help with question

- lilg132

help with question

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- 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

lanaaa to the rescue :p

- lalaly

which course are u taking lilg/?

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## More answers

- lilg132

you know already ee can you help me?

- lalaly

differentiation?

- lilg132

i think it involves differentiation

- lalaly

this is a differential equation and e^-kt is the homogeneous solution

- lilg132

the value for k (s^-1) is 0.15
the value of t you can say is 1 second

- lilg132

what would it be
-kte^-kt?

- lilg132

0? :s

- lalaly

I still dont get what ur question is, why did u differentiate that? write down the full question lilg

- lilg132

its really long one minute

- lalaly

Take ur time im not leavin:)

- 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

anyone?

- anonymous

do u k anything abt Euler's formula ?

- lilg132

no

- anonymous

e^x=e^z(cosy + i siny)
this is the formula

- lalaly

Jop euler cant be used here:S

- lilg132

http://en.wikipedia.org/wiki/Euler%27s_formula

- anonymous

WHY?

- lilg132

e^2.00(1 second) = cos(1) + 2sin(1)

- anonymous

did it work with u?

- lilg132

that would give me -1.223244275
(btw i made a mistake it should -2sin(1))

- lilg132

so
\[{9.8 \over 2.00} (-1.223244275)\]

- lilg132

but shoudn't it be
\[{9.8^{-2} \over 2.00^{-1}}(-1.223244275)\]

- lilg132

which gives you -0.02547364171

- lilg132

-0.02547364171^-1

- lilg132

is that right?

- lilg132

which is -39.25626384

- anonymous

i think it makes sense
But i'm nt sure

- lilg132

but i also forgot to take in account that 2.00 should 2.00^s-1 :S that confuses me the most

- anonymous

Is your question really just what e is?

- 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

if someone can show me how they would write the equation out with the values that would help me understand alot better

- experimentX

http://www.wolframalpha.com/input/?i=y+%3D+10%281+-+e^%28-2t%29%29+from+t%3D0+to+5

- lilg132

where did you get 10 from?
and what about when it says k is measured in s^-1?

- 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

ok thanks for that smoothmath

- lilg132

also g = 9.8 ms^-2

- 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

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

{lilg scratches his head}

- anonymous

1) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist

- 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

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

just 9.8 =)

- 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

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

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

Yup. Don't worry too much about the units here =)

- lilg132

ok :)

- 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

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

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

can you show me how you would plus in the values in the equation please it will help me understand much better

- lilg132

use the value of k = 0.15 and t = 1 second

- anonymous

are u asking the value of e? it is approximately equal to 2.71..

- anonymous

Sure.
k = .15 t=1
\[\frac{g}{k}(1-e^{-kt}) = \frac{9.8}{0.15}(1-e^{-0.15*1})\]

- anonymous

Easy peasy. Plug in and solve. Make sure to use the correct order of operations and keep track of signs =)

- anonymous

Still confused?

- lilg132

do i not use eulers formula then?

- lilg132

when i have e^-0.15*1
what do i do with that?

- anonymous

You don't need to, so I don't see why you should.

- anonymous

\[a^{-x} = (\frac{1}{a})^x\]

- lilg132

so would it be
\[({1 \over e})^{-0.15}\]

- 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

Not quite.
\[(\frac{1}{e})^{0.15}\]

- lilg132

oh ok yes my mistake

- lilg132

so it would be
\[(1 - ({1 \over e})^{0.15})\]

- anonymous

Yessir.

- lilg132

thanks alot mate appreciate your great help

- anonymous

My pleasure =)

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