Question

there's this one vertical motion word problem: A falling object encounters air resistance that is proportional to its velocity. The acceleration due to gravity is -9.8 meters per second per second. The net change in velocity is dy/dt=kv-9.8.
(a) find the velocity of the object as a function of time if the intial velocity is v(subscript "0").
(b)Use the result of part (a) to find the limit of the velocity as t approaches infinity.
(c) Intergrate the velocity function found in part (a) to find the position function "s".

Answer 1

@ganeshie8

Answer 2

\(\large \frac{dv}{dt}=kv-9.8\)

Answer 3

like that ?

Answer 4

so, I would integrate the dy/dt=kv-9.8 to get the velocity equation for a...so, v(subscript 0)= kv^2 -9.8 v? because the K is just a constant, even if divided by 2, wouldn't really show up... and yesh, exactly like that ^^

Answer 5

You will get an exponential approach to a terminal
velocity v = (vfinal)(1-exp(-kt))

Answer 6

Alright, you're mixing y's and v's

Answer 7

so, there are NO y's in the question anywhere right ?

Answer 8

its all about v's

Answer 9

?

Answer 10

oh! yeah~ hehe whoops many apologies~

Answer 11

\(\large \frac{dv}{dt}=kv-9.8 \)

Answer 12

:) just making sure we'rent missing anything... thats all :)

Answer 13

okay, Velocity and Time are variables here

Answer 14

separate variables

Answer 15

\(\large \frac{dv}{dt}=kv-9.8 \)

\(\large \frac{dv}{kv-9.8}= dt \)

Answer 16

Integrate

Answer 17

\(\large \int \frac{dv}{kv-9.8}= \int dt \)

Answer 18

and you're right, k is just a constant

Answer 19

oh so then the ln rule thing, right?

Answer 20

yup!

Answer 21

and so,....oh...hold on... so then where does the k go??? wud it go outside the ln like this? kln|kv-9.8|+c?

Answer 22

differentiate kln|kv-9.8|+c,
and check if u get back ur 1/(kv-9.8)

Answer 23

if u get back,
then ur integration is correct, otherwise - think again :)

Answer 24

oh, uhm, would the ln be divided by k?

Answer 25

you got it :)

Answer 26

ok so then the answer to part a would be ln|kv(subscript 0)-9.8|/k +c =V

Answer 27

\(\large \frac{dv}{dt}=kv-9.8\)


\(\large \frac{dv}{kv-9.8}= dt\)


\(\large \int \frac{dv}{kv-9.8}= \int dt \)


\(\large \frac{\ln |kv-9.8|}{k}= t + c \)


\(\large \ln |kv-9.8|= kt + c \)



\(\large kv-9.8= e^{kt + c} \)


\(\large kv-9.8= ce^{kt} \)

Answer 28

solve \(\large v\)

Answer 29

and see that, i have change the place of \(c\) watever way i felt good

Answer 30

ohh ok~ so then\[v=\frac{ ce^{kt}+9.8 }{ k }\]

Answer 31

Correct,
lets find the value of \(c\) now :)

Answer 32

read part a) question again
we're given some initial value right ?

Answer 33

(a) find the velocity of the object as a function of time if the intial velocity is v(subscript "0").

Answer 34

initial velocity = \(V_0\)
wat does that mean ?

Answer 35

is that supposed to be given?

Answer 36

that means, when time, \(t = 0\), the velocity of object \(v = V_0\)

Answer 37

oh, rght~XD

Answer 38

so, \((0, V_0)\) is a point on out velocity function \(\large v\)

Answer 39

plugit in and find the value of \(c\)

Answer 40

\(\large \frac{dv}{dt}=kv-9.8\)


\(\large \frac{dv}{kv-9.8}= dt\)


\(\large \int \frac{dv}{kv-9.8}= \int dt \)


\(\large \frac{\ln |kv-9.8|}{k}= t + c\)


\(\large kv-9.8= e^{kt + c}\)


\(\large kv-9.8= ce^{kt}\)


\(\large v= \frac{ce^{kt}+9.8}{k}\)

plugin \((0, V_0)\)

\(\large V_0 = \frac{ce^{0t}+9.8}{k} \implies c = V_0k - 9.8\)

So, the particular velocity function is :

\(\large v= \frac{(V_0k-9.8)e^{kt}+9.8}{k}\)

Answer 41

we're done with part a)

Answer 42

let me knw if smthng doesnt make sense...

Answer 43

ohok got it XP haha redid some stuff to make sure i got it ^^

Answer 44

good :)

Answer 45

before moving to part b)
let me ask u a simple question

Answer 46

ok go for it ^^

Answer 47

lets assume climbing up is positive,

here is the question :
when you're FALLING DOWN,
whats the "sign" of ur velocity ?

Answer 48

OH negaitve!

Answer 49

good, now look at our final velocity equation :

\(\large v= \frac{(V_0k-9.8)e^{kt}+9.8}{k}\)

Answer 50

since the object is always falling down, can we say \(\large v\) needs to be negative always ?

Answer 51

yes~ and so would we have to make k a negative number?

Answer 52

you got it !!

Answer 53

\(k\) is a negative number,
having that knowledge we're completely ready to pursue part b)

Answer 54

(b)Use the result of part (a) to find the limit of the velocity as t approaches infinity.

Answer 55

\(\large \lim \limits_{t\to \infty }v= \lim \limits_{t\to \infty }\frac{(V_0k-9.8)e^{kt}+9.8}{k}\)

Answer 56

the exponent of e gets infintely negative, and so that would mean that Vk-9.8 is etting huge,

Answer 57

what do u get ?

Answer 58

er, small, srry

Answer 59

what is the value of \(e^{-\infty}\) ?

Answer 60

i meant, what value it approaches..

Answer 61

1/something super big, so it's approaching 0?

Answer 62

ohhhhh hehe whoops omg i need sleep after this problem~

Answer 63

\(\large \lim \limits_{t\to \infty }v= \lim \limits_{t\to \infty }\frac{(V_0k-9.8)e^{kt}+9.8}{k}\)


\(\large ~~~~~~~~~= \frac{(V_0k-9.8)(0) +9.8}{k}\)


\(\large ~~~~~~~~~= \frac{9.8}{k}\)

Answer 64

oh wait yeah I was right, actually ^^

Answer 65

yes you're right :)

Answer 66

btw, its an amazing result, if you haven't thought of it before

Answer 67

9.8/k is a constant !

the free falling body's velocity will not continue to increase forever, it reaches a limit and stays fixed there

Answer 68

wattttttt!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! so there's a natural limit to how fast we can fall?

Answer 69

that is interesting indeed...uhm... so onto part c, or?

Answer 70

Suppose ur plane crashed and you're falling down freely; then your velocity will increase in the start... but after some point it approaches a fixed value :


air pushes u up
gravity pulls u down

how much force air exerts on u depends on ur velocity,

so after u reach certain speed, the air and gravity effects balance each other out

Answer 71

(c) Intergrate the velocity function found in part (a) to find the position function "s".

Answer 72

\(\large v= \frac{(V_0k-9.8)e^{kt}+9.8}{k}\)


Integrating velocity gives u position


\(\large \int v dt= \int \frac{(V_0k-9.8)e^{kt}+9.8}{k} dt\)


\(\large s = \frac{1}{k} \left( \int (V_0k-9.8)e^{kt}+9.8 ~dt \right)\)

Answer 73

piece of cake for u

Answer 74

is it (1/k) (vk^2 e^(kt) - 9.8 ke^kt + 9.8)?

Answer 75

whoops, last one is 9.8t

Answer 76

NONO I did that WRONG

Answer 77

it's actually v(sub 0 ) e^kt -9.8 e^kt + 9.8t all divided by k

Answer 78

middle term is divided by k

Answer 79

\(\large s = \frac{1}{k} \left( \int (V_0k-9.8)e^{kt}+9.8 ~dt \right) \)


\(\large s = \frac{1}{k} \left( (V_0k-9.8)\frac{e^{kt}}{k}+9.8t + c \right) \)

Answer 80

yesh~ eexcept I forgot the c...hehe

Answer 81

^^constant comes up... cuz we dont knw from where we have started falling down...
we're not given any initial position... so \(c\) sticks there

Answer 82

ohh ok ^^ thank you lots!!! oh hey btw b4 we part, I have a random question~

Answer 83

shoot

Answer 84

Have you lived in Alabama(US) before?

Answer 85

nooo... i never moved out of india till now
why do u ask ha ?

Answer 86

ohhh hahah you sounded like one of my teachers that I knew quite a while back ^^ woah india? You'e in india?

Answer 87

is that where that "ha" comes from?

Answer 88

lol xD yes

Answer 89

just curio~ XD

Answer 90

you're in Alabama now is it... must be very late over there lol u should sleep now... otherwise u wont be fresh for exam tomorrow

Answer 91

^^ actually I'm in New Mexico hehe :D ok then~ thank you~ I'll cya around XP

Answer 92

have good sleep :) good luck for exam !! bye byeeeeee

Answer 93

^^ I probs wont go to slp until later because I have to rememorize the all of the trig and inverse trig identities for deriv and integrals~ sigh :) thnx have a good afternoon XP

Answer 94

okay you're a machine lol !! i can give a hand.. when u need me just holler... il be around :)

Answer 95

awesome ^^