# Sunday night physics problems

1. Jan 27, 2008

### ax_xaein

Sunday night physics problems :(

Alright, so I'm supposed to find the displacement equation for a particle fired vertically under a constant gravitational field, where the resisting force is proportional to the instantaneous velocity of the particle. Here's where I'm at:

Fup = ma; Fdown = -mg - kmv; Fnet = (ma) + (-mg - kmv) // Where k is a constant

ma = (ma) + (-mg - kmv) // Force = Fup + Fdown
dv/dt = -g - kv // m's cancel, differentiate wrt v
dv * 1/(-g -kv) = -dt

// Integrating...
1/k ln(kv+ g) = -t + c0 // Where c0 represents initial velocity
ln(kv + g) = -tk + kc0
kv + g = e^(-tk + kc0)
v = [ e^(-tk + kc0) - g ] / k

Integrate wrt t, to obtain position ( v(t) )
v(t) = -g/k + -------> ???? <----------

The answer is supposed to simplify to: v(t) = [-g/k] + [ (kc0 + g) / k] * e^(-kt)
I have no idea how. I have no idea where the g in the second term came from in the first place. Thanks ahead of time!

2. Jan 28, 2008

### ayoshi

you have Fdown = -mg - kmv; (k a constant), so I guess k will have units [1/seconds]?

also, you wrote: dv/dt = -g - kv // m's cancel, differentiate wrt v

I can see the m's cancel, but then you have (before differentiating)

a = a + (-g-kv)

but a IS dv/dt, so i don't know how you can get:
dv/dt = -g - kv // m's cancel, differentiate wrt v

AFTER differentiating wrt v... (also, dont u usually diff wrt to t??)

3. Jan 28, 2008

### ax_xaein

Yeah, in order for the units to match up i believe k would technically need to have units of 1/t.

And yes, a is dv/dt. So instead of writing:
a = ..., I've substituted dv/dt = ...

This way, we can say dv = blah (partial derivative dt)
Integrating both sides, we then obtain the velocity equation.