Velocity/Acceleration relation w/ constants

[snovak216]

Homework Statement

The velocity of a particle is related to its position by: v² = w² (A² - x²) where w and A are constants. Show that the acceleration is given by: a=-w²x[/B]

Homework Equations

The Attempt at a Solution

a= v* dv/dt

v=(A²w²-x²w²)^1/2

dv/dt= 1/2 (A²w²-x²w²)^-1/2 * -2xw²

v * dv/dt = 1/2*-2xw² * (A²w²-x²w²)^1/2/ (A²w²-x²w²)^-1/2

-w²x

I think this is the proper way of showing the acceleration, but I'm not 100% with my chain rule and cancellations.

 

[haruspex]

a= v* dv/dt

I think not.

 

[snovak216]

I think not.

Sorry, I looked at v* dv/dx. Now definitely not sure how to go about it.

 

[haruspex]

Sorry, I looked at v* dv/dx. Now definitely not sure how to go about it.

What do you get if you simply differentiate the given equation wrt time?

 

[snovak216]

What do you get if you simply differentiate the given equation wrt time?

Wouldn't we have to differentiate wrt x? As t doesn't appear in the equation.

 

[haruspex]

Wouldn't we have to differentiate wrt x? As t doesn't appear in the equation.

Either will work, but you're right that differentiating wrt x is quicker here - so do it.

 

[snovak216]

After eliminating the ^2,

v= w * √(a^2 - x^2)

dv/dx= -xw/ √(a^2-x^2)

 

[haruspex]

After eliminating the ^2,

No need - just differentiate as is.

 

[snovak216]

I think I understand the derivation, but I do not understand the notation for v^2 and a

v^2*dv/dx
2v/dx? = 2w(a^2-x^2)+ w^2(2a -2x)

but since 2w and 2a are constants,

so 2v/dx?= w^2*-2x

dividing by 2

v/dx= -xw^2

and, however notation allows, the v/dx? is equivalent to a.

therefore a= -xw^2

 

[haruspex]

v^2*dv/dx

No, you want $\frac d{dx} v^2$. Apply the chain rule.

 

[snovak216]

2v * dv/dx= 0*(a^2-x^2)+ w^2*(0 -2x)= w^2*-2x

2v dv/dx = -2x*w^2

v dv/dx = -x*w^2

where
v dv/dx= a?

 

[haruspex]

2v * dv/dx= 0*(a^2-x^2)+ w^2*(0 -2x)= w^2*-2x

2v dv/dx = -2x*w^2

v dv/dx = -x*w^2

where
v dv/dx= a?

Yes.

 

[snovak216]

I appreciate the help, it's nice to be walked through the proper notation.