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.
a. To calculate the velocity of the body as a function of time, vx(t), we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, there is only one force acting on the body, which is the drag force given by Fx = (-1/2)(CD * rhoair * S * vx^2). We can equate this drag force to mass times acceleration:
(-1/2)(CD * rhoair * S * vx^2) = m * dvx/dt,
where dvx/dt is the derivative of vx with respect to time.
Rearranging the equation, we have:
dvx/(vx^2) = -2*(CD * rhoair * S)/(m) * dt.
Now we can integrate both sides:
∫ dvx/(vx^2) = ∫ -2*(CD * rhoair * S)/(m) * dt.
Integrating the left side gives:
-1/vx = -2*(CD * rhoair * S)/(m) * t + C,
where C is the constant of integration.
Simplifying the equation, we have:
1/vx = 2*(CD * rhoair * S)/(m) * t + C.
To solve for vx(t), we can rearrange the equation:
vx(t) = 1 / (2*(CD * rhoair * S)/(m) * t + C).
b. To determine if the body comes to rest in a finite time, we need to analyze the behavior of vx(t) as t approaches infinity. From the equation derived in part a, we can see that as t approaches infinity, the denominator (2*(CD * rhoair * S)/(m) * t + C) increases without bound. Therefore, the velocity of the body, vx(t), approaches zero as t approaches infinity. This indicates that the body comes to rest in a finite time.
c. To calculate the position of the body as a function of time, x(t), we can integrate the velocity function vx(t) obtained in part a with respect to time:
∫vx(t) dt = ∫(1 / (2*(CD * rhoair * S)/(m) * t + C)) dt.
Integrating both sides gives:
x(t) = (m / (2*(CD * rhoair * S))) * ln(2*(CD * rhoair * S)/(m) * t + C) + D,
where D is the constant of integration.
d. To determine if the body comes to rest in a finite distance, we need to analyze the behavior of x(t) as t approaches infinity. From the equation derived in part c, we can see that as t approaches infinity, the term (2*(CD * rhoair * S)/(m) * t + C) in the natural logarithm increases without bound. Therefore, the position of the body, x(t), also increases without bound. This indicates that the body does not come to rest in a finite distance; it covers an infinite distance as time progresses.