dV/dS*v=a now at the higest point when we throw a ball. v=0 which implies a=0 but that is npot true....any expanations?
welcome to pf! hi carhah! welcome to pf! no, v= 0 does not imply dv/dt = 0 (draw a graph of v against t)
hi carhah! (you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in ) but dv/ds = ∞ (draw the graph of v against s) eg, if a is constant, and if s_{o} = v_{o} = 0, then s = 1/2 at^{2} and v = at, so v = √(2as), so dv/ds|_{t=0} = √(a/s_{o})= √(a/0) = ∞
Re: dV/dS*v=a thanks tim :D BUT DOES that mean that we cannot apply that formula a=dv/ds*v in case of constant acceleration. :)