# Integrate Acceleration to Find Time?

• Trapezoid
In summary, the problem involves a particle being attracted towards a fixed point with a force inversely proportional to its distance from the point. The particle is released from rest at a certain distance and the question asks for the time it takes to reach the fixed point. The solution involves solving a differential equation and using a technique called "quadrature" to integrate. The final integral can be solved by substituting u = L/x and using the gamma function.
Trapezoid
This question is from my calculus assignment but I apologize if it belongs on the a physics board regardless.

## Homework Statement

A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If a particle is released from rest at a distance L from 0, how long will it take to reach 0?

2. The attempt at a solution
$$F = ma = \frac{k}{d}$$
where d is the distance from 0 and k is a proportionality constant. Therefore, $$a = \frac{d^2x}{dt^2} = \frac{k}{md}$$
I know that integration is required but I don't see how to use it to find time. Could somebody give me a tip as to how to proceed?

Thanks,
Trapezoid

Well, imagine that the question asked you for an equation that gave the position of the particle as a function of time. E.g. $s(t) = a_0 + a_1t + a_2t^2$. You know how to do this, right? Now, once you have the equation, just put the distance traveled on the LHS and solve for $t$.

Trapezoid said:
This question is from my calculus assignment but I apologize if it belongs on the a physics board regardless.

## Homework Statement

A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If a particle is released from rest at a distance L from 0, how long will it take to reach 0?

2. The attempt at a solution
$$F = ma = \frac{k}{d}$$
where d is the distance from 0 and k is a proportionality constant. Therefore, $$a = \frac{d^2x}{dt^2} = \frac{k}{md}$$
I know that integration is required but I don't see how to use it to find time. Could somebody give me a tip as to how to proceed?

Thanks,
Trapezoid
$$x\frac{d^2x}{dt^2} = \frac{k}{m}$$
d is x. Is that what you were forgetting?

You say "d is the distance from 0 and k is a proportionality constant" but you don't say what "x" is! I think you mean x to be also the distance from 0 at time t so your differential equation is really
$$\frac{d^2x}{dt^2}-= \frac{k}{mx}$$

You will have to solve the differential equation for x, then set it equal to 0 and solve for t.

Now, that is not easy to integrate directly but you can use a technique called "quadrature".
Let v= dx/dt so that $d^2x/dt^2= dv/dt$. By the chain rule, $dv/dt= (dv/dx)(dx/dt)= v dx/dt$. So your equation is
$$v\frac{dv}{dx}= \frac{k}{mx}$$
a separable equation.

$$v dv= \frac{k}{m}\frac{dx}{x}$$
Integrating,
$$(1/2)v^2= \frac{k}{m}ln(|x|)+ C$$
Since the object was released "distance L" (x= L) at rest (v= 0) we have
$$(1/2)0^2= \frac{k}{m}ln(L)+ C$$
so
$$C= -\frac{k}{m}ln(L)$$
and the equation for v becomes
$$\frac{1}{2}v^2= \frac{k}{m}ln(x)- \frac{k}{m}ln(L)= \frac{k}{m}ln\left(\frac{x}{L}\right)$$
(I am assuming that L is positive and as x changes from L to 0, it will be positive so we don't need the absolute value.)
$$v= \frac{dx}{dt}= \sqrt{\frac{2k}{m}ln\left(\frac{x}{L}\right)}$$

Integrate that, solve for x as a function of t, set it equal to 0, and solve for t.

Thanks for all the help everybody. I'll take a closer look to see if I can't find the answer from here.

I thought it would just be
$$x\frac{d^2x}{dt^2} = \frac{k}{m}$$
$$\int \int x dxdx = \frac{k}{m} \int \int dtdt$$
$$\frac{x^3}{6}=\frac{k}{m}\frac{t^2}{2}+c$$
$$x^3=3\frac{k}{m}t^2+c$$
Here is a nice junction to find c (assuming x = 0 is at the point and x = L is the initial condition, L > 0):
$$L^3 = 0 + c \rightarrow c = L^3$$
then
$$x^3=3\frac{k}{m}t^2+L^3$$
$$x(t)=\left ( 3\frac{k}{m}t^2+L^3 \right )^{1/3}$$

You then solve for the time for which x(t) = 0. Note, k must be < 0 for such a solution.

Thanks HallsofIvy and RoshanBBQ,

Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral $\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}$ which I cannot even begin to solve. With regards to RoshanBBQ's solution, it seems odd to me that k would be negative since constants in examples done in class are always specified to be positive. That being said, a negative k would imply that the force acts in the negative direction, which seems consistent with the problem. Is my thinking here correct?

Trapezoid

Trapezoid said:
Thanks HallsofIvy and RoshanBBQ,

Unless I've misunderstood, the method outlined by HallsofIvy leaves me with the integral $\int \frac{dx}{\sqrt{\ln(\frac{x}{L})}}$ which I cannot even begin to solve. With regards to RoshanBBQ's solution, it seems odd to me that k would be negative since constants in examples done in class are always specified to be positive. That being said, a negative k would imply that the force acts in the negative direction, which seems consistent with the problem. Is my thinking here correct?

Trapezoid

It just depends on how you set up the problem.I'm not entirely comfortable with the answer I found, though. Consider when t -> infinity. You would expect x -> 0. My solution has x become the 3rd root of an increasingly negative number...

Last edited:
Hi all,

Sorry to bump but we got the solutions to the assignment so I thought I'd post some more guidance in case anybody else is having trouble with a similar question.

The method outlined by HallsofIvy is correct. Without giving too much away, to solve the last integral, substitute $u = \frac{L}{x}$ and use the gamma function.

## 1. What is acceleration?

Acceleration is a measure of how quickly the velocity of an object changes over time. It is the rate of change of velocity, and is typically measured in meters per second squared (m/s^2).

## 2. How is acceleration related to time?

Acceleration is directly related to time through the equation a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time. This means that as time increases, acceleration also increases, assuming a constant change in velocity.

## 3. What is integration?

Integration is a mathematical process of finding the area under a curve. In the context of acceleration and time, it involves finding the area under an acceleration vs. time graph to determine the change in velocity over a specific time period.

## 4. How do you integrate acceleration to find time?

To integrate acceleration to find time, you must first have an acceleration vs. time graph. Then, you can use the formula Δv = aΔt to solve for Δt, which represents the change in time. This will give you the time it takes for the change in velocity to occur.

## 5. Why is integrating acceleration to find time important?

Integrating acceleration to find time is important because it allows us to determine how long it takes for a change in velocity to occur. This can be useful in various fields of science, such as physics and engineering, as it helps in understanding the motion of objects and predicting their behavior.

Replies
3
Views
1K
Replies
20
Views
971
Replies
12
Views
3K
Replies
3
Views
625
Replies
3
Views
1K
Replies
1
Views
914
Replies
1
Views
1K
Replies
9
Views
2K
Replies
6
Views
2K
Replies
1
Views
1K