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

• thejohan
In summary: Now what do you do?In summary, the conversation discusses a problem of finding v(t) from the equation Sigma F = m*a, where gravity force and resistive force are acting on a particle. The conversation also includes a discussion on how to approach the problem and provides a possible solution using integration and substitution.
thejohan
<Moderator's note: Moved from a technical forum and thus no template.>

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
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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thejohan said:
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
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??

Your style is a little confusing and not quite mathematically correct here and there.

thejohan said:
(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)
$$\frac{dv}{g - (\frac{b}{m})v^2} = dt$$

thejohan said:
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.

thejohan said:
(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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## 1. What is Newton's Second Law?

Newton's Second Law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration, or F=ma. In other words, the greater the force applied to an object, the greater its acceleration will be.

## 2. How does Newton's Second Law relate to finding v(t)?

Newton's Second Law can be used to find the velocity (v) of an object at a specific time (t). By rearranging the equation F=ma to solve for a, we can then integrate the resulting equation to find the velocity function v(t).

## 3. What is a differential equation?

A differential equation is an equation that relates a function to its derivatives. In the context of finding v(t) from Newton's Second Law, the differential equation involves the acceleration function, which is the second derivative of the position function.

## 4. How do you solve a differential equation to find v(t)?

To solve a differential equation for v(t), you must first integrate the acceleration function to find the velocity function. This can be done using various methods, such as separation of variables or integrating factors, depending on the specific equation.

## 5. Can Newton's Second Law and the resulting differential equation be used to find the position function?

Yes, by integrating the velocity function found from Newton's Second Law and the resulting differential equation, we can find the position function. This is known as the process of "double integration." However, additional initial conditions or information about the object's motion may be needed to fully determine the position function.

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