TITLE: Equations for an object moving linearly but with air resistance?
QUESTION [1 upvotes]: If F be the independent force acting on the object and D be the velocity dependent force acting in the opposite direction of F. The net force accelerating the object is;
$$F − D = ma$$
where $D = \frac{1}{2}CA\rho v^2$, if both the object and air are in motion and in opposite direction then $D = \frac{1}{2}A\rho(v_{air}^2 + v_{object}^2$) resulting into;
$$\frac{F}{m} - \frac{[CA\rho(v_{air}^2 + v_{object}^2)]}{2m} = \frac{\mathrm{d}v_{object}}{\mathrm{d}t}$$
Is this a correct model? and how to manipulate this expression for  $v(t)$ ?

REPLY [1 votes]: Actually I just want to leave comments but my reputation is too low... Im in favor with your idea of considering stag pressure in equation. In fact, during my undergrad I also raised that same question regarding the equation of motion of a parachute.
On the otherhand, with regards to your original equation, I think it must be;
$$\frac{F}{m} - \frac{[CAρ(v_{air}^2 + v_{object}^2)]}{2m} = \frac{\mathrm{d}v_{object}}{\mathrm{d}t}$$
Since your object is a valve which is usually used for closing therefore the velocity of air will not remained constant. Anyway either the velocity of air is constant or not, the velocity of your object in still air must be known so that;
$$v_{object} = v_{object\ still\ air} - v_{air}$$
I don't know exactly your situation, but incorporating this equation to your original equation which will lead you to partial frac and with known values you can determine v(t).


text:
{\bf Problem.} The coefficients of the polynomial
\[a_{10} x^{10} + a_9 x^9 + a_8 x^8 + \dots + a_2 x^2 + a_1 x + a_0 = 0\]are all integers, and its roots $r_1,$ $r_2,$ $\dots,$ $r_{10}$ are all integers.  Furthermore, the roots of the polynomial
\[a_0 x^{10} + a_1 x^9 + a_2 x^8 + \dots + a_8 x^2 + a_9 x + a_{10} = 0\]are also $r_1,$ $r_2,$ $\dots,$ $r_{10}.$  Find the number of possible multisets $S = \{r_1, r_2, \dots, r_{10}\}.$

(A multiset, unlike a set, can contain multiple elements.  For example, $\{-2, -2, 5, 5, 5\}$ and $\{5, -2, 5, 5, -2\}$ are the same multiset, but both are different from $\{-2, 5, 5, 5\}.$  And as usual, $a_{10} \neq 0$ and $a_0 \neq 0.$)
{\bf Level.} Level 5
{\bf Type.} Intermediate Algebra
{\bf Solution.} Let $r$ be an integer root of the first polynomial $p(x) = a_{10} x^{10} + a_9 x^9 + a_8 x^8 + \dots + a_2 x^2 + a_1 x + a_0 = 0,$ so
\[a_{10} r^{10} + a_9 r^9 + \dots + a_1 r + a_0 = 0.\]Since $a_0$ is not equal to 0, $r$ cannot be equal to 0.  Hence, we can divide both sides by $r^{10},$ to get
\[a_{10} + a_9 \cdot \frac{1}{r} + \dots + a_1 \cdot \frac{1}{r^9} + a_0 \cdot \frac{1}{r^{10}} = 0.\]Thus, $\frac{1}{r}$ is a root of the second polynomial $q(x) = a_0 x^{10} + a_1 x^9 + a_2 x^8 + \dots + a_8 x^2 + a_9 x + a_{10} = 0.$  This means that $\frac{1}{r}$ must also be an integer.

The only integers $r$ for which $\frac{1}{r}$ is also an integer are $r = 1$ and $r = -1.$  Furthermore, $r = \frac{1}{r}$ for these values, so if the only roots of $p(x)$ are 1 and $-1,$ then the multiset of roots of $q(x)$ are automatically the same as the multiset of roots of $p(x).$  Therefore, the possible multisets are the ones that contain $k$ values of 1 and $10 - k$ values of $-1,$ for $0 \le k \le 10.$  There are 11 possible values of $k,$ so there are $\boxed{11}$ possible multisets.

set_name: MATH