# DV/dS*v = a

1. Jan 8, 2013

### carhah

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?

2. Jan 8, 2013

### tiny-tim

welcome to pf!

hi carhah! welcome to pf!
no, v= 0 does not imply dv/dt = 0

(draw a graph of v against t)

3. Jan 8, 2013

### carhah

Re: dV/dS*v=a

but if we put v=0 in equation-dV/dS*v=a

we get a=0?

i am in doubt...

4. Jan 8, 2013

### carhah

Re: welcome to pf!

but if we put v=0 in equation-dV/dS*v=a

5. Jan 8, 2013

### tiny-tim

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 so = vo = 0, then s = 1/2 at2 and v = at,

so v = √(2as), so dv/ds|t=0 = √(a/so)= √(a/0) = ∞

6. Jan 8, 2013

### carhah

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. :)

7. Jan 8, 2013

### tiny-tim

no, you can always apply that formula …

but if it says a = ∞*0, that doesn't help very much!

8. Jan 8, 2013

### carhah

Re: dV/dS*v=a

can u draw graph of dv/ds ? and prove it is infinity?

9. Jan 8, 2013

### tiny-tim

you can draw a graph of v against s (as i've already asked you to) …

where it's vertical, dv/ds = ∞​