How to find the Acceleration when Velocity depends on position?

• zvjaginsevfan
In summary: Yes, right at the start. You can't differentiate f(x) by writing it as ##x\frac{f(x)}x## then just differentiating the leading x to get ##\frac{f(x)}x##.
zvjaginsevfan
Homework Statement
An object is moving through x-axis with velocity v = 1/x. Find the time elapsed and accelartion of the object at the moment when x = 4m.
Relevant Equations
v = 1/ x
This is a homework question from my friend, I found the time but a tough differential equation occurred when I was trying to find accelaration, is there a simple solution for this?

Try finding ## a=\frac{dv}{dt} ## using the calculus chain rule. It is actually quite simple. ## \\ ## Edit: I also did it by solving the differential equation, but the first way is slightly easier.

Last edited:
zvjaginsevfan said:
Homework Statement: An object is moving through x-axis with velocity v = 1/x. Find the time elapsed and accelartion of the object at the moment when x = 4m.
Homework Equations: v = 1/ x

This is a homework question from my friend, I found the time but a tough differential equation occurred when I was trying to find accelaration, is there a simple solution for this?
Welcome to the PF.

Try finding ## a=\frac{dv}{dt} ## using the calculus chain rule. It is actually quite simple.
I used that formula. At the end I got dv = axdx, and as a also depends on x, I could not figure out how to find the a from here. Can you explain?

zvjaginsevfan said:
I used that formula. At the end I got dv = axdx, and as a also depends on x, I could not figure out how to find the a from here. Can you explain?

haruspex said:
I want to differentiate it w.r.t t but I think my calculus skills are rusty after holiday, so I went with this way:
1- dv/dx = dv/dt * dt / dx
2- v = 1/x
3-differentiating w.r.t x, dv/dx = -1/x^2 *dx
4- as dv/dt = a, I found -1/x^2*dx = a*dt/dx

-1/x^2*dx = a*dt/dx is where I stuck. I could not go further.

berkeman said:
Welcome to the PF.

1- dv/dx = dv/dt * dt / dx
2- v = 1/x
3-differentiating w.r.t x, dv/dx = -1/x^2 *dx
4- as dv/dt = a, I found -1/x^2*dx = a*dt/dx

-1/x^2*dx = a*dt/dx is where I stuck. I could not go further.

Sorry for flood, but I also found something with chain rule, but it turns out accelaration is a positive number in my case, which is illogical as 1/x is a decreasing function.

zvjaginsevfan said:
dv/dx = dv/dt * dt / dx
No, that's differentiating wrt x. Just do it wrt t.

haruspex said:
No, that's differentiating wrt x. Just do it wrt t.
I stated it above, I could not do it due to problem in retrieving calculus knowledge. It felt like 1/x is gone and there is 0 at the right side of the equation but probably it is not.

zvjaginsevfan said:
I stated it above, I could not do it due to problem in retrieving calculus knowledge. It felt like 1/x is gone and there is 0 at the right side of the equation but probably it is not.
Just use the chain rule, remembering that x is a variable. ##\frac{dx}{dt}=v##.

haruspex said:
Just use the chain rule, remembering that x is a variable. ##\frac{dx}{dt}=v##.
Four message above, there is an attached photo. I used chain rule and I used the equation you wrote but it turns out accelaration is a positive number in my case, which is illogical as 1/x is a decreasing function.

zvjaginsevfan said:
Four message above, there is an attached photo. I used chain rule and I used the equation you wrote but it turns out accelaration is a positive number in my case, which is illogical as 1/x is a decreasing function.
In post #8, though it is somewhat fuzzy, I still read dv/dx on the left. Just do d/dt to both sides of v=1/x.
And please do not post algebra as images. It is against forum guidelines.

haruspex said:
In post #8, though it is somewhat fuzzy, I still read dv/dx on the left. Just do d/dt to both sides of v=1/x.
And please do not post algebra as images. It is against forum guidelines.
So applying d/dt to both sides of v = 1/x
1- dv/dt = d/dt * x/x^2
2- a = dx/dt * 1/x ^2
3- a = v / x^2
4- a = 1/x^3
for x = 4, a is found 1/256 m/s^2 here. But shouldn't accelaration be negative, did I make a mistake in these steps?

zvjaginsevfan said:
But shouldn't accelaration be negative, did I make a mistake in these steps?
Yes, right at the start. You can't differentiate f(x) by writing it as ##x\frac{f(x)}x## then just differentiating the leading x to get ##\frac{f(x)}x##.
What is the derivative wrt x of x-1?

haruspex said:
Yes, right at the start.
What is the derivative wrt x of x-1?
It's -1/x^2 but you said you should do it wrt to t, am I missing something?

zvjaginsevfan said:
It's -1/x^2 but you said you should do it wrt to t, am I missing something?
You are applying the chain rule. ##\frac{d}{dt}f(x)=\frac{dx}{dt}\frac{d}{dx}f(x)##.

haruspex said:
You are applying the chain rule. ##\frac{d}{dt}f(x)=\frac{dx}{dt}\frac{d}{dx}f(x)##.
I finally solved it, thank you very much!

1. What is acceleration and how is it related to velocity and position?

Acceleration is the rate of change of velocity over time. It is a vector quantity, meaning it has both magnitude and direction. When velocity depends on position, acceleration is affected by both the change in position and the change in time.

2. How do you calculate acceleration when velocity depends on position?

To calculate acceleration in this scenario, you can use the formula a = (v2 - v1) / (x2 - x1), where v2 and v1 are the final and initial velocities, and x2 and x1 are the final and initial positions.

3. Can acceleration be negative when velocity depends on position?

Yes, acceleration can be negative when velocity depends on position. This indicates that the object is decelerating or slowing down in the direction of motion.

4. How can we determine the direction of acceleration when velocity depends on position?

The direction of acceleration can be determined by looking at the direction of change in velocity and position. If both are increasing in the same direction, then the acceleration is positive. If they are decreasing in the same direction, the acceleration is negative.

5. Can we find acceleration at a specific point when velocity depends on position?

Yes, acceleration can be calculated at a specific point when velocity depends on position. This can be done by taking the derivative of the velocity function with respect to position at that point. Alternatively, you can use the average acceleration formula by choosing two points on either side of the desired point and calculating the acceleration between them.

• Introductory Physics Homework Help
Replies
9
Views
426
• Introductory Physics Homework Help
Replies
13
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
19
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
828
• Introductory Physics Homework Help
Replies
17
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
1K