Recent content by snovak216

I appreciate the help, it's nice to be walked through the proper notation.

2v * dv/dx= 0*(a^2-x^2)+ w^2*(0 -2x)= w^2*-2x 2v dv/dx = -2x*w^2 v dv/dx = -x*w^2 where v dv/dx= a?

I think I understand the derivation, but I do not understand the notation for v^2 and a v^2*dv/dx 2v/dx? = 2w(a^2-x^2)+ w^2(2a -2x) but since 2w and 2a are constants, so 2v/dx?= w^2*-2x dividing by 2 v/dx= -xw^2 and, however notation allows, the v/dx? is equivalent to a. therefore a= -xw^2

After eliminating the ^2, v= w * √(a^2 - x^2) dv/dx= -xw/ √(a^2-x^2)

Wouldn't we have to differentiate wrt x? As t doesn't appear in the equation.

Sorry, I looked at v* dv/dx. Now definitely not sure how to go about it.

Homework Statement The velocity of a particle is related to its position by: v2 = w2 (A2 - x2) where w and A are constants. Show that the acceleration is given by: a=-w2x[/B]Homework EquationsThe Attempt at a Solution a= v* dv/dt v=(A2w2-x2w2)1/2 dv/dt= 1/2 (A2w2-x2w2)-1/2 * -2xw2 v *...

I appreciate the help, thanks everyone!

If my thinking is correct, in order to get a dimensionless exp^(), B must have units of [T]^2. As well as, in order to resolve At^2 to a dimension of [L], A must have a dimension of [L]/[T]^2

Are A and B meant to resolve the variable of time into a unit of distance?

Well, arbitrarily speaking, we can say x has a unit of meters and t has a unit of seconds, so the equation gives meters as a function of time squared.

Homework Statement If x and t represent position and time, respectively, and A and B are constants in x = At2{1-exp(-t2/B)}, what are the dimensions of A and B? 2. The attempt at a solution I just slept through my first class and found that I have many homework problems related to...