# Deriving displacement, velocity and acceleration projecticle motion

1. Aug 20, 2009

### 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 whats 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 therfore 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]

Last edited: Aug 20, 2009
2. Aug 20, 2009

### Staff: Mentor

I'm not tracking this part:

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

3. Aug 20, 2009

### TheYoungFella

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

mdv/(mg+kv)=-dt

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 whats going on here.

I believe its a separable equation yes no

Last edited: Aug 20, 2009
4. Aug 20, 2009

### Staff: Mentor

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.

5. Aug 21, 2009

$$\frac{mdv}{mg+kv}=-dt[\tex] [tex]\int\frac{mdv}{mg+kv}=\int-dt[\tex] [tex]\frac{m}{k}\ln{mg+kv}=-t+c[\tex] [tex]mg+kv=\exp^\frac{-k}{m}t+c[\tex] ah I cant for the life od me figure this fancy way of displaying the functions like you would on paper so I will gracefully withdraw. 6. Aug 22, 2009 ### Redbelly98 Staff Emeritus Use a forward slash in the [noparse]$$[/noparse] tag:

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

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

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

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

7. Aug 22, 2009

### TheYoungFella

Thanks for the tip man

8. Aug 22, 2009

### TheYoungFella

$$F=ma$$

$$F=ma=(-mg-kv)$$

$$a=\frac{dv}{dt}$$

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

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

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

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

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

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

$$v=v0$$ $$t=0$$

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

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

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

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

Last edited: Aug 22, 2009
9. Aug 22, 2009

### 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

Last edited: Aug 22, 2009