Recent content by maunder

dont worry, i got it. a=\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=v\frac{dv}{ds} \frac{dv}{ds}=\frac{-4s}{\sqrt{A-4s^{2}}} v\frac{dv}{ds}=-4s=a I'd tried this before, but i must of stuffed it up. Thanks guys. I can't beleave how easy it was.

i don't understand?? how would i express s in terms of t

the integral i get is \frac{-ln(2s-\sqrt{A})}{4\sqrt{A}}+\frac{ln(2s+\sqrt{A})}{4\sqrt{A}} by using integration by partial fractions or \frac{1}{2}tan^{-1}\frac{2s}{\sqrt{a-4s^{2}}} The function isn't wrong, the velocity is given in terms of its position. I'm thinking what will happen is...

Homework Statement The question is: The velocity of a particle is given as a function of its position s: v = (A - 4 s2)1/2 where s is displacement (in m), and v is velocity (in m/s). A is a constant but you do not need to know its value to answer this question. What is the acceleration (in...