A particle moves along a wire that follows the curve y = 4(e^2x + e^-2x) at a constant velocity of 4 m/s. What is the normal acceleration of the particle when x = 0?

Partical at x = 0

Since the velocity is constant, there is no acceleration tangent to its path. However, there is acceleration normal (perpendicular) to the path as given by,

      a_n = v²/ρ

Need to determine, ρ
     

     dy/dx = d[ 4(e^2x + e^-2x) ] / dx
              = 8 (e^2x - e^-2x )

     d²y/dx² = d[ 8(e^2x - e^-2x ) ] / dx
                 = 16 (e^2x + e^-2x )

     (dy/dx)_x=0 = 8(1 - 1) = 0
     (d²y/dx²)_x=0 = 16(1 + 1) = 32

Substituting back into the acceleration equation,
     a_n = v²/ρ = 4² (32)

a_n = 512 ft/s²