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

In summary: 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
  • #1
Anthony Salls
3
0

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?
 
Physics news on Phys.org
  • #2
Anthony Salls said:
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
Anthony Salls said:

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
mfb said:
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.

LCKurtz said:
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 said:
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 said:
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
Anthony Salls said:
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
Anthony Salls said:
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 to Is the Hypothesized Law of Gravity for a Falling Body's Speed Accurate?

1. What is the equation for calculating the position of a falling body?

The equation for calculating the position of a falling body is:
x(t) = x0 + v0t + (1/2)at2
where x(t) is the position at time t, x0 is the initial position, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time.

2. How is the velocity of a falling body affected by air resistance?

The velocity of a falling body is affected by air resistance, also known as drag, in the opposite direction of motion. As the body falls, it accelerates due to gravity until the air resistance force increases to match the force of gravity. At this point, the body reaches its terminal velocity, which is constant and depends on the body's shape and size.

3. What is the difference between free fall and controlled fall?

In free fall, a body is falling under the sole influence of gravity and there is no air resistance or other forces acting upon it. In a controlled fall, there is some external force, such as air resistance or propulsion, that affects the body's motion. Examples of controlled falls include skydiving and bungee jumping.

4. How does the mass of a falling body affect its acceleration?

The mass of a falling body does not affect its acceleration due to gravity. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In the case of a falling body, the only force acting on it is gravity, which is constant regardless of the body's mass. Therefore, all objects in free fall will accelerate at the same rate, regardless of their mass.

5. Can the trajectory of a falling body be affected by external forces?

Yes, the trajectory of a falling body can be affected by external forces such as air resistance or a push or pull from another object. These forces can cause the body to deviate from a straight free-fall trajectory and result in a curved or angled path. Additionally, the angle at which the body is dropped can also affect its trajectory.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
475
  • Calculus and Beyond Homework Help
Replies
1
Views
434
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Classical Physics
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
Replies
3
Views
995
Replies
0
Views
509
  • Calculus and Beyond Homework Help
Replies
10
Views
6K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
Replies
7
Views
3K
Back
Top