Question

let me answer with explanation

Answer 1

t = (x²-1)^(1/2)

squaring both sides

t² = x²-1

x² = t²+1

diffrentiate both side wrt t

2x (dx/dt) = 2t -----1

(dx/dt) = t/x =( (x²-1)½)/x

now again diffrentiate eqn 1 both sides wrt t

2(dx/dt)² +2x (dx²/(dt)²) = 2

since (dx²/(dt)²) = acceleration (a)

thus
(dx/dt)²+x(a) = 1

(((x²-1)½)/x)² +x(a) = 1

((x²-1)/x²) + x(a) = 1

x(a) = 1-((x²-1)/x²)

x(a) = (x²-x²+1)/x²

(a) = 1/x³

thus acceleration in terms of X = 1/x³