How Do You Calculate Stopping Distance with Velocity-Dependent Resistive Force?

[mgeoghe2]

dynamics problem!

this is a problem about a resistive force that is a function of v, not x.
a block with an initial velocity goes through a medium that resists motion with force proportional to the block's velocity, with proportionality constant K. so basically the resistive force = -Kx (i choose the pos x dir to be in the dir of initial velocity)
i am supposed to find the stopping distance by finding v as a function of x and then i am supposed to find it again by finding v as a function of t.
i cruised wikipedia because it helped me on an escape velocity problem, and it gave me some help, but i don't know how they got what they did and I am not entirely sure if its what i want.

they said:
-Kx = ma = m dv/dt which i have and understand
they say it integrates to
v = v(initial) * e^(-Kt/m)
can anyone explain how they got that? i am missing some math skillz here.
any other insights to the problem would be greatly appreciated.

 

[Astronuc]

The problem statement indicates that a resistive force a function of velocity, i.e. is proportional to velocity, then

force = -Kv, or -K$\dot{x}$, and not force = -Kx.

So the force balance becomes,

m dv/dt = -Kv which is a first order linear differential equation,

which gives the solution

v = v(initial) * e^(-Kt/m) for v=v(initial) at t=0.

 

[piercas]

I can offer some insights and explanations for this dynamics velocity problem. First of all, the resistive force being proportional to the block's velocity is known as a viscous drag force. This type of force is commonly seen in fluids, such as air or water, where the resistance to motion increases as the velocity of an object increases.

To solve this problem, we can use Newton's second law, F=ma, where F is the net force acting on the object, m is its mass, and a is its acceleration. In this case, the net force is the resistive force, -Kx, and the acceleration is dv/dt, the rate of change of velocity over time.

To integrate this equation, we can use the power rule of integration, which states that ∫xn dx = (1/n+1)xn+1 + C. Applying this to our equation, we get ∫dv = ∫-Kx/m dt. Integrating both sides gives us v = (-Kx/m)t + C. We can then solve for C by plugging in the initial conditions, where v= v(initial) when t=0. This gives us C= v(initial).

Thus, our final equation is v = v(initial) - (K/m)x. We can see that as x (or the distance) increases, v decreases, which makes sense since the resistive force is opposing the motion and slowing the object down.

To find the stopping distance, we can use the fact that when an object stops, its velocity is 0. So we can set v=0 in our equation and solve for x, giving us x= v(initial)/K. This tells us that the stopping distance is dependent on the initial velocity and the proportionality constant K.

Another way to approach this problem is by using the concept of exponential decay. This is where the velocity decreases exponentially over time, which is represented by the equation v= v(initial) * e^(-Kt/m). This equation is derived from the same integration process as before, but it is more commonly used in physics and engineering applications.

Overall, this dynamics velocity problem involves understanding the concept of resistive forces, using Newton's second law and the power rule of integration, and applying the concept of exponential decay. I hope this helps to clarify the problem and provides some insights into the solution.