This problem through me for a loss. Really, I am lost here, and if anyone could help me on these parts it'd go along way to me finally starting to grasp this stuff.


The net force on a particle with mass m is:

F(v) = -k(v - a)^3

k>0, a>0, initial velocity = v0


(a) Write the equation of motion

(b ) What is the terminal velocity?

(c ) Find t(v)

(d) Find v(t)




For (a), I was a bit confused.

I decided that (dv/dt) = -(k/m) (v - a)^3

is the equation of motion, because if F(v) is the total force, then F(v) = ma = -k(v-a)^3. Is this wrong? If so, why?



For (b ) I decided that the net force acting on a particle is zero at the terminal velocity, so I figured that v(ter) = a. Is that right?


For (c ), I had no idea.

I just did this:

dv/dt = -(k/m) (v - a)^3

(-m/k) dv/( (v - a)^3) = dt

t(v) = (-m/k) ∫ dv/( (v - a)^3)


And of course, I couldn't figure out the integral. /sigh

1) I don't even know if my method is right

2) Even if it is, I am pretty sure I messed up the integral.

I had

t(v) = (-m/k) [ 1/{2(v -a)^2} ] from initial velocity to v.


Then to find v(t), I tried to solve the above for v, which of course was a nightmare.




If someone can help me out so I can gouge how badly I missed this, I'd appreciate it.

dv/dt = -(k/m)(v -a)^3

Set up as ODE: (v -a)^-3 dv = -(k/m) dt

Integrate: -(1/2)(v - a)^-2 = -(k/m) t + C

Solve for C at time t=0: -(1/2)(v - a)^-2 = -(k/m) t - (1/2)(v_0 - a)^-2

Solve for t as function of v: [(1/2)(v_0 - a)^-2 -(1/2)(v - a)^-2 ](-m/k) = t(v) = (m/2k)(1/(v - a)^2 - 1/(v_0 - a)^2)

Solve for v-a: (v-a) = ±1/sqrt(2 kt/m + (v_0 - a)^-2)

Solve for v as function of t: v(t) = a ± 1/sqrt(2 kt/m + (v_0 - a)^-2)

Find cases:

v(t) = a + 1/sqrt((1/(v_0-a)^2 + 2 k t/m)) if v_0 >= a
v(t) = a - 1/sqrt((1/(v_0-a)^2 + 2 k t/m)) if v_0 < a

Thank you as always rpener!


PS- I'm aware I used the wrong "threw" in the OP