Deriving acceleration from velocity

AI Thread Summary
The discussion focuses on deriving the acceleration of an asteroid moving toward the sun, given its velocity as a function of distance from the sun. Participants clarify that to find acceleration, one must apply the chain rule, recognizing that velocity is a function of position, which in turn is a function of time. The correct approach involves differentiating velocity with respect to position and then using the relationship between velocity and position to express acceleration. Confusion arises regarding the relationship between velocity and time, but it is established that all quantities are functions of time, even if not explicitly stated. Ultimately, the acceleration can be expressed in terms of position, leading to the conclusion that a = -c2 / 2x².
uchicago2012
Messages
74
Reaction score
0

Homework Statement


An asteroid moves toward the sun along a straight line with a velocity given by v = -(c1+c2/x)½ , where x is the distance from the center of the sun. Use the chain rule for derivatives to show that the asteroid's acceleration is a = -c2 /2x2.

The Attempt at a Solution


I'm not sure what c1 and c2 are, I assumed constants, but just throwing that out there. I also assumed I was supposed to derive the velocity with respect to x in order to get the given acceleration. Unfortunately, I don't get the given acceleration. I suspect it is a simplification problem, or perhaps I've just forgotten how to do calculus.

v = -(c1 + c2/ x)1/2
dv/dt = -1/2(c1 + c2/ x)-1/2 * ((x-c1 +c2)/x2)
 
Physics news on Phys.org
uchicago2012 said:

Homework Statement


The Attempt at a Solution


I'm not sure what c1 and c2 are, I assumed constants

Yeah, exactly

uchicago2012 said:
I also assumed I was supposed to derive the velocity with respect to x in order to get the given acceleration.

Differentiate the velocity with respect to x. And yes, that is a necessary step, but that alone does not give you the acceleration.

A couple of things to note:

v is a function of x, which can be written v(x)
x is a function of time x(t).

Hence, the chain rule says:

\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt}​

As a first step, we need to calculate dv/dx (which you've attempted). Since the expression v(x) can itself be expressed as a compound function v(u(x)), you need to use the chain rule on that, meaning compute the derivative of the "outside function" v = -u1/2, and then multiply it by the derivative of the "inside function" (u(x) = c1 + c2/x). The derivative of the outside function, you already got:

-1/2(c1 + c2/x)-1/2

What's the derivative of the inside function? (You screwed this up)
Hints:

1. The derivative of a sum is the sum of the derivatives
2. The derivative of a constant is zero
 
Ah, calculus.

So I differentiated again and now the derivative of the velocity I got was:

c2/2x2 (c1 + c2/x)-1/2

I'm fairly certain that derivative is correct now. Isn't the derivative of velocity the acceleration? I don't understand the relation between the derivative of velocity that I got and the acceleration given.
 
I've thought about it further and I think I've only further confused myself.

cepheid said:
v is a function of x, which can be written v(x)
x is a function of time x(t).

which means I found dv/dx, which is decidedly not acceleration, since that is dv/dt, according to the definition

But v isn't a function of time and it's called a velocity, which is defined as dx/dt, so I'm confused on that part. Also, how is x a function of time? Do you just mean in general? Because time isn't mentioned in this problem, which confuses me.

I see how acceleration = dv/dt = dv/dx * dx/dt, according to the chain rule, but I'm not sure what to do with it. As far as I can tell, I've found the dv/dx part of the equation by differentiating the given equation with respect to x and I apparently need the dv/dt part for the acceleration, which would imply that I should then integrate the given equation to find x(of something unclear) but that integration is unhelpful.
 
uchicago2012 said:
which means I found dv/dx, which is decidedly not acceleration, since that is dv/dt, according to the definition

Yes

uchicago2012 said:
But v isn't a function of time and it's called a velocity, which is defined as dx/dt, so I'm confused on that part.

Velocity is a function of time here. You just aren't given the functional form (i.e. you don't know HOW it depends upon time, which doesn't matter, because you don't need to know that.)

Yes, v = dx/dt. What are you confused about specifically?

uchicago2012 said:
Also, how is x a function of time? Do you just mean in general? Because time isn't mentioned in this problem, which confuses me.

Well maybe time isn't mentioned explicitly, but the problem does say that the asteroid MOVES TOWARD THE SUN, which means that its position would depend upon when you look. Its position is changing with time. The point is, all of these quantities, x, v, and a, are functions of time, and you don't know ANY of these functions i.e. you don't know how any of those quantities depends upon time. But it doesn't matter, because you're given how velocity depends upon position, which means, with the help of the chain rule, you can express everything in terms of position as the independent variable. Observe:

a = dv/dt = dv/dx * dx/dt

Now, dx/dt is just v. And we know how v depends on x, v(x), because it is given in the problem. We also know dv/dx, since we computed it. Therefore, we know everything in terms of x, and we can write:

a(x) = dv/dx * v(x)

Our final answer is going to express how the acceleration of the asteroid depends upon its position.
 
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 '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