Resistance proportional with velocity problem

[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.

Homework Equations

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...

 

[Ray Vickson]

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.

Homework Equations

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 $x(t)$.

 

[Levilaon]

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.

 

[Ray Vickson]

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?

 

[Levilaon]

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.

 

[Ray Vickson]

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.

 

[Levilaon]

I think you got something wrong.

 

[Ray Vickson]

I think you got something wrong.

Show your work, so we can decide.

 

[Levilaon]

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.

 

[Levilaon]

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

 

[Ray Vickson]

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.

 

[Levilaon]

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.
I quote from your post:
"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.

 

[Ray Vickson]

Ok I see, but I don't see mass in your equation.
I quote from your post:
"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.