Tonyt88 Messages 62 Reaction score 0 Thread starter Oct 11, 2006 #1 A particle of mass m is subject to a force F(v) = -m(alpha)v^2. The initial position is zero, and the initial speed is v nought. Find x(t).
A particle of mass m is subject to a force F(v) = -m(alpha)v^2. The initial position is zero, and the initial speed is v nought. Find x(t).
quasar987 Science Advisor Homework Helper Gold Member Messages 4,796 Reaction score 32 Oct 11, 2006 #2 It's only a matter of solving the differential equation [tex]-m\alpha v^2=m\frac{dv}{dt}[/tex] and then [tex]v=\frac{dx}{dt}[/tex]
It's only a matter of solving the differential equation [tex]-m\alpha v^2=m\frac{dv}{dt}[/tex] and then [tex]v=\frac{dx}{dt}[/tex]
Tonyt88 Messages 62 Reaction score 0 Oct 11, 2006 #3 Sorry perhaps I'm not catching on, so I have: -1/v = -(alpha)t Do I merely say that 1/(alpha t) = dx/dt or perhaps I'm missing something. Or is my first step incorrect?
Sorry perhaps I'm not catching on, so I have: -1/v = -(alpha)t Do I merely say that 1/(alpha t) = dx/dt or perhaps I'm missing something. Or is my first step incorrect?
quasar987 Science Advisor Homework Helper Gold Member Messages 4,796 Reaction score 32 Oct 11, 2006 #4 You forgot the constant of integration. The solution to the differential equation for v is -(alpha)t = -1/v +C And plugging v(0)=[itex]v_0[/itex] gives C=1/[itex]v_0[/itex]. So [tex]v(t)=\frac{1}{\alpha t}+\frac{1}{v_0}[/tex] And now your have to solve the differential equation. [tex]\frac{dx}{dt}= \frac{1}{\alpha t}+\frac{1}{v_0}[/tex] with initial condition x(0)=0 to find x(t). Makes sense?
You forgot the constant of integration. The solution to the differential equation for v is -(alpha)t = -1/v +C And plugging v(0)=[itex]v_0[/itex] gives C=1/[itex]v_0[/itex]. So [tex]v(t)=\frac{1}{\alpha t}+\frac{1}{v_0}[/tex] And now your have to solve the differential equation. [tex]\frac{dx}{dt}= \frac{1}{\alpha t}+\frac{1}{v_0}[/tex] with initial condition x(0)=0 to find x(t). Makes sense?
Tonyt88 Messages 62 Reaction score 0 Oct 11, 2006 #5 Okay, so I got: x(t) = (ln(t)/alpha) + (t/v nought) + C and C = 0 at x(0) So I have x(t) = (ln(t)/alpha) + (t/v nought) Is that correct or did I once again miss something?
Okay, so I got: x(t) = (ln(t)/alpha) + (t/v nought) + C and C = 0 at x(0) So I have x(t) = (ln(t)/alpha) + (t/v nought) Is that correct or did I once again miss something?
quasar987 Science Advisor Homework Helper Gold Member Messages 4,796 Reaction score 32 Oct 11, 2006 #6 It looks fine.