Question: velocity and acceleration. (1 Viewer)

[cj_bridle]

velocity of a particle is given by:

v = (root2x)e^(-x^2)

Show the particle has acceleration given by:

a = e^(-2x^2) [ 1 - 4x^2]

i can do it by using dv/dt = dv/dx.dx/dt

but i tried doing it by converting velocity to v^2/2 and then differentiating to find accelaration... but when i differentiate an extra 'x' comes out in the wash and it doesnt work? can anyone show me how to do it the v^2/2 way?

 

[acmilan]

v^2 = 2xe^(-2x^2)
(v^2)/2 = xe^(-2x^2)

a = d[(v^2)/2)/dx
a = e^(-2x^2) - 4x.xe^(-2x^2) [Note: . means multiplication]
a = e^(-2x^2) [1 - 4x^2]

 

[cj_bridle]

ahh i see what i did... silly mistake.. thanx milan..