For part i) of the question c original post, captain black has integrated to give a in terms of x we actually want to go the other way therefore answer should be 2x - 4
For part i) of the question c original post, captain black has integrated to give a in terms of x we actually want to go the other way therefore answer should be 2x - 4
He used the expression a=dv/dt =dv/dx * dx/dt to calculate acceleration. He never used integration(he used integration only for calculating x in terms of t).
I have just reworked the whole question. The answers given for a) and b) in the original post are incorrect. As I mentioned before, the answers CaptainBlack came up with originally are correct.
v = (2-x)^2
expanding gives
v = 4 - 4x + x^2
a= dv/dx = 2x -4
or
v = (2-x)^2
a = dv/dx = using d/dx[f(x)^n] = f'(x)n[f(x)]^(n-1)
a = ((-1)(2))(2-x)^(2-1)
a = -2(2-x)
a = 2x - 4
v = (2-x)^2
expanding gives
v = 4 - 4x + x^2
a= dv/dx = 2x -4
or
v = (2-x)^2
a = dv/dx = using d/dx[f(x)^n] = f'(x)n[f(x)]^(n-1)
a = ((-1)(2))(2-x)^(2-1)
a = -2(2-x)
a = 2x - 4