Expert: Ahmed Salami - 1/21/2006

Question
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Followup To
Question -
s m(D2x/Dt2)dt = s w – k (Dx/Dt)2 dt

s= integration sign
D= Delta for differentiation

Note: All the 2 ment to be square of the leter or braket befor them.
Answer -
Hi Lyzo,
Have you written $(d^2x/dt^2) dt = $k(dx/dt)^2 dt  ?
And what exactly are we doing?
Get back to me.
Regards.


the equation is massmultoplied by accelearation (D SQUARED X OVER D T SQUARED) = WEIGHT MINUS THE VELOCITY SQUARED MULTIPLIED BY K(constant). AND VELOCITY IS DELTA (D)x/ delta t. intergrate all sides making v and t the subjects


Answer
Hi lyzo,
Sorry for the time it took.
I'm terribly sorry for the delay, there was a mix up and
i thought i had your solution sent to you.
So if we'll move on, we have got
m(d^2x/dt^2) = w - k(dx/dt)^2
which can be re written as
m(dv/dt) = w - kv^2
We can then rearrange to give
[m/(w - kv^2)]dv = dt
(m/w)[1/1 - (k/w)v^2]dv = dt
let k/w = p^2, we the have
(m/w)[1/1 - (pv)^2]dv = dt
we now have to integrate both sides with respect to v
and t respectively.
But first, by partial fractions
1/1 - (pv)^2 = 1/2[1/1+pv + 1/1-pv]
Therefore,
$[1/1 - (pv)^2]dv = $1/2[1/1+pv + 1/1-pv] dv
= 1/2[(1/p)ln(1+pv) - (1/p)ln(1-pv)]
= 1/2p[ln(1+pv) - ln(1-pv)]
= (1/2p)ln[(1+pv)/(1-pv)]
Returning to our equation, by integrating both sides
(m/w)(1/2p)ln[(1+pv)/(1-pv)] = t + c(constant)
(m/2wp)ln[(1+pv)/(1-pv)] = t + c
again, for simplicity let m/2wp = 1/U and t + c = T, so
ln[(1+pv)/(1-pv)] = UT
[(1+pv)/(1-pv)] = e^(UT)
1 + pv = (1 - pv)e^(UT)
1 + pv = e^(UT) - pve^(UT)
pv + pve^(UT) = e^(UT) - 1
pv(1 + e^(UT)) = e^(UT) - 1
v = [e^(UT) - 1]/[p(e^(UT) + 1)]
v = e^[2wp(t+c)/m] - 1  /  p(e^[2wp(t+c)/m] + 1)
with v now written as a function of t.
You can always go and simplify further.
I hope i have helped. You can always get back to me.
Regards.