How to find the Acceleration when Velocity depends on position?

[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?

 

[Charles Link]

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.

 

[berkeman]

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.

Please show your best effort at working the problem. Once you show your work, we will be able to help you a lot more. Thanks.

 

[zvjaginsevfan]

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?

 

[haruspex]

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?

Please post your working. You started by differentiating v=1/x, yes?

 

[zvjaginsevfan]

Please post your working. You started by differentiating v=1/x, yes?

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.

 

[zvjaginsevfan]

Welcome to the PF.

Please show your best effort at working the problem. Once you show your work, we will be able to help you a lot more. Thanks.

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.

 

[zvjaginsevfan]

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.

 

[haruspex]

dv/dx = dv/dt * dt / dx

No, that's differentiating wrt x. Just do it wrt t.

 

[zvjaginsevfan]

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.

 

[haruspex]

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. $\frac{dx}{dt}=v$.

 

[zvjaginsevfan]

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.

 

[haruspex]

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.

 

[zvjaginsevfan]

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?

 

[haruspex]

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?

 

[zvjaginsevfan]

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?

 

[haruspex]

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. $\frac{d}{dt}f(x)=\frac{dx}{dt}\frac{d}{dx}f(x)$.

 

[zvjaginsevfan]

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!