Displacement vs time graph for a given equation of acceleration

[dk_ch]

Homework Statement

The accompanying graph of position x versus time t represents the motion of a particle. If p and q are both
positive constants, the expression that best describes the acceleration a of the particle is

Homework Equations

(A) a = – - p – - qt (B) a = - –p + qt (C) a = p + qt (D) a = p – - qt

The Attempt at a Solution

option D is mentioned correct in a given solution of the problem. If it is correct then dv/dt = p - qt on itegration will give v = pt -0.5qt^2 +c., for inflexion pts on x-t curve v =0 ,this gives qt^2-2pt-2c=0 , one of the two roots of which becoming negative, which is not realistic with the given graph. I need help to know the real fault made by me.

 

[haruspex]

for inflexion pts on x-t curve v =0 ,

You mean extremum pts, right?

this gives qt^2-2pt-2c=0 , one of the two roots of which becoming negative, which is not realistic with the given graph. I need help to know the real fault made by me.

You are assuming c is positive? What is the sign of v at t=0?