Question

help with question

Answer 1

\[v = {g \over k} (1 - e ^{-kt})\]

v = velocity
g = gravitational constant which is \[9.8 ms ^{-2}\]

k = (measured in \[s^{-1}\]

t = seconds


what is e? in the equation

Answer 2

lanaaa to the rescue :p

Answer 3

which course are u taking lilg/?

Answer 4

you know already ee can you help me?

Answer 5

differentiation?

Answer 6

i think it involves differentiation

Answer 7

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

Answer 8

the value for k (s^-1) is 0.15

the value of t you can say is 1 second

Answer 9

what would it be

-kte^-kt?

Answer 10

0? :s

Answer 11

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

Answer 12

its really long one minute

Answer 13

Take ur time im not leavin:)

Answer 14

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

Answer 15

anyone?

Answer 16

do u k anything abt Euler's formula ?

Answer 17

no

Answer 18

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

Answer 19

Jop euler cant be used here:S

Answer 20

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

Answer 21

WHY?

Answer 22

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

Answer 23

did it work with u?

Answer 24

that would give me -1.223244275

(btw i made a mistake it should -2sin(1))

Answer 25

so

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

Answer 26

but shoudn't it be

\[{9.8^{-2} \over 2.00^{-1}}(-1.223244275)\]

Answer 27

which gives you -0.02547364171

Answer 28

-0.02547364171^-1

Answer 29

is that right?

Answer 30

which is -39.25626384

Answer 31

i think it makes sense
But i'm nt sure

Answer 32

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

Answer 33

Is your question really just what e is?

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

where did you get 10 from?

and what about when it says k is measured in s^-1?

Answer 38

Okay okay. Well, the question of what e is is not too hard. e is a mathematical constant. Its value is approximately 2.71828

Answer 39

ok thanks for that smoothmath

Answer 40

also g = 9.8 ms^-2

Answer 41

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)

Answer 42

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!!

Answer 43

{lilg scratches his head}

Answer 44

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

Answer 45

I mean... I'm not fantastic at spreadsheets, but this is a simple problem as long as you are inputting the equation correctly.

Answer 46

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

Answer 47

just 9.8 =)

Answer 48

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

Answer 49

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."

Answer 50

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

Answer 51

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

Answer 52

ok :)

Answer 53

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)

Answer 54

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
\[\frac{m}{s}\]

Answer 55

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.

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

Still confused?

Answer 62

do i not use eulers formula then?

Answer 63

when i have e^-0.15*1

what do i do with that?

Answer 64

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

Answer 65

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

Answer 66

so would it be

\[({1 \over e})^{-0.15}\]

Answer 67

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.

Answer 68

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

Answer 69

oh ok yes my mistake

Answer 70

so it would be

\[(1 - ({1 \over e})^{0.15})\]

Answer 71

Yessir.

Answer 72

thanks alot mate appreciate your great help

Answer 73

My pleasure =)