respect displace - find Accel?)

Printable View

Apr 6th 2009, 08:28 PM
dtb

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) = $\sqrt{25-x^2}$ 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 $\frac{dv}{dt}$ 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 = $\frac{dv}{dt}$ = $\frac{dv}{dx}$ x $\frac{dx}{dt}$

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

(btw, using trial and error on my graphics calculator, I think the Acceleration in this case is actually inverse to the displacement).
Apr 6th 2009, 08:33 PM
TheEmptySet

Quote:

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) = $\sqrt{25-x^2}$ 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 $\frac{dv}{dt}$ 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 = $\frac{dv}{dt}$ = $\frac{dv}{dx}$ x $\frac{dx}{dt}$

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

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

$\frac{dx}{dt}=v$ The derivative of position w.r.t is velcoity (Cool)
Apr 6th 2009, 08:57 PM
dtb

Thanks TES.

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