Question

A particle moves in a straight line and its position
x at time t is given by x² = 2+ t. Its acceleration
is given by

(1) -2/x³
(2) -1/4x³
(3) -1/4x²
(4) 1/x²​

Answer 1

x=(2+t)^1/2

x^3=(2+t)^3/2...i)

dx/dt=1/{2(2+t)^1/2 }. (using chain rule we /differentiate it)

v=1/{2(2+t)^1/2}

v={(2+t)^-1/2}/2

dv/dt=(1/2)(-1/2)(2+t)^-1-1/2)

a=(-1/4)(2+t)^-3/2

a=(-1/4)/(2+t)^3/2

from eq i)

put

(2+t)^3/2=x^3

a=-1/4x^3

so option 2)