Originally Posted by

**GoldPhoenix**
I think you thought you put more information into the question than you did, but the correct answer is:

Let the x-position as a function of time be denoted by x = f(t), s.t. f(0) = x_{0}. Then the position 2-vector is given by:

(x(t), y(t) ) == ( f(t) , m f(t) + b ) .

In other words there's a whole twice-differentiable function's worth of information that's unspecified. Then the velocity vector is given by

v(t) = (u,v) == ( dx/dt , dy/dt ) = ( d/dt [ f(t) ] , m d/dt[f(t)] ) .

The direction that you're looking for is parallel to the line in question. The component along that direction is the total norm of the function, however. So, while enforcing positivity as requested (physically this means it is right-moving, i.e. df/dt > 0), the the norm at time, t=0, gives us

|v(0)| == sqrt{ u^{2} + v^{2} }= sqrt{ 1+m^{2}} df/dt(0),

The acceleration may be computed similarly:

a(t) = (a_{x}, a_{y}) == ( du/dt, dv/dt ) = d^{2}f/dt^{2} (1,m)

And the norm at time, t = 0, is given as:

|a(0)| == sqrt{a_{x}^{2}, a_{y}^{2}} = sqrt{ 1+m^{2}} | d^{2}f/dt^{2}| (0)

(We must enforce positivity on the second derivative of f(t) because nothing prevents it from being negative but the norm takes the absolute value, obviously)

To make Clive happy:

1.) f is a twice-differentiable, real-valued function: f: R -> R.

2.) p = (x,y) is a vector living in Euclidean 2-space, RxR, with Euclidean norm d(,y): RxR -> R*, and additive identity 0 for the vector space, such that the Euclidean norm is defined by

d(p_{1},p_{2}) = sqrt{ (x_{1} - x_{2})^2 + (y_{1} - y_{2}) }

with p_{1 }= (x_{1}, y_{1}) and p_{2} = (x_{2}, y_{2}).

3.) Finally, define the unary operator |x| = d(x,0).
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