Resistance proportional with velocity problem

• Levilaon
In summary: Can you try calculate time it would take it to come to 0 velocity. If your technique is valid you should get error or infty.EDIT: ok I did something bad, it never comes to infinity it goes to limes.Can you try calculate time it would take it to come to 0 velocity. If your technique is valid you should get error or infty.Ok with my calculations it comes again that time is infinite until velocity reach 0, what did you get?I solved it like this: v=f(t), v₁=50, v₂=10, m=10, Fr=f(t)²f(t)=v
Levilaon

Homework Statement

If we have object with mass 10 kg traveling at starting velocity of 50 km/s and one force of resistance that is equal to v^2 of object's velocity how can we calculate distance and time in which that object travels until it gets to velocity of 10 km/s.

The Attempt at a Solution

Kinetic energy at start - energy lost due to work of resistance = final kinetic energy
1/2 m v0^2 - ? = 1/2 m vf^2

As you can see I have trouble calculating work done by resistance. At start it will be starting velocity squared times distance but it changes every moment as velocity changes...

Levilaon said:

Homework Statement

If we have object with mass 10 kg traveling at starting velocity of 50 km/s and one force of resistance that is equal to v^2 of object's velocity how can we calculate distance and time in which that object travels until it gets to velocity of 10 km/s.

The Attempt at a Solution

Kinetic energy at start - energy lost due to work of resistance = final kinetic energy
1/2 m v0^2 - ? = 1/2 m vf^2

As you can see I have trouble calculating work done by resistance. At start it will be starting velocity squared times distance but it changes every moment as velocity changes...
Force = mass × acceleration, so if the force of resistance is ##k v^2## (for some constant ##k##) then we have ##m\, dv/dt = -k v^2##, a simple differential equation to determine ##v = v(t).## When you know ##v(t)## you can integrate to get ##x(t)##.

I solved it, first I wrote function v=f(t) then found ▲t by simply putting in final velocity. Then I integrated that function with domain ▲t to get distance.

Levilaon said:
I solved it, first I wrote function v=f(t) then found ▲t by simply putting in final velocity. Then I integrated that function with domain ▲t to get distance.

For the sake of interest, what did you get?

Ray Vickson said:
For the sake of interest, what did you get?

I got that time it took object to slow from 50 to10 km/s is 4 s and distance it passed during that time is 62.76 m.

Levilaon said:
I got that time it took object to slow from 50 to10 km/s is 4 s and distance it passed during that time is 62.76 m.

I got ##t = 4/5## sec and distance = ##10 \ln(5) \doteq 16.09## m.

I think you got something wrong.

Levilaon said:
I think you got something wrong.

Show your work, so we can decide.

Ray Vickson said:
I got ##t = 4/5## sec and distance = ##10 \ln(5) \doteq 16.09## m.
Can you try calculate time it would take it to come to 0 velocity. If your technique is valid you should get error or infty.
EDIT: ok I did something bad, it never comes to infinity it goes to limes.

Ray Vickson said:
Show your work, so we can decide.

Ok with my calculations it comes again that time is infinite until velocity reach 0, what did you get?

I solved it like this: v=f(t), v₁=50, v₂=10, m=10, Fr=f(t)²

f(t)=v₁-a·∆t => f(t)=v₁- (f(t)²/m)·∆t => (∆t/m)⋅f(t)²+f(t)-v₁ = 0

So when velocity is f(t)=v₂ then: ∆t=(-10⋅f(t)+500)/f(t)² = 4s

And for distance: s=∫{0 to 4}(-1+√(1+4⋅(∆t/m)⋅v₁)/(2⋅∆t/m) d∆t ≈ 62.76

Levilaon said:
Ok with my calculations it comes again that time is infinite until velocity reach 0, what did you get?

I solved it like this: v=f(t), v₁=50, v₂=10, m=10, Fr=f(t)²

f(t)=v₁-a·∆t => f(t)=v₁- (f(t)²/m)·∆t => (∆t/m)⋅f(t)²+f(t)-v₁ = 0

So when velocity is f(t)=v₂ then: ∆t=(-10⋅f(t)+500)/f(t)² = 4s

And for distance: s=∫{0 to 4}(-1+√(1+4⋅(∆t/m)⋅v₁)/(2⋅∆t/m) d∆t ≈ 62.76

Your solution makes no sense and is incorrect. Do you know what is meant by a differential equation? Do you know how to solve the differential equation for v(t) in this problem? That was given in post #2.

If you do not know how to solve a differential equation, you can develop a finite-difference scheme to solve the problem approximately. For small ##\Delta t > 0## we have ##v(t+\Delta t) \doteq v(t) - \Delta t \cdot v(t)^2## (which becomes more and more nearly exact as ##\Delta t## gets smaller and smaller), so
$$\begin{array}{rcl} v(\Delta t) & \doteq &50 - \Delta t \cdot 50^2 \\ v(2 \Delta t)&\doteq& v(\Delta t) - \Delta t \cdot v(\Delta t)^2 \\ v(3 \Delta t) &\doteq& v(2 \Delta t) - \Delta t \cdot v(2 \Delta t)^2 \\ \vdots &\vdots & \vdots \\ v(N \Delta t) &\doteq & v((N-1) \Delta t) - \Delta t \cdot v((N-1) \Delta t)^2 \end{array}$$
Be warned, however, that we need to take ##\Delta t ## very small to get any kind of decent accuracy. For example, we can take ##\Delta t = 1/1000 = 0.001## and ##N = 1000##, to get a good numerical solution over the interval ##0 \leq t \leq 1##, that is, at values ##t = 0, 0.001, 0.002, \ldots, 0.999, 1.000.## You could do all this in a spreadsheet.

There is also a simple exact formula for the solution ##v(t)##, but I think PF rules forbid me from telling you what it is.

Below is a plot of the 1000-point curve of ##v(t)##, and the constant function v=10; you can see that ##v(t)## reaches 10 at about ##t = 0.8## sec. That is exact when we use the true formula for the exact solution.

BTW: the solution never reaches 0 exactly, just as you claimed.

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Ray Vickson said:
Your solution makes no sense and is incorrect. Do you know what is meant by a differential equation? Do you know how to solve the differential equation for v(t) in this problem? That was given in post #2.

If you do not know how to solve a differential equation, you can develop a finite-difference scheme to solve the problem approximately. For small ##\Delta t > 0## we have ##v(t+\Delta t) \doteq v(t) - \Delta t \cdot v(t)^2## (which becomes more and more nearly exact as ##\Delta t## gets smaller and smaller), so
$$\begin{array}{rcl} v(\Delta t) & \doteq &50 - \Delta t \cdot 50^2 \\ v(2 \Delta t)&\doteq& v(\Delta t) - \Delta t \cdot v(\Delta t)^2 \\ v(3 \Delta t) &\doteq& v(2 \Delta t) - \Delta t \cdot v(2 \Delta t)^2 \\ \vdots &\vdots & \vdots \\ v(N \Delta t) &\doteq & v((N-1) \Delta t) - \Delta t \cdot v((N-1) \Delta t)^2 \end{array}$$
Be warned, however, that we need to take ##\Delta t ## very small to get any kind of decent accuracy. For example, we can take ##\Delta t = 1/1000 = 0.001## and ##N = 1000##, to get a good numerical solution over the interval ##0 \leq t \leq 1##, that is, at values ##t = 0, 0.001, 0.002, \ldots, 0.999, 1.000.## You could do all this in a spreadsheet.

There is also a simple exact formula for the solution ##v(t)##, but I think PF rules forbid me from telling you what it is.

Below is a plot of the 1000-point curve of ##v(t)##, and the constant function v=10; you can see that ##v(t)## reaches 10 at about ##t = 0.8## sec. That is exact when we use the true formula for the exact solution.

BTW: the solution never reaches 0 exactly, just as you claimed.
Ok I see, but I don't see mass in your equation.
"v(Δt)≐50−Δt⋅50^2
v(2Δt)≐v(Δt)−Δt⋅v(Δt)^2
..."
I believe it should be modified like this, after all acceleration is force/mass.
v(Δt)≐50−Δt⋅(50^2)/m
v(2Δt)≐v(Δt)−Δt⋅(v(Δt)^2)/m
...
Tell me if I am mistaken.

Levilaon said:
Ok I see, but I don't see mass in your equation.
"v(Δt)≐50−Δt⋅50^2
v(2Δt)≐v(Δt)−Δt⋅v(Δt)^2
..."
I believe it should be modified like this, after all acceleration is force/mass.
v(Δt)≐50−Δt⋅(50^2)/m
v(2Δt)≐v(Δt)−Δt⋅(v(Δt)^2)/m
...
Tell me if I am mistaken.

You are correct.

The graph I attached is correct because it used the correct finite-difference method.

1. How does resistance change with velocity?

The resistance is directly proportional to the velocity. This means that as the velocity increases, the resistance also increases. This relationship is represented by the equation R = kv, where k is a constant value determined by the properties of the medium through which the object is moving.

2. Why does resistance increase with velocity?

The increase in resistance with velocity is due to the increased amount of energy needed to overcome the drag force of the medium. As an object moves faster, it creates more disturbance in the medium, resulting in a larger amount of drag force acting on it. This requires the object to use more energy to maintain its velocity, resulting in a higher resistance.

3. How does the medium affect resistance and velocity?

The properties of the medium, such as density and viscosity, directly affect the resistance and velocity of an object moving through it. A denser medium will create more drag force and therefore increase resistance, while a less viscous medium will result in lower resistance and allow for higher velocities.

4. What is the relationship between resistance and acceleration?

Resistance does not directly affect acceleration. However, since resistance is proportional to velocity, it can indirectly affect acceleration. As an object accelerates, its velocity increases, resulting in a higher resistance. This means that more energy is needed to maintain the acceleration, potentially slowing down the object's acceleration rate.

5. How does resistance impact the motion of an object?

Resistance can significantly affect the motion of an object, especially at high velocities. The higher the resistance, the more energy is needed to overcome it and maintain the object's motion. This means that a higher resistance can slow down an object's velocity and potentially bring it to a stop. Additionally, resistance can also cause objects to change direction or experience turbulence, further impacting their motion.

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