Determine the final result after integration

  • Thread starter Kampret
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  • #1
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Homework Statement


acceleration of moving particle is described by
a=-kv^1,5 where k is a constant. if the condition when t=0 is v=v0 and x=0 prove that xt = √(vv0).t

Homework Equations


dv/dt=a, dx/dt=v

The Attempt at a Solution



dv/dt=a
dv/v^1,5=-k dt
v^-1,5 dv = -k dt ← integrating both sides
-2(v^-½)= -kt + C or
(v^-½) = ½kt + C then plug the initial condition : since when t = 0 v = v0 this turned into
(v0^-½) = ½k(0) + C

so the value of C is equal to v0^-½ continuing by insert the value of C into general equation:

v^-½= ½kt + v0^-½ and for value of v :

v = (½kt + v0^-½)^-²

then move into x

dx/dt = v
dx = v dt
dx = (½kt + v0^-½)^-² dt ← integrating both sides
x = -2/k (½kt + v0^-½)^-¹ + C
the given condition is when t is equal to zero x is also zero thus,
0 = -2/k ( 0 + v0^-½)^-1 + C then re-prashe
(v0^-½)^-1 into v0^½
it would became
0 = -2/k v0^½ + C and the value of C is: 2/k v0½

and i go back to general equation
x = -2/k (½kt + v0^-½)^-¹ + 2/k v0½

the question demand for xt = √(vv0)t and i somewhat know that (½kt + v0^-½)^-¹ is equal to √v

back at
v^-½= ½kt + v0^-½ so this implies v^½ or √v is equal to (½kt + v0^-½)^-¹ so i rewrite the equation into
x = -2/k v½ + 2/k v0½

but, i cant go any further than this i just cannot understand how xt=√(vv0)t obtained, and i did not thoughts that i made mistakes in my integrating process, so can anyone guide me to past this ? and i'm sorry i don't know the suitable title for this, since i not familiar with physics terms i english, so please excuse me.
 

Answers and Replies

  • #2
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Can you turn the equations into the latex form. Its so hard to read and its also hard to help you like this.
 
  • #3
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Can you turn the equations into the latex form. Its so hard to read and its also hard to help you like this.
i'm truly sorry but i never use latex before, so i'm not sure from where i should start, and since it quite complicated i will try learn it from the basic on very near future
 
  • #5
2,016
173

Homework Statement


acceleration of moving particle is described by
##a=-kv^{1.5}## where ##k## is a constant. if the condition when ##t=0## is ##v=v_0## and ##x=0## prove that ##x_t = \sqrt {vv_0}.t##

Homework Equations


##dv/dt=a##, ##dx/dt=v##

The Attempt at a Solution



##dv/dt=a##
##dv/v^{1.5}=-k dt##
##v^{-1,5} dv = -k dt## ← integrating both sides
##-2(v^{-1/2})= -kt + C## or
##(v^{-1/2}) =(1/2)kt + C## then plug the initial condition : since when ##t = 0##, ##v = v_0## this turned into
##(v_0^{-1/2}) =(1/2)k(0) + C##

so the value of ##C## is equal to ##v_0^{-1/2}## continuing by insert the value of ##C## into general equation:

##v^{-1/2}= (1/2)kt + v_0^{-1/2}## and for value of ##v## :

##v = ((1/2)kt + v_0^{-1/2})^{-2}##

then move into ##x##

##dx/dt = v##
##dx = v dt##
##dx = (1/2kt + v_0^{-1/2})^{-2} dt## ← integrating both sides
##x = \frac {-2} {k (1/2kt + v_0^{-1/2})^{-1}} + C##
the given condition is when ##t## is equal to zero ##x## is also zero thus,
##0 = \frac {-2} {k ( 0 + v0^-½)^-1} + C## then re-prashe
##(v_0^{-1/2})^{-1}## into ##v_0^{1/2}##
it would became
##0 = -2/k v_0^{1/2} + C## and the value of ##C## is: ##2/k v_0^{1/2}##

and i go back to general equation
##x = \frac {-2} {k (1/2kt + v_0^{-1/2})^{-1}} + 2/k v_0^{1/2}##

the question demand for ##x_t = \sqrt {vv0} t## and i somewhat know that ##(1/2kt + v_0^{-1/2})^{-1}## is equal to ##\sqrt{v}##

back at
##v^{-1/2}= 1/2kt + v_0^{-1/2}## so this implies ##v^{1/2}## is equal to ##(1/2kt + v_0^{-1/2})^{-1}## so i rewrite the equation into
##x = \frac {-2} {k v^{1/2} + 2/k (v_0)^{1/2}}##

but, i cant go any further than this i just cannot understand how ##x_t=\sqrt {vv_0}t##obtained, and i did not thoughts that i made mistakes in my integrating process, so can anyone guide me to past this ? and i'm sorry i don't know the suitable title for this, since i not familiar with physics terms i english, so please excuse me.

Is this true ?
 
  • #6
verty
Homework Helper
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198
I get $$v = {4 \over (kt + c_1)^2} \\
x = -{4 \over k(kt + c_1)} \\
\sqrt{vv_0} = {2 \sqrt{v_0} \over kt + c_1}$$ and they are totally different.
 
  • #7
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Is this true ?
I'm sorry for late reply and thank you for your help but any some part of your writing is misplace when it should be nominator it become denominator in yours , so i'm on my way construct it and once i finish i will post it here
 
  • #8
33
0
I get $$v = {4 \over (kt + c_1)^2} \\
x = -{4 \over k(kt + c_1)} \\
\sqrt{vv_0} = {2 \sqrt{v_0} \over kt + c_1}$$ and they are totally different.
thanks , i will check it later
 
  • #9
33
0
Is this true ?
$$\frac {dv} {dt} = a$$
$$\frac {dv} {dt} = -kv^{1,5}$$
$$\frac {dv} {v^{1,5}} = -k {dt}$$
$$\ {v^{-1,5}}{dv} = -k {dt}$$
$$\int {v^{-1,5}}{dv} = - \int k{dt}$$
$$\frac {v^{-0,5}} {-0,5} = -{kt} + C$$
$$\ {-2} {v^{-0,5}} = -{kt} + C$$
Now for the initial condition when t=0 v=v0
$$\ {-2} {v0^{-0,5}} = C$$
So the general equation of V become
$$\ {-2} {v^{-0,5}} = -{kt} {-2} {v0^{-0,5}} $$
Simplyfing :
$$\ {v^{-0,5} = \frac 1 2 \{ kt} + {v0^{-0,5}} $$ sorry for { part i just cannot remove it
then into equation of v
$$\ v = {(\frac 1 2 \{ kt} + {v0^{-0,5})}^{-2}$$
then into x part
$$\frac {dx} {dt} = v $$
$$\frac {dx} {dt} =
{(\frac 1 2 \{ kt} + {v0^{-0,5})}^{-2}$$
$$\int {dx} = \int
{(\frac 1 2 \{ kt} + {v0^{-0,5})}^{-2} {dt} $$
$$x = \frac {-2} k
{(\frac 1 2 \{ kt} + {v0^{-0,5})}^{-1} +C$$
this part should be nominator instead of denominator
$${(\frac 1 2 \{ kt} + {v0^{-0,5})}^{-1}$$
 
  • #10
gneill
Mentor
20,925
2,866
If you go back to where you were integrating to find the velocity, you should be able to borrow an expression for ##k t## that you can substitute in.
 
  • #12
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after take a look at gneill's hint
i realized that i made a mistake in the end
before, i write like this :
$$ x = \frac {-2} {k} v^{0,5} + \frac 2 k v^{0,5}$$
but it should be
$$ x = \frac {-2} {k} v^{0,5} \frac {-2} k
v^{0,5}$$
so it will turn into
$$ \frac {xk} {-2} = \sqrt {vv0}$$
it seems little neat than before but i still can't find the demanded value :(
 
  • #13
33
0
I get $$v = {4 \over (kt + c_1)^2} \\
x = -{4 \over k(kt + c_1)} \\
\sqrt{vv_0} = {2 \sqrt{v_0} \over kt + c_1}$$ and they are totally different.
i'm sorry if my question seems little foolish, is that mean $$-{4 \over k(kt + c_1)}$$ is equal to $$\sqrt{vv_0}$$ and
$$2 \sqrt{v_0} \over kt + c_1$$
if yes would you mind to tell me how
$$-{4 \over k(kt + c_1)}$$ turn into
$$\sqrt{vv_0}$$ and
$$2 \sqrt{v_0} \over kt + c_1$$ i'm trying to subtitute both k and t from v general equation but it's only become more complicated
 
  • #14
verty
Homework Helper
2,164
198
i'm sorry if my question seems little foolish, is that mean $$-{4 \over k(kt + c_1)}$$ is equal to $$\sqrt{vv_0}$$ and
$$2 \sqrt{v_0} \over kt + c_1$$
if yes would you mind to tell me how...

No, I was trying to show that it doesn't. There must be something wrong with the question.
 
  • #15
33
0
No, I was trying to show that it doesn't. There must be something wrong with the question.
thank you it's is really reassuring hearing it from you, in addition i want to ask about my final result (post number #12) and the process of it is written on post number #9 what do you think about it ?
i've been restless these past few days thought about this problem since my math is not very good.
 
  • #16
2,016
173
Wait I don't get it. You asked us to find an answer to a wrong question ???
 
  • #17
33
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Wait I don't get it. You asked us to find an answer to a wrong question ???
i like to apologized, i do not intend to do such behaviour , i am also do not know that something wrong in this question and i can't prove the answer by myself so i tried brought my question here, what i typed here is exactly like written in the book. once again please forgive me
 
  • #18
2,016
173
i like to apologized, i do not intend to do such behaviour , i am also do not know that something wrong in this question and i can't prove the answer by myself so i tried brought my question here, what i typed here is exactly like written in the book. once again please forgive me
I am not angry or something. I was just suprised that when said "the quetion is wrong" you were happy and accepted it.
There might be typo or maybe theres something that we are missing. You dont have to say forgive me...
 
  • #19
33
0
I am not angry or something. I was just suprised that when said "the quetion is wrong" you were happy and accepted it.
There might be typo or maybe theres something that we are missing. You dont have to say forgive me...
Well, i've been try my best to solve that problem and i do not thought that i made mistake so i've been restless these few days wondering how solve this, and if the problem is really the one which wrong i am really glad iam worried over nothing , but if that value $$ x = \sqrt{vv0}t$$ value really obtainable , i would like to ask yours assistance and i'm worried if you upset since you already fix my question (latex) and go as far as tried solving this problem so i hope you don't misunderstand me
 
  • #20
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173
Well, i've been try my best to solve that problem and i do not thought that i made mistake so i've been restless these few days wondering how solve this, and if the problem is really the one which wrong i am really glad iam worried over nothing , but if that value $$ x = \sqrt{vv0}t$$ value really obtainable , i would like to ask yours assistance and i'm worried if you upset since you already fix my question (latex) and go as far as tried solving this problem so i hope you don't misunderstand me
I am not upset.. I ll also share the parts that I have done. I can understand you..It's okay.
 
  • #21
2,016
173
Member warned against providing complete solutions
I derived it
First $$dv/dt=-kv^{1.5}$$
and by doing that we are getting
$$\frac {-2} {\sqrt{v}} = -kt -C_1$$ where ##C_1## is just a constant.
When we put ##t=0## we should have ##v=v_0## so lets put it.
we have ##\frac {2} {\sqrt{v_0}} = C_1## Let us put it aside.

Now lets take out ##v## from here.

from some algebra $$v=\frac {4} {(kt+C_1)^2}$$ Now we should take another integral to find ##x##.

$$ x = \frac {-4} {k^2t+C_1k} + C_2$$ from the second condition we know that for ##t=0##, ##x=0##.

So $$C_2=\frac {4} {C_1k}$$

Finally we have

$$x = \frac {-4} {k^2t+C_1k}+\frac {4} {C_1k}$$

Then $$ x = \frac {-4C_1k+4k^2t+4C_1k} {(C_1k)(k^2t+C_1k)}$$

Also canceling the ##k##

$$x = \frac {4t} {(C_1)(kt+C_1)}$$

we know ##C_1=\frac {2} {\sqrt{v_0}}## and ##kt=2(\frac{1} {\sqrt{v}}-\frac{1} {\sqrt{v_0}})##

Put those values and you ll see that it works.
 
  • #22
33
0
I derived it
First $$dv/dt=-kv^{1.5}$$
and by doing that we are getting
$$\frac {-2} {\sqrt{v}} = -kt -C_1$$ where ##C_1## is just a constant.
When we put ##t=0## we should have ##v=v_0## so lets put it.
we have ##\frac {2} {\sqrt{v_0}} = C_1## Let us put it aside.

Now lets take out ##v## from here.

from some algebra $$v=\frac {4} {(kt+C_1)^2}$$ Now we should take another integral to find ##x##.

$$ x = \frac {-4} {k^2t+C_1k} + C_2$$ from the second condition we know that for ##t=0##, ##x=0##.

So $$C_2=\frac {4} {C_1k}$$

Finally we have

$$x = \frac {-4} {k^2t+C_1k}+\frac {4} {C_1k}$$

Then $$ x = \frac {-4C_1k+4k^2t+4C_1k} {(C_1k)(k^2t+C_1k)}$$

Also canceling the ##k##

$$x = \frac {4t} {(C_1)(kt+C_1)}$$

we know ##C_1=\frac {2} {\sqrt{v_0}}## and ##kt=2(\frac{1} {\sqrt{v}}-\frac{1} {\sqrt{v_0}})##

Put those values and you ll see that it works.
thanks, that was truly beautiful! . i guess keeping the variable as short as possible until the last equation really help to prevent confusion...
 
  • #23
2,016
173
thanks, that was truly beautiful! . i guess keeping the variable as short as possible until the last equation really help to prevent confusion...
Yes indeed. I was also opening it before the taking the integral but then I decided not to and it worked well.
 

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