How Does Acceleration as a Function of Position Affect Speed and Travel Time?

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Homework Help Overview

The problem involves understanding the acceleration of an object described by a function of position, specifically a(x) = bx, where b is a constant. The original poster seeks to determine the speed at a specific position and the time taken to travel between two points.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and position, with some referencing differential equations. There is mention of using sine and cosine functions in relation to simple harmonic motion. Questions arise regarding the application of the chain rule and its relevance to the problem.

Discussion Status

The discussion is ongoing, with participants exploring various mathematical approaches and clarifying concepts. Some guidance has been offered regarding the use of trigonometric functions and the chain rule, but no consensus has been reached on the best method to apply.

Contextual Notes

Participants note the requirement to utilize the chain rule in their solutions, which has led to questions about its relevance and application in this specific context. There is also a focus on the need for initial conditions to determine constants in the proposed solutions.

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Homework Statement




Acceleration of an object is described by a function of x, a(x) = bx.
(b is a constant 2 sec^-2)

1. If the speed at x=1m is zero, what is the speed at x=3m?

2. How long does the travel from x = 1m to x = 3m takes?


Homework Equations



chain rule : dr/dt = dr/ds*ds/dt


The Attempt at a Solution



I have thought of spring in a simple harmonic motion.
But I could not take any further steps... Somebody help me please!
 
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The first formula you give says that
x''(t) = b x(t).

This is a differential equation. Do you know of any functions who are their own derivative (maybe up to a constant numerical factor?)
 
I thought of sine and cosine fuction in a simple harmonic motion.
But how can I relate them to the chain rule?
 
Sine and cosine sounds good.
I don't know what you want to do with the chain rule, but you can just use
x(t) = A cos(t) + B sin(t)
then differentiate to get the velocity x'(t) and acceleration y''(t) and use that to determine the constants A and B.
 
CompuChip said:
Sine and cosine sounds good.
I don't know what you want to do with the chain rule, but you can just use
x(t) = A cos(t) + B sin(t)
then differentiate to get the velocity x'(t) and acceleration y''(t) and use that to determine the constants A and B.
What do you mean by y''(t)? Did you mean x(t)" ??

I also don't know what the chain rule has to do with this problem but the instruction was to utilize this rule when solving the problem =<
Could you please think a little bit more about the relevance of the chain rule to this problem?

Also,
Is there any reason that you included both cosine and sine fuction in the fuction of x(t)?
Originally I intended to use x(t) = Asin([tex]\omega[/tex]t + [tex]\phi[/tex])...
If there is any further advices or recommendations, please let me know

Thanks!
 
fdsadsfa said:
What do you mean by y''(t)? Did you mean x(t)" ??
Yes, that was a typo. I apologise.

fdsadsfa said:
I also don't know what the chain rule has to do with this problem but the instruction was to utilize this rule when solving the problem =<
Could you please think a little bit more about the relevance of the chain rule to this problem?
Hmm, maybe you have to use the chain rule with something like
[tex]\frac{da}{dx} \frac{dx}{dt}[/tex]

fdsadsfa said:
Also,
Is there any reason that you included both cosine and sine fuction in the fuction of x(t)?
Originally I intended to use x(t) = Asin([tex]\omega[/tex]t + [tex]\phi[/tex])...
Well, my reason was the following (rather technical point): x'' = a x is a second order equation (because there are two derivatives), so there must be two independent solutions. In this case, they can be easily found to be sin(x) and cos(x). The general solution to the equation is then A sin(x) + B cos(x), with A and B two constants which have to be determined from initial conditions (for example, at t = 0, x must equal 23 and x' must equal 0). You need two conditions to fix A and B.
In the expression you gave, there are also two undetermined constants, [itex]\omega[/itex] and [itex]\phi[/itex]. If you take [itex]\sin(\omega t + \phi)[/itex] you can use the two conditions to fix these constants.
It doesn't really matter which of the two you take, in the end they are equivalent. In fact, a cosine is nothing more than a sine which is shifted a bit horizontally so you can always write cos(t) in the form [itex]\sin(t + \phi)[/itex] by choosing [itex]\phi[/itex] properly.

I don't know how much you know about differential equations and related topics, but I hope that makes sense. I'll stick to [itex]\sin(\omega t + \phi)[/itex] :smile:
 

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