A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0\[v = {g \over k} (1  e ^{kt})\] 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
 2 years ago
Best ResponseYou've already chosen the best response.0lanaaa to the rescue :p

lalaly
 2 years ago
Best ResponseYou've already chosen the best response.0which course are u taking lilg/?

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0you know already ee can you help me?

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0i think it involves differentiation

lalaly
 2 years ago
Best ResponseYou've already chosen the best response.0this is a differential equation and e^kt is the homogeneous solution

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0the value for k (s^1) is 0.15 the value of t you can say is 1 second

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0what would it be kte^kt?

lalaly
 2 years ago
Best ResponseYou've already chosen the best response.0I still dont get what ur question is, why did u differentiate that? write down the full question lilg

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0its really long one minute

lalaly
 2 years ago
Best ResponseYou've already chosen the best response.0Take ur time im not leavin:)

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0It 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

JopHP
 2 years ago
Best ResponseYou've already chosen the best response.0do u k anything abt Euler's formula ?

JopHP
 2 years ago
Best ResponseYou've already chosen the best response.0e^x=e^z(cosy + i siny) this is the formula

lalaly
 2 years ago
Best ResponseYou've already chosen the best response.0Jop euler cant be used here:S

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0e^2.00(1 second) = cos(1) + 2sin(1)

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0that would give me 1.223244275 (btw i made a mistake it should 2sin(1))

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0so \[{9.8 \over 2.00} (1.223244275)\]

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0but shoudn't it be \[{9.8^{2} \over 2.00^{1}}(1.223244275)\]

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0which gives you 0.02547364171

JopHP
 2 years ago
Best ResponseYou've already chosen the best response.0i think it makes sense But i'm nt sure

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0but i also forgot to take in account that 2.00 should 2.00^s1 :S that confuses me the most

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Is your question really just what e is?

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0thats 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
 2 years ago
Best ResponseYou've already chosen the best response.0if someone can show me how they would write the equation out with the values that would help me understand alot better

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=y+%3D+10%281++e^%282t%29%29+from+t%3D0+to+5

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0where did you get 10 from? and what about when it says k is measured in s^1?

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Okay okay. Well, the question of what e is is not too hard. e is a mathematical constant. Its value is approximately 2.71828

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0ok thanks for that smoothmath

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Yup. 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
 2 years ago
Best ResponseYou've already chosen the best response.0just 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
 2 years ago
Best ResponseYou've already chosen the best response.0{lilg scratches his head}

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.11) use a spreadsheet to plot the changing velocity over the 0 to 3 seconds for a parachutist

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1I mean... I'm not fantastic at spreadsheets, but this is a simple problem as long as you are inputting the equation correctly.

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0what 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

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0and 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

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1When 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
 2 years ago
Best ResponseYou've already chosen the best response.0also 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

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Yup. Don't worry too much about the units here =)

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0soooo 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)

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Those 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 \[\frac{m}{s}\]

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0can you show me how you would plus in the values in the equation please it will help me understand much better

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0use the value of k = 0.15 and t = 1 second

rs32623
 2 years ago
Best ResponseYou've already chosen the best response.0are u asking the value of e? it is approximately equal to 2.71..

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Sure. k = .15 t=1 \[\frac{g}{k}(1e^{kt}) = \frac{9.8}{0.15}(1e^{0.15*1})\]

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Easy peasy. Plug in and solve. Make sure to use the correct order of operations and keep track of signs =)

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0do i not use eulers formula then?

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0when i have e^0.15*1 what do i do with that?

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1You don't need to, so I don't see why you should.

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1\[a^{x} = (\frac{1}{a})^x\]

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0so would it be \[({1 \over e})^{0.15}\]

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1That'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.

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Not quite. \[(\frac{1}{e})^{0.15}\]

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0so it would be \[(1  ({1 \over e})^{0.15})\]

lilg132
 2 years ago
Best ResponseYou've already chosen the best response.0thanks alot mate appreciate your great help
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.