The correct option is C t>1s
As the curve is parabolic let us assume a second order equation for velocity to fit in this curve
Let v=at2+bt+c(1)
Apply boundary condition to above equation to find a,b and c
At t=0 s, v=0 ms2
0=(a×02)+(b×0)+c
c=0
At t=1 s, v=−3 ms2
−3=(a×12)+(b×1)+0
−3=a+b(2)
At t=2 s, v=0 ms2
0=(a×22)+(b×2)
b=−2a(3)
Substitute (3) in equation (2) we get,
−3=a−2a
a=3
Substitute value of a in equation (2)
−3=3+b
b=−6
Substitute these a, b and c in equation (1)
v=3t2−6t
dvdt=6t−6
For t≤1, dvdt≤0
For t>1, dvdt>0
Acceleration changing its direction after 1 s
∴t>1s
Acceleration at any point is the slope of v−t graph at that point. From t=0 to 1 s, slope is negative but after t=1 s, slope of v−t graph becomes positive, which means direction of acceleration changes after t=1 s.