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

  • Thread starter Thread starter fdsadsfa
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion centers on solving a physics problem involving acceleration as a function of position, specifically a(x) = bx, with b being a constant. Participants explore how to determine the speed at x=3m given that the speed at x=1m is zero, and how long the travel takes between these points. The conversation highlights the use of differential equations and the chain rule, with suggestions to apply sine and cosine functions to find solutions. Clarifications are made regarding the notation used and the relevance of the chain rule in the context of the problem. The thread emphasizes the importance of initial conditions in solving the differential equation.
fdsadsfa
Messages
4
Reaction score
0

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!
 
Physics news on Phys.org
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(\omegat + \phi)...
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
\frac{da}{dx} \frac{dx}{dt}

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(\omegat + \phi)...
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, \omega and \phi. If you take \sin(\omega t + \phi) 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 \sin(t + \phi) by choosing \phi 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 \sin(\omega t + \phi) :smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top