How can I Easily Integrate F = ma into My Solution?

[Gaidzahg]

Homework Statement

The following experiment is performed in a spacecraft that is at rest where the net gravitational field is zero. A small sphere is injected into a viscous medium with initial velocity v-naught. The sphere experiences a resistive force R = -bv. Find the velocity of the sphere as a function of time.

The solution I was able to come up with was v = e^-(b/m)t, using an indefinite integral. The given solution is v = v-naught*e^-(b/m)t. I'm assuming that it was achieved using a definite integral, but I can't seem to figure what limits of integration will get me the given solution.

Relevant Equations

F = ma --> -bv = ma
b is a constant

F = ma
-bv = ma
-bv/m = a
-bv/m = dv/dt
dt = -mdv/bv
∫dt = -m/b ∫dv/v
t = -m/b ln v
-(b/m)t = ln v
e^-(b/m)t = v

 

[PeroK]

What does your solution gives for the velocity at time $t = 0$?

What does the book solution give at $t = 0$?

 

[PeroK]

PS a definite integral will give you a displacement (integral of velocity over time). But, an indefinite integral ought to have a constant of integration!

 

[Gaidzahg]

What does your solution gives for the velocity at time $t = 0$?

What does the book solution give at $t = 0$?

The book only says that the initial velocity at time zero is v-naught (v₀). The books solution is literally just v=v₀e^-(b/m)t

 

[PeroK]

The book only says that the initial velocity at time zero is v-naught (v₀). The books solution is literally just v=v₀e^-(b/m)t

What does that solution give at time $t = 0$?

What does your solution give at time $t = 0$?

 

[Gaidzahg]

Maybe I'm not understanding you, or you are not understanding me. There is no solution given at time t = 0. The whole point of the problem is deriving the equation that shows the velocity as a function of time.

 

[PeroK]

Maybe I'm not understanding you, or you are not understanding me. There is no solution given at time t = 0. The whole point of the problem is deriving the equation that shows the velocity as a function of time.

Okay, but "solution as a function of time" is not some abstract thing, so if you plug in $t = 0$ you better get $v_0$. You must get $v(t=0) = v_0$.

You solution gives $v(0) = e^{-0} = 1$, which can't be right at all.

The book's solution gives $v(0) = v_0e^{-0} = v_0$, which is correct.

 

[PeroK]

Anyway, you also forgot about the constant of integration.

 

[Gaidzahg]

Exactly. What I'm trying to understand is how is the equation derived so that the v₀ appears in it. You are correct that the answer I came up with gives a nonsensical answer.

 

[PeroK]

Exactly. What I'm trying to understand is how is the equation derived so that the v₀ appears in it. You are correct that the answer I came up with gives a nonsensical answer.

You seem to have a blind spot when it comes to the constant of integration!

 

[PeroK]

∫dt = -m/b ∫dv/v
t = -m/b ln v

This is the step where the constant of integration is missing.

 

[Mark44]

∫dt = -m/b ∫dv/v
t = -m/b ln v

What @PeroK is more-or-less saying is that $\int \frac {dv}v = \ln(v) + \text{ constant }$.
Your 2nd line above does not include the constant of integration, and that is why your answer is coming out wrong.

 

[Gaidzahg]

Okay, so then after integration we get

t = -m/b ln v + C
t - C = -m/b ln v
-(t - C)b/m = ln v
(C-t)b/m = ln v
e^(C-t)b/m = v
(e^Cb/m) * (e^-tb/m) = v

does that look right?

 

[Mark44]

Okay, so then after integration we get

t = -m/b ln v + C
t - C = -m/b ln v
-(t - C)b/m = ln v
(C-t)b/m = ln v
e^(C-t)b/m = v
(e^Cb/m) * (e^-tb/m) = v

does that look right?

It's close but what you have is more complicated than is necessary. You can replace $e^{\frac{Cb}m}$ with $e^{C^*}$, where $C^* = \frac{Cb}m$. After all, a constant times b/m is just a different constant. There's no need to bring b/m along.

Since $v(0) = v_0$, can you simplify it some more?

Also, I would have done the integration a different way.
Starting with $\frac{-bv}m = \frac{dv}{dt}$, I would have separated the two sides like this:
$-\frac b m dt = \frac{dv}v$
$\Rightarrow \int \frac{dv}v = \int -\frac b m dt$ (Switching the sides of the equation above...)
$\Rightarrow \ln v = -\frac b m t + C$

I've assumed that $v(t) \ge 0$, so I didn't use absolute values for the antiderivative of dv/v. If that's an incorrect assumption, then absolute values would be needed.
$\Rightarrow v(t) = e^{-\frac b m t} \cdot e^C$

 

[PeroK]

Okay, so then after integration we get

t = -m/b ln v + C
t - C = -m/b ln v
-(t - C)b/m = ln v
(C-t)b/m = ln v
e^(C-t)b/m = v
(e^Cb/m) * (e^-tb/m) = v

does that look right?

It's not right because you are given $v(0) = v_0$ and (as in all such problems) you must use the initial conditions to specify the constant of integration. So, $v(0) = e^{Cb/m} = v_0$,

Also, you already know the correct solution is $v(t) = v_0e^{-tb/m}$. That tells you that $e^{Cb/m} = v_0$.

 

[Gaidzahg]

It's close but what you have is more complicated than is necessary. You can replace $e^{\frac{Cb}m}$ with $e^{C^*}$, where $C^* = \frac{Cb}m$. After all, a constant times b/m is just a different constant. There's no need to bring b/m along.

Since $v(0) = v_0$, can you simplify it some more?

Also, I would have done the integration a different way.
Starting with $\frac{-bv}m = \frac{dv}{dt}$, I would have separated the two sides like this:
$-\frac b m dt = \frac{dv}v$
$\Rightarrow \int \frac{dv}v = \int -\frac b m dt$ (Switching the sides of the equation above...)
$\Rightarrow \ln v = -\frac b m t + C$

I've assumed that $v(t) \ge 0$, so I didn't use absolute values for the antiderivative of dv/v. If that's an incorrect assumption, then absolute values would be needed.
$\Rightarrow v(t) = e^{-\frac b m t} \cdot e^C$

Well, since v₀ = v(0), then wouldn't v₀ = e^C? I guess that makes sense to me now. Thank you.

On another note, where can I learn how to use the hashtags and other code you used to make the equations look nice?

 

[Mark44]

Well, since v0 = v(0), then wouldn't v0 = e^C?

Yes, from the work that I showed.

On another note, where can I learn how to use the hashtags and other code you used to make the equations look nice?

At the lower left corner of the pane you type into, there's a link that says LaTeX Guide. It shows you what you need to get up and running.

 

[PeroK]

On another note, where can I learn how to use the hashtags and other code you used to make the equations look nice?

If you reply to any post in this thread (or any post in any thread on this site), you get the raw Latex that the poster typed in. If you like it, you can copy and paste that and then discard the reply!

 

[vela]

If you reply to any post in this thread (or any post in any thread on this site), you get the raw Latex that the poster typed in. If you like it, you can copy and paste that and then discard the reply!

You can also right-click on an equation and choose "Show Math As".