Is the Hypothesized Law of Gravity for a Falling Body's Speed Accurate?

[Anthony Salls]

Homework Statement

The speed of a falling body might be based on the observation that the velocity of a falling object seems to increase the further it has fallen. Model the hypothesis "The speed of a falling object is proportional to the distance it has fallen" as a differential equation initial value problem. By analyzing the predictions of your model, explain why this "law of gravity" could not be correct.

Homework Equations

N/A

The Attempt at a Solution

So, obviously, v(t)=v0+k*d(t), where k is the proportionality constant.
d(t) = (v0+v(t)/2)*t
But plugging in d(t) and solving yields -(2*v0 + k v[0])/(-2 + 1*k) which if v0 = 0 is always 0 and there are restrictions on k due to the numerator. Also, it's not a differential equation and I'm not sure how to get one. I tried setting dv/dt = a and solving that way but it yields a nonsensical equation for v(t).

So my question is, how do I set up the model?

 

[mfb]

d(t) = (v0+v(t)/2)*t

How did you get this?

v0=0 is a problematic starting condition. I would exclude it here and consider other cases. You don't need it for the differential equation anyway.

 

[LCKurtz]

Homework Statement

The speed of a falling body might be based on the observation that the velocity of a falling object seems to increase the further it has fallen. Model the hypothesis "The speed of a falling object is proportional to the distance it has fallen" as a differential equation initial value problem. By analyzing the predictions of your model, explain why this "law of gravity" could not be correct.

Homework Equations

N/A

The Attempt at a Solution

So, obviously, v(t)=v0+k*d(t), where k is the proportionality constant.
d(t) = (v0+v(t)/2)*t
But plugging in d(t) and solving yields -(2*v0 + k v[0])/(-2 + 1*k) which if v0 = 0 is always 0 and there are restrictions on k due to the numerator. Also, it's not a differential equation and I'm not sure how to get one. I tried setting dv/dt = a and solving that way but it yields a nonsensical equation for v(t).

So my question is, how do I set up the model?

If you call the position at time $t$ by $s(t)$, then $v(t) = s'(t)$. Start by expressing "The speed of a falling object is proportional to the distance it has fallen" as an equation in $s(t)$.

 

[Anthony Salls]

How did you get this?

v0=0 is a problematic starting condition. I would exclude it here and consider other cases. You don't need it for the differential equation anyway.

It's the average velocity which would have units of distance/ time multiplied by time so it would be the distance as a function of time.

If you call the position at time $t$ by $s(t)$, then $v(t) = s'(t)$. Start by expressing "The speed of a falling object is proportional to the distance it has fallen" as an equation in $s(t)$.

So s(t) = c(v(t)-v(0)) (rearrangement of v(t) = v(0) + k*s(t) with 1/k = c )
Alright, so if I'm understanding this correctly ds(t)/dt = v(t), and therefore ∫ds =∫v(t)dt so v(t) = v(0)+c_inte^(c*t) then log[v(t)] = log[v(0)]+c*t and therefore v(t) = v(0) + e^(c*t)
and the reason this is a poor model is because the acceleration is not constant because it is tied to e^t and therefore overestimates the acceleration of the object.

 

[LCKurtz]

If you call the position at time $t$ by $s(t)$, then $v(t) = s'(t)$. Start by expressing "The speed of a falling object is proportional to the distance it has fallen" as an equation in $s(t)$.

So s(t) = c(v(t)-v(0))

That is not a correct rendition of the above quote in terms of $s$. Try again.

 

[Anthony Salls]

I don't understand why it isn't. Forget the initial velocity? But that doesn't work either. ||v(t)|| ∝ s(t)-s(0),
so v(t) = k*(s(t)-s(0)). -> v(t)/k +s(0) = s(t), s'(t) = v(t). -> d/dt(v(t)/k + s(0)) = v(t), v(t) -> 0

EDIT: okay so I took the integral of v(t) and set it equal to v(t)/k +s(0) and solved for v(t), which = k*s(0)/(-1+k*t) but this has weird things going on like the initial position cannot equal 0 or else the velocity is 0, and k*t cannot equal 1

 

[LCKurtz]

I don't understand why it isn't. Forget the initial velocity? But that doesn't work either. ||v(t)|| ∝ s(t)-s(0),
so v(t) = k*(s(t)-s(0)). -> v(t)/k +s(0) = s(t), s'(t) = v(t). -> d/dt(v(t)/k + s(0)) = v(t), v(t) -> 0

Remember $v(t) = s'(t)$ so the equation I have highlighted in red can be written $s'(t) = k(s(t)-s(0))$. Solve that DE for $s(t)$ and you will also know $v(t)$. Then you will have the correct solution to analyze.

EDIT: okay so I took the integral of v(t) and set it equal to v(t)/k +s(0) and solved for v(t), which = k*s(0)/(-1+k*t) but this has weird things going on like the initial position cannot equal 0 or else the velocity is 0, and k*t cannot equal 1

That makes no sense to me. How can you take the integral of v(t) when you don't know what it is?

 

[mfb]

It's the average velocity which would have units of distance/ time multiplied by time so it would be the distance as a function of time.

It is not the average velocity, in particular not if you don't even know the acceleration profile. There is an easy formula for a constant acceleration (this formula is different from what you wrote), but you do not have a constant acceleration.