9.8 \over 0.15}(1 + \cos(1) - 0.15\sin(1))\] or is it - cos(1)
anonymous
  • anonymous
Those are units of measurement. If I ask you how much something costs, and you answer "20," I will get angry and ask you "20 WHAT?" Same thing here. Velocity is measured in meters per second, which can be written as \[\frac{m}{s}\]
anonymous
  • anonymous
Oh my god no no no. Ignore that fancy pants crap those other people were giving you. That's just making the problem complicated. Just plug in the right values to the original equation and solve.
lilg132
  • lilg132
can you show me how you would plus in the values in the equation please it will help me understand much better
lilg132
  • lilg132
use the value of k = 0.15 and t = 1 second
anonymous
  • anonymous
are u asking the value of e? it is approximately equal to 2.71..
anonymous
  • anonymous
Sure. k = .15 t=1 \[\frac{g}{k}(1-e^{-kt}) = \frac{9.8}{0.15}(1-e^{-0.15*1})\]
anonymous
  • anonymous
Easy peasy. Plug in and solve. Make sure to use the correct order of operations and keep track of signs =)
anonymous
  • anonymous
Still confused?
lilg132
  • lilg132
do i not use eulers formula then?
lilg132
  • lilg132
when i have e^-0.15*1 what do i do with that?
anonymous
  • anonymous
You don't need to, so I don't see why you should.
anonymous
  • anonymous
\[a^{-x} = (\frac{1}{a})^x\]
lilg132
  • lilg132
so would it be \[({1 \over e})^{-0.15}\]
anonymous
  • anonymous
That's the rule for negative exponents. Take the reciprocal and make the exponent positive. Alternatively, you could just put in that expression to a calculator and it'll do the work for you.
anonymous
  • anonymous
Not quite. \[(\frac{1}{e})^{0.15}\]
lilg132
  • lilg132
oh ok yes my mistake
lilg132
  • lilg132
so it would be \[(1 - ({1 \over e})^{0.15})\]
anonymous
  • anonymous
Yessir.
lilg132
  • lilg132
thanks alot mate appreciate your great help
anonymous
  • anonymous
My pleasure =)

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