Modeling a falling body

  • #1

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?
 

Answers and Replies

  • #2
34,946
11,127
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.
 
  • #3
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766

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)##.
 
  • #4
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)+cinte^(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.
 
  • #5
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
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.
 
  • #6
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
 
Last edited:
  • #7
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766
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?
 
Last edited:
  • #8
34,946
11,127
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.
 

Related Threads on Modeling a falling body

  • Last Post
Replies
7
Views
2K
Replies
1
Views
17K
  • Last Post
Replies
10
Views
9K
Replies
7
Views
10K
Replies
3
Views
2K
Replies
7
Views
555
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
9
Views
5K
  • Last Post
Replies
8
Views
9K
Top