ODE for free fall in a medium

[Albert86]

<Moderator's note: Moved from a technical forum and thus no template.>

An object with mass 96 kg is given an initial downward velocity −3m/s in a medium that exerts a resistive force with magnitude proportional to the square of the speed. The resistance is 60 N when the velocity is −2m/s. Use
g=10m/s^2.

a. Write out a differential equation in terms of the velocity v, and acceleration a

b. Find the velocity v(t)

mv'=-mg-kv^2
96v'=-960-15v^2
-32/5v^2+320 d/dt=1
-32*1/40arctan(1/8 v(t)=t
V(t)=-8tan(5/4(c+t))

I need help with
V(t)=-8tan(5/4(c+t)) but I can solve for c
for separable first order
can't figure where I am going wrong

 

[kuruman]

There should be no c involved. Your integrals should have lower and upper limits. When t = 0, v = v₀; when t = t, v = v.

 

[haruspex]

initial downward velocity −3m/s

Taken literally, that means an upward velocity of +3m/s.
I assume they mean a velocity of -3m/s, taking up as positive. Maybe that was not the exact wording?

mv'=-mg-kv^2

Careful with signs. If up is positive then you are taking g as a positive constant, so -mg is negative. Since -kv² is also negative your equation says the drag is helping gravity accelerate the mass downwards. That would be true when the velocity is positive, but it never will be in this question.

 

[Albert86]

Taken literally, that means an upward velocity of +3m/s.
I assume they mean a velocity of -3m/s, taking up as positive. Maybe that was not the exact wording?

Careful with signs. If up is positive then you are taking g as a positive constant, so -mg is negative. Since -kv² is also negative your equation says the drag is helping gravity accelerate the mass downwards. That would be true when the velocity is positive, but it never will be in this question.

That is how the problem was presented and I asked if I had the right formula which I was told yes but now I am 100% lost

 

[haruspex]

That is how the problem was presented

Ok.

I asked if I had the right formula which I was told yes

Who said that? You can easily see that it is wrong by dropping out gravity. You get mv'=-kv². That formula is correct whilst v is positive, but with v negative it will produce a velocity of ever increasing magnitude.

 

[kuruman]

... I asked if I had the right formula ...

Here is another way to see why the formula you posted is incorrect. On the right side you have all the forces acting on the mass. These are gravity and the resistive force. Gravity points down, always. The resistive force in this case is up, opposite to the velocity which is down. Therefore, the two forces on the F_net side of the equation must have opposite signs regardless of which direction, up or down, you define as positive. Your equation shows the forces with the same sign.

My personal preference is to write differential equations of this sort with the positive axis chosen always in the same direction as the velocity. Then v becomes the speed and the resistive force comes in always with a negative sign while gravity is either positive or negative depending on whether it is in the same or opposite direction as the velocity. I solve the differential equation for the speed and, if I want the velocity, I tack on the appropriate unit vector in the end.

 

[Albert86]

Would mv'=-mg +kv^2 be the correct equation to start with

 

[haruspex]

Would mv'=-mg +kv^2 be the correct equation to start with

For the cases in which the velocity is negative (as it is here), yes.