Have dv/dx need to find dx/dt ? (have velocity w/ respect displace - find Accel?)

Hi all,
I'm trying to figure out this problem related to motion of a spring. I know the solution but I don't know how to make it work.
Thanks in advance (Nod)

I'm given velocity (v) = $\displaystyle \sqrt{25-x^2}$ where 'x' is displacement

Normally velocity is a function of time, so this one has really thrown me.

I want to find a formula for Acceleration $\displaystyle \frac{dv}{dt}$ which would normally be quite easy, but the above formula is Velocity as a function of x.

I've done a bit of related rates & differentiating w/ respect other variables, but I can't figure that out for this one....

If Acceleration = $\displaystyle \frac{dv}{dt}$ = $\displaystyle \frac{dv}{dx}$ x $\displaystyle \frac{dx}{dt}$

and I know $\displaystyle \frac{dv}{dx}$ (by differentiating the provided equation: ---> $\displaystyle \frac{dv}{dx} = \frac{-x}{\sqrt{25-x^2}}$
then how do I find $\displaystyle \frac{dx}{dt}$ please?

(btw, using trial and error on my graphics calculator, I think the Acceleration in this case is actually inverse to the displacement).

Quote:

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Originally Posted by dtb

Hi all,
I'm trying to figure out this problem related to motion of a spring. I know the solution but I don't know how to make it work.
Thanks in advance (Nod)

I'm given velocity (v) = $\displaystyle \sqrt{25-x^2}$ where 'x' is displacement

Normally velocity is a function of time, so this one has really thrown me.

I want to find a formula for Acceleration $\displaystyle \frac{dv}{dt}$ which would normally be quite easy, but the above formula is Velocity as a function of x.

I've done a bit of related rates & differentiating w/ respect other variables, but I can't figure that out for this one....

If Acceleration = $\displaystyle \frac{dv}{dt}$ = $\displaystyle \frac{dv}{dx}$ x $\displaystyle \frac{dx}{dt}$

and I know $\displaystyle \frac{dv}{dx}$ (by differentiating the provided equation: ---> $\displaystyle \frac{dv}{dx} = \frac{-x}{\sqrt{25-x^2}}$
then how do I find $\displaystyle \frac{dx}{dt}$ please?

(btw, using trial and error on my graphics calculator, I think the Acceleration in this case is actually inverse to the displacement).

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$\displaystyle \frac{dx}{dt}=v$ The derivative of position w.r.t is velcoity (Cool)

Thanks TES.

I think this is when I realise I've been looking at this for too long (Headbang)