Understanding Derivatives, velocity and acceleration

[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??

 

[Delphi51]

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

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.

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

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

derivative of V(average) = instant acceleration..

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

 

[Doc Al]

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 ?

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

V_ave = Δx/Δt

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

 

[intenzxboi]

o ok thank i think i got it

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