# Help with question mass/displacement

1. Apr 24, 2012

### djames1009

A particle of mass 2kg is subject to a drag force from the air: F(v)= -0.5v^2. The initial displacement of the mass is zero and the initial speed is 1ms. Find an expression for x(t) and hence work out x(4)

Ive got this question in one of my text books and cant seem to get it to work, could anybody help me with the answer at all?

I have tried a few things, and i know i have to integrate, but i am getting a bit confused :s. If somebody could show me how its done, then i can look back at it and understand all the steps, as there is other questions like this i plan to eventually answer.
Thanks Alot :D

2. Apr 24, 2012

### djames1009

Or if anybody could just help me start off that would be great also!

3. Apr 24, 2012

### EmittingLight

From the initial conditions we know that:
$x(0)=0 m$
$v(0)=1 m/s$

I'm going to keep using the notations v(t) and x(t) to try to remind you that v and x are functions of t, which is very important when doing calculus.

For a constant mass:
$F(v(t))=ma=m\frac{dv(t)}{dt}=-0.5v(t)^2$
So:
$\frac{dv(t)}{dt}=\frac{-0.5v(t)^2}{m}=\frac{-v(t)^2}{4}$
Erm... which is a separable ordinary differential equation... this is a lot higher leveled than I thought, is this like a university leveled maths question? You're going to need someone else to confirm the answer since I'm not sure if it was meant to turn out like this lol. Anyway, separating gives:
$-\int\frac{1}{v^2}dv=\int\frac{1}{4}dt$
Integrating gives:
$\frac{1}{v(t)}=\frac{t}{4}+c=\frac{t+4c}{4}$
$\Rightarrow v(t)=\frac{4}{t+C}$
Since, from the second initial condition:
$v(0)=\frac{4}{0+C}=1 m/s$
Then:
$C=4$
Giving:
$v(t)=\frac{dx(t)}{dt}=\frac{4}{t+4}$
Integrating with respect to t:
$\int\frac{dx(t)}{dt}dt=\int\frac{4}{t+4}dt$
Gives:
$x(t)=4ln(t+4)+b$
Since, from the first initial condition:
$x(0)=4ln(0+4)+b=0$
Then:
$b=-4ln(4)$
Which gives the final equation for displacement, x(t):
$x(t)=4(ln(t+4)-ln(4))=4ln(\frac{t+4}{4})$
And:
$x(4)=4ln(\frac{4+4}{4})=4ln(2) m$

Last edited: Apr 24, 2012
4. Apr 24, 2012

### djames1009

Okay thanks alot, i'll post it onto an advanced section and see what other people get aswell, thanks :D

5. Apr 24, 2012

### djames1009

And sorry yeah it does mention about seperating by variables, so this does seem correct. The only thing i have seen is when you have put -v(t)^2/m you have put m as 4, should it be 2 or have i got it wrong ha?

6. Apr 24, 2012

### EmittingLight

$0.5=\frac{1}{2}$

So:
$a=-0.5v(t)^2/m=-\frac{1}{2}\frac{v(t)^2}{m}$

7. Apr 24, 2012

### djames1009

Ah i see my fault ha XD

8. Apr 24, 2012

### djames1009

$-\int\frac{1}{v^2}dv=\int\frac{1}{4}dt$

Hi i've just realised at this point the limits should be,
for $-\int\frac{1}{v^2}dv$ should be V and 1.

and for $-\int\frac{1}{4}dt$ should be t and 0.

Could you give me any help here?

9. Apr 25, 2012

### EmittingLight

What I did there is called an indefinite integral. There are no limits, and is also the reason why I had to add a constant to the answer (I had to add one to both sides of the equation but I shifted both constants to one side to give the single c; another subtlety you might have noticed was that I converted the 4c into a single big C).