## lilg132 Group Title help with question 2 years ago 2 years ago

1. lilg132 Group Title

$v = {g \over k} (1 - e ^{-kt})$ v = velocity g = gravitational constant which is $9.8 ms ^{-2}$ k = (measured in $s^{-1}$ t = seconds what is e? in the equation

2. lilg132 Group Title

lanaaa to the rescue :p

3. lalaly Group Title

which course are u taking lilg/?

4. lilg132 Group Title

you know already ee can you help me?

5. lalaly Group Title

differentiation?

6. lilg132 Group Title

i think it involves differentiation

7. lalaly Group Title

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

8. lilg132 Group Title

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

9. lilg132 Group Title

what would it be -kte^-kt?

10. lilg132 Group Title

0? :s

11. lalaly Group Title

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

12. lilg132 Group Title

its really long one minute

13. lalaly Group Title

Take ur time im not leavin:)

14. lilg132 Group Title

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

15. lilg132 Group Title

anyone?

16. JopHP Group Title

do u k anything abt Euler's formula ?

17. lilg132 Group Title

no

18. JopHP Group Title

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

19. lalaly Group Title

Jop euler cant be used here:S

20. lilg132 Group Title
21. JopHP Group Title

WHY?

22. lilg132 Group Title

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

23. JopHP Group Title

did it work with u?

24. lilg132 Group Title

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

25. lilg132 Group Title

so ${9.8 \over 2.00} (-1.223244275)$

26. lilg132 Group Title

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

27. lilg132 Group Title

which gives you -0.02547364171

28. lilg132 Group Title

-0.02547364171^-1

29. lilg132 Group Title

is that right?

30. lilg132 Group Title

which is -39.25626384

31. JopHP Group Title

i think it makes sense But i'm nt sure

32. lilg132 Group Title

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

33. SmoothMath Group Title

Is your question really just what e is?

34. lilg132 Group Title

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

35. lilg132 Group Title

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

36. experimentX Group Title
37. lilg132 Group Title

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

38. SmoothMath Group Title

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

39. lilg132 Group Title

ok thanks for that smoothmath

40. lilg132 Group Title

also g = 9.8 ms^-2

41. SmoothMath Group Title

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)

42. experimentX Group Title

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!!

43. lilg132 Group Title

44. SmoothMath Group Title

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

45. SmoothMath Group Title

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

46. lilg132 Group Title

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

47. SmoothMath Group Title

just 9.8 =)

48. lilg132 Group Title

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

49. SmoothMath Group Title

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."

50. lilg132 Group Title

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

51. SmoothMath Group Title

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

52. lilg132 Group Title

ok :)

53. lilg132 Group Title

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)

54. SmoothMath Group Title

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 $\frac{m}{s}$

55. SmoothMath Group Title

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.

56. lilg132 Group Title

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

57. lilg132 Group Title

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

58. rs32623 Group Title

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

59. SmoothMath Group Title

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

60. SmoothMath Group Title

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

61. SmoothMath Group Title

Still confused?

62. lilg132 Group Title

do i not use eulers formula then?

63. lilg132 Group Title

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

64. SmoothMath Group Title

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

65. SmoothMath Group Title

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

66. lilg132 Group Title

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

67. SmoothMath Group Title

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.

68. SmoothMath Group Title

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

69. lilg132 Group Title

oh ok yes my mistake

70. lilg132 Group Title

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

71. SmoothMath Group Title

Yessir.

72. lilg132 Group Title

thanks alot mate appreciate your great help

73. SmoothMath Group Title

My pleasure =)