- #1
zenterix
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- Homework Statement
- Refer to example 2 of section 8.6. Use the chain rule to write
$$\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=v\frac{dv}{ds}$$
and thus show that the differential equation in the example can be expressed as follows
$$\frac{ds}{dv}=\frac{bv}{c-v}$$
where ##b=m/k## and ##c=gm/k##. Integrate this equation to express ##s## in terms of ##v##. Check your result with the formulas for ##v## and ##s## derived in the example
- Relevant Equations
- The cited example 2 is a bit large to be shown here in all its steps.
Here are the main points and equations.
A body of mass ##m## is dropped from rest from a great height in the earth's atmosphere. Assume it falls in a straight line and that the only forces acting on it are the earth's gravitational attraction and a resisting force due to air resistance which is proportional to its velocity.
Newton's second law tells us
$$ma=mg-kv$$
where ##k## is some positive constant and ##-kv## is the force due to air resistance.
$$mv'=mg-kv$$
is a first-order equation in velocity ##v##.
We can write this in the form
$$v'+\frac{k}{m}v=g$$
which we can solve using an integrating factor to obtain (assuming v(0)=0)
$$v(t)=e^{-kt/m}\int_0^t ge^{ku/m}du=\frac{mg}{k}(1-e^{-kt/m})$$
we can differentiate to find acceleration
$$a(t)=ge^{-kt/m}$$
We can also integrate to obtain position
$$s(t)=\frac{mg}{k}t+\frac{gm^2}{k^2}e^{-kt/m}+C$$
and if ##s(0)=0## we have
$$s(t)=\frac{mg}{k}t+\frac{gm^2}{k^2}(e^{-kt/m}-1)$$
Here is my solution to this problem. Unfortunately, I can't check it because it is not contained in the solution manual.
$$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}$$
$$\frac{ds}{dv}=\frac{v}{v'}=\frac{v}{ge^{-kt/m}}$$
$$=\frac{\frac{m}{k}v}{\frac{gm}{k}e^{-kt/m}}$$
$$=\frac{\frac{m}{k}v}{\frac{gm}{k}-\frac{gm}{k}+\frac{gm}{k}e^{-kt/m}}$$
$$=\frac{\frac{m}{k}v}{\frac{gm}{k}-\frac{gm}{k}(e^{-kt/m}-1)}$$
$$=\frac{bv}{c-v}$$
My main question is about the integration of this expression to obtain ##s## in terms of ##v##.
$$\int_0^v \frac{ds}{dv}dv=\int_{s(0)}^{s(v)} ds = s(v)-s(0)=\int_0^v\frac{bv}{c-v}dv$$
$$=bc(\ln{(c)}-\ln{(c-v)})-bv$$
$$s(v)=s(0)+bc(\ln{(c)}-\ln{(c-v)})-bv$$
I don't recall seeing this relationship very often and so I am not sure if this is correct. The problem says to check this result with the equations derived in the cited Example 2. But that example derived equations for ##v## and ##s## relative to ##t##. How would I go about using those equations to check my result of ##s## as a function of ##v##?
$$\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}$$
$$\frac{ds}{dv}=\frac{v}{v'}=\frac{v}{ge^{-kt/m}}$$
$$=\frac{\frac{m}{k}v}{\frac{gm}{k}e^{-kt/m}}$$
$$=\frac{\frac{m}{k}v}{\frac{gm}{k}-\frac{gm}{k}+\frac{gm}{k}e^{-kt/m}}$$
$$=\frac{\frac{m}{k}v}{\frac{gm}{k}-\frac{gm}{k}(e^{-kt/m}-1)}$$
$$=\frac{bv}{c-v}$$
My main question is about the integration of this expression to obtain ##s## in terms of ##v##.
$$\int_0^v \frac{ds}{dv}dv=\int_{s(0)}^{s(v)} ds = s(v)-s(0)=\int_0^v\frac{bv}{c-v}dv$$
$$=bc(\ln{(c)}-\ln{(c-v)})-bv$$
$$s(v)=s(0)+bc(\ln{(c)}-\ln{(c-v)})-bv$$
I don't recall seeing this relationship very often and so I am not sure if this is correct. The problem says to check this result with the equations derived in the cited Example 2. But that example derived equations for ##v## and ##s## relative to ##t##. How would I go about using those equations to check my result of ##s## as a function of ##v##?