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Physics 16 Online
OpenStudy (anonymous):

How do I prove that a = v * dv/dx? My book says that a =dv/dt = dv/dx * dx/dt = dv/dx * v That looks all ok but according to this equation it seems that x(displacement) is a function of t whereas x=v * t. So how can we differentiate x without differentiating v which leads to circular reasoning?

OpenStudy (anonymous):

you do have a point, but if we replace backwards. i.e, we write x as function of v rather than v as function of x, it wouldn't be right. x and t are the final or say independent variables here . So, if a term comes dx/dt ,we simply write it = v instead of putting x = v*t . And , the more important fact is that in this case, x is not equal to simply v*t , instead., dx = vdt

OpenStudy (anonymous):

basically, the glitch is : assuming x = vt

OpenStudy (anonymous):

I guess we are getting somewhere..... equation in the book: dv/dt = dv/dx * dx/dt .....(1) According to it,it seems like t is the only independent variable and not x! Heres my thinking: v=w o u (v is a composite function) x=u(t) => dx/dt v=w(x) => dv/dx so dv/dt = dx/dt * dv/dx and so we get equation (1) I want you to tell me that is my process right or i am mistaken. Also my conclusion is that t is the only independent variable here. Am i right?

OpenStudy (anonymous):

there is more of differential calculus here, than simply functions . If y = f(x) and x = g(t) , then while differentiating y with respect to x, we can use the given function of x and can go a step forward and find the result in t-derivative.

OpenStudy (anonymous):

maybe, I'm unable to explain this, but I can tell you for sure, it's not simply multiplying dv/dx and dx/dt ,

OpenStudy (anonymous):

what is your real question here. lets discuss it the not calculus way if thats what your looking for.

OpenStudy (anonymous):

x is not displacement x is position :P v= dx/dt is the rate of change of position..

OpenStudy (anonymous):

correct mashy. don't forget that acceleration is the slope of velocity vector if you use the graphical analysis approach

OpenStudy (anonymous):

The above equation mentioned in the question is perfect.. and ll tell you when to use it if your velocity function is given in terms of displacement.. for example v(x) = 3x (m/s) then find the acceleration at x = 2 m so we can do a = dv/dx times v = 3 *2 = 6m/s^2

OpenStudy (anonymous):

m sorry that is 3*6= 18m/s^2 :D

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