A body of mass m moves in the +x direction, subject only to drag force: Fx= (-1/2)(CD*rhoair*S*vx^2)
With drag coefficient CD, air density rho air, cross sectional area S, and velocity component vx. At time t=0 the body is at postion x=0 and has velocity of vx=v0 where v0 is some positive value.
a. Calculate the velocity of the body as a function of time, vx(t) for t >=0:
b. Does the body come to rest in a finite time? Remember you must justify your answer.
c. Calculate the position of the body as a function of time, x(t) for t>=0
d. Does the body come to rest in a finite distance, i.e. cover only a finite distance? Remember you must justify your answer.
m dv/dt = -[(1/2) rho Cd S] v^2
dv/dt = - [(1/2) rho Cd S/m] v^2
let k = [(1/2) rho Cd S/m]
then
dv/dt = -k v^2
dv/v^2 = -k dt
-1/v = -kt - c
v = 1/(kt+c)
when t = 0, v = Vo
Vo = 1/c
so c = 1/Vo
so
v = 1/(kt +1/Vo)
when does v = 0?
only when t = oo
dx/dt = 1/(kt+r) where r = 1/Vo
dx = dt/(kt+r)
x = (1/k)ln(kt+r)+c2
when t = 0, x = 0
0 = (1/k) ln r + c2
c2 = -(1/k) ln r
so
x = (1/k)(ln(kt+r)-ln r)
x = (1/k) ln [ (kt+r)/r) ]
remember r = 1/Vo
x = (1/k) ln [Vo kt + 1} that does not become constant
check details, I did that fast but perhaps you get the idea
To calculate the velocity of the body as a function of time, vx(t), we need to solve the differential equation that represents the motion of the body. The drag force acting on the body is given by Fx = (-1/2)(CD * rhoair * S * vx^2). Newton's second law states that F = ma, where F is the net force, m is the mass of the body, and a is the acceleration. Since the body is moving only in the +x direction, the net force acting on it is equal to the drag force, Fx. Therefore, we can write the equation of motion as:
ma = (-1/2)(CD * rhoair * S * vx^2)
Rearranging the equation, we get:
a = (-1/2m)(CD * rhoair * S * vx^2)
Now, we can use the chain rule to express the acceleration a in terms of velocity vx and time t:
dvx/dt = (-1/2m)(CD * rhoair * S * vx^2)
Separating variables and integrating, we get:
∫(1/vx^2) dvx = - (∫(CD * rhoair * S)/(2m) dt
Solving the integrals, we obtain:
-1/vx = (CD * rhoair * S * t)/(2m) + C1
Multiplying both sides by -1 and rearranging, we have:
vx = -2m/(CD * rhoair * S * t + C1)
Since at t=0, the body has an initial velocity of v0, we can substitute this value into the equation and solve for C1:
v0 = -2m/(CD * rhoair * S * 0 + C1)
v0 = -2m/C1
Therefore, C1 = -2m/v0. Substituting this value back into the equation, we get:
vx = -2m/(CD * rhoair * S * t - 2m/v0)
This is the velocity of the body as a function of time, vx(t), for t >= 0.
To determine if the body comes to rest in a finite time, we need to find the limit of the velocity as t approaches infinity. Taking the limit as t approaches infinity in the equation for vx, we have:
lim(t->∞) vx = lim(t->∞) (-2m/(CD * rhoair * S * t - 2m/v0))
As t approaches infinity, the first term in the denominator dominates and the second term becomes negligible compared to it. Therefore, the limit simplifies to:
lim(t->∞) vx = lim(t->∞) (-2m/(CD * rhoair * S * t))
Since the denominator of the expression is increasing with t, this limit approaches zero. Therefore, the body comes to rest in a finite time.
To calculate the position of the body as a function of time, x(t), we need to integrate the velocity function vx(t) with respect to time. The initial position of the body is x=0 at t=0, so we can integrate as follows:
∫vx dt = ∫(-2m/(CD * rhoair * S * t - 2m/v0)) dt
Integrating, we get:
x = ∫(-2m/(CD * rhoair * S * t - 2m/v0)) dt
Simplifying the integral, we have:
x = (-2m/v0) ln(CD * rhoair * S * t - 2m/v0) + C2
Since the initial position of the body is x=0 at t=0, we can substitute these values into the equation and solve for C2:
0 = (-2m/v0) ln(CD * rhoair * S * 0 - 2m/v0) + C2
0 = (-2m/v0) ln(- 2m/v0) + C2
Therefore, C2 = (-2m/v0) ln(- 2m/v0). Substituting this value back into the equation, we get:
x = (-2m/v0) ln(CD * rhoair * S * t - 2m/v0) + (-2m/v0) ln(- 2m/v0)
This is the position of the body as a function of time, x(t), for t >= 0.
To determine if the body comes to rest in a finite distance, we need to examine the limit of the position function as t approaches infinity. Taking the limit as t approaches infinity, we have:
lim(t->∞) x = lim(t->∞) [(-2m/v0) ln(CD * rhoair * S * t - 2m/v0) + (-2m/v0) ln(- 2m/v0)]
As t approaches infinity, the natural logarithm term ln(CD * rhoair * S * t - 2m/v0) diverges to negative infinity. Therefore, the limit simplifies to:
lim(t->∞) x = lim(t->∞) (-2m/v0) ln(- 2m/v0)
Since the denominator of the expression is a negative value, the natural logarithm term is undefined and the limit does not exist. This indicates that the body does not come to rest in a finite distance.