# Velocity/Acceleration relation w/ constants

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1. Feb 16, 2015

### snovak216

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

2. Feb 16, 2015

### haruspex

I think not.

3. Feb 16, 2015

### snovak216

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

4. Feb 16, 2015

### haruspex

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

5. Feb 16, 2015

### snovak216

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

6. Feb 16, 2015

### haruspex

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

7. Feb 16, 2015

### snovak216

After eliminating the ^2,

v= w * √(a^2 - x^2)

dv/dx= -xw/ √(a^2-x^2)

8. Feb 16, 2015

### haruspex

No need - just differentiate as is.

9. Feb 16, 2015

### 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

10. Feb 16, 2015

### haruspex

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

11. Feb 16, 2015

### 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?

12. Feb 16, 2015

### haruspex

Yes.

13. Feb 17, 2015

### snovak216

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