Find v(t) from Newton's Second Law and Differential Equation

  • #1
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Is what I have done correct ?
I want to find v(t) from Sigma F = m*a. I have gravity force mg pointing downward with positive direction and resistive force R = -b*v^2 pointing upwards with negative direction are acting on a particle. So
( a = dv / dt) , m ,b and g are constants
I start with : Sigma F = m *a
mg - b*v^2 = m*a
g - b/m * v^2 = a
dv/dt + b/m v^2 = g
(dv/dt) / (g - (b/m) * v ^2) = 1
(1 / g - (b/m) * v ^2) * (dv/dt)= 1

integral (1 / g - (b/m) * v ^2) * (dv/dt) * dt = integral 1 dt
I use substitution u = g - (b/m) * v^2
du/ dt = -(b/m) * v
du =(2b/m)* v * dt
(2b/m)* v * du = dt

(2b/m)*v integral 1 / u du = integral 1 dt

(2b/m ) * v * ln |u| = t + c

(2b/m ) * v * ln |g - (b/m) * v^2 | = t +c
 
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  • #3
nrqed
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Is what I have done correct ?
I want to find v(t) from Sigma F = m*a. I have gravity force mg pointing downward with positive direction and resistive force R = -b*v^2 pointing upwards with negative direction are acting on a particle. So
( a = dv / dt) , m ,b and g are constants
I start with : Sigma F = m *a
mg - b*v^2 = m*a
g - b/m * v^2 = a
dv/dt + b/m v^2 = g
(dv/dt) / (g - (b/m) * v ^2) = 1
How did you get that last line??
 
  • #4
RPinPA
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Your style is a little confusing and not quite mathematically correct here and there.

(1 / g - (b/m) * v ^2) * (dv/dt)= 1
integral (1 / g - (b/m) * v ^2) * (dv/dt) * dt = integral 1 dt
I would just say
$$\frac{1}{g - (\frac{b}{m})v^2} \frac{dv}{dt} = 1$$
And this differential equation is separable (v's on one side, t's on the other)
$$\frac{dv}{g - (\frac{b}{m})v^2} = dt$$

I use substitution u = g - (b/m) * v^2
du/ dt = -(b/m) * v
du =(2b/m)* v * dt
No, no, no! You are not defining u(t), u as a function of t, and you aren't differentiating u with respect to t. If you were you'd have to differentiate v with respect to t also.
When you are doing a substitution for an integral dv, you are defining u(v) and you want du in terms of dv.
$$u = g - (\frac{b}{m})v^2$$
$$du = -(\frac{2b}{m})v dv$$
$$dv = -(\frac{m}{2b}) \frac{du}{v}$$
which means when you make the substitution for dv, you're still going to have a v under the integral. You could express that in terms of u but that introduces a square root. So this is not too helpful as a substitution.

(2b/m)* v * du = dt

(2b/m)*v integral 1 / u du = integral 1 dt
That first line absolutely is not equivalent to your previous integral, and neither is your second line. I can't see how you got there. When you do a subsitution, you're finding an expression for f(v) dv in terms of g(u) du, replacing the dv and f(v) by du and expressions that do not contain v, and obtaining a new function in terms of u which is supposed to be easier to integrate. This is not it.

I think you'll have better luck writing ##\frac{1}{g - (\frac{b}{m})v^2}## in terms of partial fractions. Define ##r^2 = g## and ##s^2 = \frac{b}{m}## and expand as $$\frac{1}{r^2 - s^2v^2} = \frac{A}{r + sv} + \frac{B}{r - sv}$$.

Integrating that will give you two log terms. On the right you'll have t + C as you did in your solution.
 
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