A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration and then stops. The duration of uniform acceleration is

Let acceleration is a and retardation is $-2a.$
Then, for accelerated motion,

            $t_1 = \frac{v}{a}$                        …(i)    

For retarding motion,

            $t_2 = \frac{v}{2a}$                      …(ii)

Given,   $t_1+t_2=9\Rightarrow \frac{v}{a}+\frac{v}{2a}=9\Rightarrow \frac{3v}{2a}=9\Rightarrow \frac{v}{a}=6$

Hence, duration of acceleration, $t_1 = \frac{v}{a} = 6\text{ s}.$