Who Can Find the Time for a Falling Object with Resistive Force?

  • Thread starter Thread starter Fiasal teslla
  • Start date Start date
  • Tags Tags
    Differentiate
Fiasal teslla
Messages
2
Reaction score
0
For a falling object, who can find the time in terms of the velocity for this journey( before reaching terminal velocity) with considering the resistive force?
 
Physics news on Phys.org
Force equals mass times acceleration
Choose what resistive force to use, inverse velocity or inverse square velocity
Write down the differential equation
Solve it
 
I can"! But are you clear on the terminology? Your title asked "how can differentiate this one" implying you have a function to differentiate but that doesn't appear to be the case. You are just asking about setting up the dynamic equation.

As g edgar says, "Force equals mass times acceleration" so ma= m dv/dt= -g- f(v) where "f(v)" is the resistive force. That can be a very complicated function of the velocity depending on the situation. I do not agree with g edgar's "inverse" formulas. Typically, the faster something is going, the greater the drag, not the other way around. Normally, the drag is simplified to either -kv or -kv2 where k is the constant of proportionallity and v is the speed.
 
HallsofIvy said:
I can"! But are you clear on the terminology? Your title asked "how can differentiate this one" implying you have a function to differentiate but that doesn't appear to be the case. You are just asking about setting up the dynamic equation.

As g edgar says, "Force equals mass times acceleration" so ma= m dv/dt= -g- f(v) where "f(v)" is the resistive force. That can be a very complicated function of the velocity depending on the situation. I do not agree with g edgar's "inverse" formulas. Typically, the faster something is going, the greater the drag, not the other way around. Normally, the drag is simplified to either -kv or -kv2 where k is the constant of proportionallity and v is the speed.
When I tried to solve it, I started with this equation
F-R=ma, where F= mg, snd R is the risitive force which equals bv (b is a constant)
So, mg-bv=ma
mg-bv=m dv/dt then we separate the variables to get the time in terms of the velocity

Am I correct with that?
 
Yes, in that case, with resistive force proportional to v, you get a fairly simple equation:m dv/dt= mg- bv, a separable equation. But be careful about signs. If you are taking "upward" to be positive, then it is m dv/dt= -mg- bv. Since drag always acts opposite to velocity, the coefficient of v is always negative.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
10
Views
2K
Replies
3
Views
2K
Replies
8
Views
2K
Replies
5
Views
1K
Replies
48
Views
2K
Replies
15
Views
2K
Replies
12
Views
3K
Back
Top