Could somebody confirm this answer for me please?

[djames1009]

Could somebody confirm this answer for me please? :)

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)
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:
$\frac{dv(t)}{dt}=\frac{-0.5v(t)^2}{m}=\frac{-v(t)^2}{4}$

$-\int\frac{1}{v^2}dv=\int\frac{1}{4}dt$
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:
$x(4)=4ln(\frac{4+4}{4})=4ln(2) m$

 

[grindfreak]

Well everything looks great, except in the second line where you said 0.5 = 1/4.

 

[fluidistic]

Well everything looks great, except in the second line where you said 0.5 = 1/4.

Skipping units I think he said 0.5/m=1/4. Since m=2 kg this looks reasonable IMO.