Deriving displacement, velocity and acceleration projecticle motion

[TheYoungFella]

1. An object of mass, M, is projected into the air with an initial vertical component of velocity, Vo. If the air resistance is proportional to the instantaneous velocity, with the constant of proportionality being, K, derive the equations for the displacement, velocity and acceleration as functions of time.

appologies for the way this is wrote if anyone has a better way of notating this it would be much appreciated.


my work
m dv/dt=(-mg-kv)

so first I got my V's on one side and my T's on the other

mdv/(mg+kv)=-dt then added the integral sign to this equation and integrated it too get

m/k ln(mg+kv)=-t+c

next I tidied it up

mg+kv=e^-((k/m)*t)+c that goop in the middle is meant to be exp to the power of what's in the brackets

mg+kv=ce^-((k/m)*t) movin the c for more tidiness

now I tried to get V on its own

V=ce^-((k/m)*t)-mg/k

sub in t=0 and V=Vo

Vo=C-mg/k therefore c is

C=Vo + mg/k

subbing back into my previous I got

v=(Vo + mg/k)e^-((k/m)*t)-mg/k that is my velocity part right

thats pretty much it for me any help would be appreciated for displacement and acceleration[/b]

 

[berkeman]

1. An object of mass, M, is projected into the air with an initial vertical component of velocity, Vo. If the air resistance is proportional to the instantaneous velocity, with the constant of proportionality being, K, derive the equations for the displacement, velocity and acceleration as functions of time.

appologies for the way this is wrote if anyone has a better way of notating this it would be much appreciated.


my work
m dv/dt=(-mg-kv)

so first I got my V's on one side and my T's on the other

mdv/(mg+kv)=-dt then added the integral sign to this equation and integrated it too get

m/k ln(mg+kv)=-t+c

next I tidied it up

mg+kv=e^-((k/m)*t)+c that goop in the middle is meant to be exp to the power of what's in the brackets

mg+kv=ce^-((k/m)*t) movin the c for more tidiness

now I tried to get V on its own

V=ce^-((k/m)*t)-mg/k

sub in t=0 and V=Vo

Vo=C-mg/k therefore c is

C=Vo + mg/k

subbing back into my previous I got

v=(Vo + mg/k)e^-((k/m)*t)-mg/k that is my velocity part right

thats pretty much it for me any help would be appreciated for displacement and acceleration[/b]

I'm not tracking this part:

mdv/(mg+kv)=-dt then added the integral sign to this equation and integrated it too get

m/k ln(mg+kv)=-t+c

Integrated with respect to what? You have a differential equation... what method are you trying to use to solve this DiffEq?

 

[TheYoungFella]

Sorry I just took that part from my notes I just have first

mdv/(mg+kv)=-dt

then adding the intergration symbol

integrate mdv/(mg+kv)= integrate -dt

So I am quessing it's with respect to time as its to be as a function of time I really have only a vague idea what's going on here.

I believe its a separable equation yes no

 

[berkeman]

Sorry I just took that part from my notes I just have first

mdv/(mg+kv)=-dt

then adding the intergration symbol

integrate mdv/(mg+kv)= integrate -dt

So I am quessing it's with respect to time as its to be as a function of time I really have only a vague idea what's going on here.

Ah, confusing lecture notes, eh? Okay, writing your original post's first equation in the traditional form for inhomogeneous differential equations:

m\frac{dv(t)}{dt} + kv(t) + mg = 0

You can then solve it using the traditional techniques discussed here:

http://hyperphysics.phy-astr.gsu.edu/hbase/Math/deinhom.html

Hope that helps.

 

[TheYoungFella]

\frac{mdv}{mg+kv}=-dt[\tex]<br /> <br /> \int\frac{mdv}{mg+kv}=\int-dt[\tex]&lt;br /&gt; &lt;br /&gt; \frac{m}{k}\ln{mg+kv}=-t+c[\tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; mg+kv=\exp^\frac{-k}{m}t+c[\tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; ah I can&amp;amp;amp;#039;t for the life od me figure this fancy way of displaying the functions like you would on paper so I will gracefully withdraw.

 

[Redbelly98]

ah I can't for the life od me figure this fancy way of displaying the functions like you would on paper so I will gracefully withdraw.

Use a forward slash in the [noparse][/tex][/noparse] tag:

\frac{mdv}{mg+kv}=-dt

\int\frac{mdv}{mg+kv}=\int-dt

\frac{m}{k}\ln{mg+kv}=-t+c

mg+kv=\exp^\frac{-k}{m}t+c

 

[TheYoungFella]

Use a forward slash in the [noparse][/tex][/noparse] tag:

\frac{mdv}{mg+kv}=-dt

\int\frac{mdv}{mg+kv}=\int-dt

\frac{m}{k}\ln{mg+kv}=-t+c

mg+kv=\exp^\frac{-k}{m}t+c

Thanks for the tip man

 

[TheYoungFella]

F=ma

F=ma=(-mg-kv)

a=\frac{dv}{dt}

\frac{mdv}{dt}=(-mg-kv)

\frac{mdv}{mg+kv}=-dt

\int\frac{mdv}{mg+kv}=\int-dt

\frac{m}{k}\ln{mg+kv}=-t+c

mg+kv=\exp^\frac{-k t}{m}+c

v=C\exp^\frac{-k t}{m}-\frac{mg}{k}

v=v0 t=0

v0=C\exp^\frac{-k}t{m}-\frac{mg}{k}

v0=C\exp^0-\frac{mg}{k}

c=v0-\frac{mg}{k}

V=(v0-\frac{mg}{k})\exp^\frac{-k (0)}{m}-\frac{mg}{k}

 

[TheYoungFella]

F=ma

F=ma=(-mg-kv)

a=\frac{dv}{dt}

\frac{mdv}{dt}=(-mg-kv)

\frac{mdv}{mg+kv}=-dt

\int\frac{mdv}{mg+kv}=\int-dt

\frac{m}{k}\ln{mg+kv}=-t+c

mg+kv=\exp^\frac{-kt}{m}+c

v=C\exp^\frac{-kt}{m}-\frac{mg}{k}

v=v0
t=0

v0=C\exp^\frac{-k(0)}{m}-\frac{mg}{k}

v0=C\exp^0-\frac{mg}{k}

c=v0+\frac{mg}{k}

V=(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k} Velocity

displacement is s=\int vdt

s=\int(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}dt

s=\frac{-m}{k}(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}t-c

s=0
t=0

0=\frac{-m}{k}(v0+\frac{mg}{k})\exp^\frac{-k (0)}{m}-\frac{mg}{k}(0)+c

0=\frac{-m}{k}(v0+\frac{mg}{k})\exp^0-0+c

0=\frac{-m}{k}(v0+\frac{mg}{k})1+c

c=\frac{m}{k}(v0+\frac{mg}{k})

s=\frac{-m}{k}(v0+\frac{mg}{k})\exp^\frac{-k t}{m}-\frac{mg}{k}t+\frac{m}{k}(v0-\frac{mg}{k})

s=\frac{-m}{k}(v0+\frac{mg}{k})(1-\exp^\frac{-k t}{m}-\frac{mg}{k}t Displacement