Solving Two Parachute Problems: Mars Probe and Rocket Car

[cgone]

Hi all,
So I have these two parachute problems that have been giving me problems any aid you can provide would be much apreciated!

Homework Statement

A)A probe is to land on the surface of Mars and I am required to find out the diameter of the parachute.
Given constance:
m=40kg
g=3.75m/s²
v₀=100m/s
v_f= 3m/s
x₀=100m
Pair=2/3*1.1774
Cd=1.4
A=pi*D²/4 (where D is the unknown diameter of the parachute)

Homework Equations

m$$\frac{dV}{dt}$$=mg-$$\frac{1}{2}$$Pair*Cd*AV²

The Attempt at a Solution

I've seprated the variables and simplified to get:

g-$$\frac{k}{m}$$*V²*dV=dt
where k=$$\frac{1}{2}$$PairCdA
$$\sqrt{}$$
So i decomposed the function for integration:
$$\int$$$$\frac{\frac{1}{\sqrt{g}}}{\sqrt{g-\sqrt{\frac{k}{m}}V}}$$+$$\frac{\frac{1}{\sqrt{g}}}{\sqrt{g+\sqrt{\frac{k}{m}}V}}$$
(ignore the second square root on top of the K/m that comes from G, it's only supposed to be on g)

and so that I've integrated to get the following:

$$\frac{1}{\sqrt{g}*\sqrt{\frac{k}{m}}}(Ln(\sqrt{g}-\sqrt{\frac{k}{m}}V))+Ln(\sqrt{g}-\sqrt{\frac{k}{m}}V))=t+c$$

At this point I understand that I can raise everything to E and cancel the +-$$\sqrt{\frac{k}{m}}V$$ but from here I am unsure (also not 100% on the integration)

Also, I have a similar problem however it's not falling, rather slowing down from a parachute (rocket car slowing down from parachute), where I need to find the distance required to stop the car.

Where I was given m$$\frac{dV}{dt}$$=-$$\frac{1}{2}$$Pair*Cd*AV²

So I integrated as a seperable (don't think this is the right way though) to get
$$\frac{-1}{kV}=t+c$$ (a check or direction for this one is all I need)

Thank you!

 

[lippyka]

A)The equation you want to solve is m\frac{dV}{dt}=mg-\frac{1}{2}Pair*Cd*AV2. You can rearrange this as \frac{dV}{dt}=\frac{mg-\frac{1}{2}Pair*Cd*AV2}{m}. To integrate, you want to make this into a separable equation by rearranging as \frac{dV}{(mg-\frac{1}{2}Pair*Cd*AV2)}=\frac{dt}{m}. Then you can integrate both sides to get \int \frac{dV}{(mg-\frac{1}{2}Pair*Cd*AV2)}=\int \frac{dt}{m}, which gives \frac{1}{\sqrt{g}}\ln|g-\frac{1}{2}Pair*Cd*AV2|+C=t+C, where C is an arbitrary constant. To find the diameter of the parachute, you need to solve for A in this equation. Rearranging and taking the exponential gives A=\frac{2}{Pair*Cd*V2}\left(\sqrt{g}e^{-Ct}-g\right). Since you know the initial velocity (v0) and the final velocity (vf), you can plug those in to solve for the diameter.B)For the second problem, you want to solve the equation m\frac{dV}{dt}=-\frac{1}{2}Pair*Cd*AV2. This is a separable equation, so you can rearrange as \frac{dV}{(-\frac{1}{2}Pair*Cd*AV2)}=\frac{dt}{m}. Integrating both sides gives \int \frac{dV}{(-\frac{1}{2}Pair*Cd*AV2)}=\int \frac{dt}{m}, giving \frac{-1}{kV}=t+C, where k=\frac{1}{2}Pair*Cd*A. To find the distance required to stop the car, you need to solve for V in this equation. Rearranging and taking the exponential gives V=\frac