Differential Eqn: Find an expression of the velocity of a rock.

In summary, a 20 kg rock falls with the force of gravity acting on it. Another force acts on the rock (same direction as gravity) that is proportional to the velocity squared, 100v^2. However, there is no information given about the units or the initial velocity. Therefore, it is not possible to accurately determine the rock's velocity at any time t.
  • #1
coolusername
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Homework Statement



A 20 kg rock falls with the force of gravity acting on it. Another force acts on the rock (same direction as gravity) that is proportional to the velocity squared, 100v^2. Come up with an expression of the rock's velocity at any time t.

Homework Equations



Hint: (d/dx)(tan^-1(x)) = 1/(1 + x^2)


The Attempt at a Solution



F(net) = ma = Fg + 100v^2

ma = mg + 100v^2
a = g +(100v^2)/m (m=20)
dv/dt = g + 5v^2
dv/dt = g((5/g)v^2 + 1)
dv/((5/g)v^2 + 1) = g(dt)
dv/((root(5/g)*v)^2 +1) = g(dt) (integrate both sides)
tan^-1(root(5/g)*v) = gt + C
root(5/g)*v = tan(gt + C)

v(t) = (root(g/5))tan(gt + C)

I want to know if I can get rid of the constant as well as if my approach to the general solution is correct. Also if my general soln is correct.

Thanks!
 
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  • #2
You cannot ignore units like that. 20 and 20kg are completely different things.
The force given as 100v^2 has mismatching units, are you sure there is nothing else given there?

You cannot get rid of the constant as you do not know the initial velocity.
 
  • #3
That's all that the question gives. There's no initial conditions given either. I assumed that the additional force would equal 100v^2 as it says it's proportional. But you're right, the units don't match up.. Could it be possible that the units are hidden within the coefficient of v^2?
 
  • #4
coolusername said:
Could it be possible that the units are hidden within the coefficient of v^2?
Yes, there has to be a dimensional coefficient with the proper units.

Normally the force proportional to v2 would be something like the drag force, which acts in the direction opposite to the velocity, hence against gravity in the example. But oh well.
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. In physics, it is commonly used to model the behavior of systems that involve changes in time or space.

2. How does a differential equation relate to the velocity of a rock?

The velocity of a rock can be described using a differential equation because it involves the rate of change of the rock's position over time. By taking the derivative of the position function, we can obtain an expression for the velocity of the rock.

3. What information do we need to find an expression for the velocity of a rock using a differential equation?

In order to find an expression for the velocity of a rock, we need to know the initial position and velocity of the rock, as well as any external forces acting on it.

4. Can we use a differential equation to find the exact velocity of a rock at a specific time?

Yes, by solving the differential equation and substituting in the initial conditions, we can obtain an expression for the velocity of the rock at any given time.

5. Are there any limitations to using a differential equation to describe the velocity of a rock?

While differential equations are a powerful tool for modeling physical systems, they are based on simplifying assumptions and may not always accurately describe real-world situations. Additionally, solving some differential equations may require advanced mathematical techniques.

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