Understanding Derivatives, velocity and acceleration

1. Jan 16, 2009

intenzxboi

I'm having a really hard time understanding this. So far this is what i think is correct.

V(average)= (X1-X0) / change in time

the derivative of V(average)= gives me the instant velocity ???

but at the same time say i have x(t), then the derivative of that is also the instant velocity???

A(average)= change in v / change in time

and if i take the derivative of A(average) i get instant Acceleration..

but then my teacher said that the derivative of V(average) = instant acceleration..

So does that mean that instant velocity= to instant acceleration??

2. Jan 16, 2009

Delphi51

No. It doesn't make sense to take the derivative of average velocity. The average would normally be over the whole time of the motion so it is a constant that does not vary with time - and its derivative would always be zero.

If you define avg V as (x(t) - x(0))/t so it DOES vary with time, then its derivative would be some very complicated quantity that doesn't have a name.

Yes, this is true. dx/dt is the instantaneous velocity or slope on the distance vs time graph.

Same problem here. Instantaneous acceleration is the derivative of instantaneous velocity. It doesn't make sense to differentiate averages.

3. Jan 16, 2009

Staff: Mentor

The average velocity (not its derivative) will equal the instantaneous velocity in the limit as Δt goes to zero.

Vave = Δx/Δt

When Δt → 0, Δx/Δt → dx/dt = V (instantaneous)

4. Jan 16, 2009

intenzxboi

o ok thank i think i got it

so unless average=to instant then the derivative does not apply in this case.