Question 1046774
{{{a = d^2x/dt^2 = 10}}}

===> {{{dx/dt = 10t+c}}}  ===> {{{x = 10*(t^2/2) +ct+d = 5t^2 +ct +d}}}

x(2) = 0 ===> {{{5*2^2 +2c +d = 0}}} ===> 2c+d = -20.

x(3) = 25 ===> {{{5*3^2 +3c +d = 25}}} ===> 3c+d = -20.

Solving simultaneously,

c = 0 and d = -20.

===> {{{x(t) =5t^2 - 20}}}  ===> x(0) = -20, or, at the start, the body is 20 meters to the left of O.