How Do You Calculate Position, Velocity, and Acceleration from d(x)/d(t) = c/x?

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In summary, the conversation discusses how to calculate position, velocity, and acceleration for a given time using the equation v=c/x and d(x)/d(t)=c/x. The conversation also includes a misunderstanding of how to integrate the equation, with the correct solution being x dx = c dt. It is suggested to use the equation ##\frac 1 2 x^2 = ct + K## to calculate x(t), v(t), and a(t).
  • #1
69911e
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1
I have the equation v=c/x
d(x)/d(t)=c/x
velocity= c(constant)/x (distance)

I need equations to calculate position, velocity and acceleration for a given time,
My attempt was:
xdx=cdt
x2ln(x)=ct
elnxx2=ekt
xx2=ekt
but I need x(t)v(t),a(t)

Any suggestions on how to calculate?
 
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  • #2
You start of well with ##xdx = cdt##
The next step is wrong. How do you justify ##\int xdx = x^2 \text{ln}(x)##?
Because that's what happens there.

If you know ##x(t)## how can you find ##v(t)## and ##a(t)##?

You've already used how you get ##v(t)##
 
  • #3
Where did the ln(x) come from??
 
  • #4
HomogenousCow said:
Where did the ln(x) come from??
I did int of XDx= x2*(Dx/x)=
x2LN(x)

That was the wrong move...

Thanks
 
  • #5
69911e said:
I have the equation v=c/x
d(x)/d(t)=c/x
Usually, posters don't include enough parentheses, but here you have more than you need, which might have led to some confusion.

The above can be written as ##\frac{dx}{dt} = \frac c x##. Separating, we get x dx = c dt. Integrating, we get ##\frac 1 2 x^2 = ct + K##.
69911e said:
velocity= c(constant)/x (distance)

I need equations to calculate position, velocity and acceleration for a given time,
My attempt was:
xdx=cdt
x2ln(x)=ct
elnxx2=ekt
xx2=ekt
but I need x(t)v(t),a(t)

Any suggestions on how to calculate?
 

FAQ: How Do You Calculate Position, Velocity, and Acceleration from d(x)/d(t) = c/x?

1. What is the meaning of "d(x)/d(t)=c/x"?

The equation "d(x)/d(t)=c/x" is a mathematical expression that represents the rate of change of a variable x with respect to time t. It indicates that the rate of change is directly proportional to the inverse of x and is equal to a constant value c.

2. How is this equation relevant to science?

This equation is relevant to science because it is commonly used to describe physical phenomena such as diffusion, radioactive decay, and other processes that involve a change in quantity over time. It is also used in various fields of science, including physics, chemistry, and biology, to model and analyze natural phenomena.

3. What does the variable x represent in this equation?

In this equation, the variable x represents the quantity that is changing over time. It could represent various physical quantities, such as concentration, temperature, or distance, depending on the specific application of the equation.

4. How is this equation solved?

This equation can be solved by taking the derivative of both sides with respect to time t. This will result in an expression with only x and t variables, which can then be rearranged to isolate for x. The solution will depend on the specific values of c and x at a given time t.

5. What is the significance of the constant c in this equation?

The constant c in this equation represents the proportionality constant between the rate of change and the inverse of x. It helps to determine the exact relationship between the two variables and can provide valuable information about the behavior of the system being studied.

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